June 2024, Week 5: June 24th - June 30th

The central concept of this solution is to leverage a sliding window approach with a deque to efficiently track k-bit flips across the array. This method allows for optimized handling of overlapping flips without modifying the original array.

• Sliding Window: A window of size k is used to manage the effect of flips on each element.

• Deque for Flip Tracking: A deque efficiently tracks whether a flip occurred at each position within the current window.

• Current Flipped State: This variable represents the cumulative effect of all flips on the current element.

• Virtual Flips: Instead of actually modifying the array, the algorithm virtually flips elements by tracking the flip state.

1. Initialization and Edge Case Handling:

   if k == 1: return nums.count(0) flip_window_deque = deque() current_flipped_state = 0 total_flips = 0

   The function starts with an optimization for k=1, simply counting the zeros. It then initializes the deque, the current flipped state, and the total flip count.

2. Main Loop - Iterating Through the Array:

   for index, num in enumerate(nums):

   This loop processes each element of the array, deciding whether to flip at each position.

3. Managing the Sliding Window:

   if index >= k: current_flipped_state ^= flip_window_deque.popleft()

   When the window size exceeds k, the effect of the oldest flip is removed using XOR operation.

4. Flip Decision and Execution:

   if current_flipped_state == num: if index + k > len(nums): return -1 flip_window_deque.append(1) current_flipped_state ^= 1 total_flips += 1 else: flip_window_deque.append(0)

   If the current state matches the original value, a flip is needed. The function checks if there's enough space for a flip, updates the deque and state, and increments the flip count. If no flip is needed, it appends 0 to the deque.

5. Result Calculation/Return:

   return total_flips

   After processing all elements, the function returns the total number of flips performed.

Time Complexity:

• , where is the length of the input array nums. This is because the algorithm processes each element of the array exactly once, performing constant-time operations for each element.

Space Complexity:

• , where is the flip window size. This is due to the deque storing at most elements to track the sliding window of flips. The space used is independent of the input array size , making it more efficient for large arrays with smaller values.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • flip_window_deque is initialized as an empty deque: deque([])

   • current_flipped_state is set to 0

   • total_flips is set to 0

2. Main Loop (Iterating through nums):

   • Iteration 1 (index 0, num = 0):

     • current_flipped_state (0) == num (0), so a flip is needed

     • Add 1 to flip_window_deque: deque([1])

     • Update current_flipped_state to 1

     • Increment total_flips to 1

   • Iteration 2 (index 1, num = 0):

     • current_flipped_state (1) != num (0), no flip needed

     • Add 0 to flip_window_deque: deque([1, 0])

   • Iteration 3 (index 2, num = 0):

     • current_flipped_state (1) != num (0), no flip needed

     • Add 0 to flip_window_deque: deque([1, 0, 0])

   • Iteration 4 (index 3, num = 1):

     • Window size reached (index >= k), remove oldest flip: deque([0, 0])

     • Update current_flipped_state to 0

     • current_flipped_state (0) != num (1), no flip needed

     • Add 0 to flip_window_deque: deque([0, 0, 0])

   • Iteration 5 (index 4, num = 0):

     • Remove oldest flip: deque([0, 0])

     • current_flipped_state (0) == num (0), so a flip is needed

     • Add 1 to flip_window_deque: deque([0, 0, 1])

     • Update current_flipped_state to 1

     • Increment total_flips to 2

   • Iteration 6 (index 5, num = 1):

     • Remove oldest flip: deque([0, 1])

     • current_flipped_state (1) == num (1), so a flip is needed

     • Add 1 to flip_window_deque: deque([0, 1, 1])

     • Update current_flipped_state to 0

     • Increment total_flips to 3

   • Iteration 7 (index 6, num = 1):

     • Remove oldest flip: deque([1, 1])

     • current_flipped_state (0) != num (1), no flip needed

     • Add 0 to flip_window_deque: deque([1, 1, 0])

   • Iteration 8 (index 7, num = 0):

     • Remove oldest flip: deque([1, 0])

     • Update current_flipped_state to 1

     • current_flipped_state (1) != num (0), no flip needed

     • Add 0 to flip_window_deque: deque([1, 0, 0])

3. Loop Termination:

   • The loop ends after processing all elements in nums

   • Final state: total_flips = 3, current_flipped_state = 1, flip_window_deque = deque([1, 0, 0])

4. Visual Aid:

   Iteration Summary Table:

   Index	Element	Flipped State	Total Flips	Flip Window
   0	0	1	1	[1]
   1	0	1	1	[1, 0]
   2	0	1	1	[1, 0, 0]
   3	1	0	1	[0, 0, 0]
   4	0	1	2	[0, 0, 1]
   5	1	0	3	[0, 1, 1]
   6	1	0	3	[1, 1, 0]
   7	0	1	3	[1, 0, 0]

5. Result Calculation/Final Steps:

   • The algorithm arrives at the final result by counting the total number of flips performed

   • The final result is the value of total_flips, which is 3

The central concept of this solution is to use an in-place marking technique with clever bit manipulation to efficiently track k-bit flips across the array. This method combines the space efficiency of in-place modifications with an optimized approach to handling overlapping flips.

• In-place Marking: The algorithm uses the value 2 to mark the start of a flip in the original array, enabling implicit storage of flip information without additional data structures.

• Active Flips Tracking: An active_flips variable keeps count of the number of active flips affecting the current element, allowing for quick flip decisions.

• Parity-based Flip Detection: The algorithm uses the parity of active_flips to determine whether the current element needs to be flipped.

• Virtual Flips: Instead of actually performing k-bit flips, the algorithm simulates their effect by tracking flip starts and active flip count.

1. Initialization and Edge Case Handling:

   if k == 1: return nums.count(0) n = len(nums) active_flips = 0 total_flips = 0

   The function starts with an optimization for k=1, simply counting the zeros. It then initializes the length of the array, active flips count, and total flips count.

2. Main Loop - Linear Scan:

   for index in range(n):

   This loop processes each element of the array, making flip decisions based on the current state and previous flips.

3. Managing Active Flips:

   if index >= k and nums[index - k] == 2: active_flips -= 1

4. Flip Decision and Execution:

   if (active_flips % 2) == nums[index]: if index + k > n: return -1 nums[index] = 2 active_flips += 1 total_flips += 1

   If the parity of active flips matches the current element's value, a flip is needed. The function checks if there's enough space for a flip, marks the flip start with 2, and updates the active and total flip counts.

5. Result Calculation/Return:

   return total_flips

   After processing all elements, the function returns the total number of flips performed.

Time Complexity:

• , where n is the length of the input array nums. This is because the algorithm processes each element of the array exactly once in the main loop, performing constant-time operations for each element.

Space Complexity:

• , as the algorithm uses only a constant amount of extra space regardless of the input size. It cleverly uses the input array itself to store flip information by marking flip starts with the value 2, avoiding the need for additional data structures that grow with the input size.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • n is set to 8 (length of nums)

   • active_flips is set to 0

   • total_flips is set to 0

2. Main Loop (Iterating through nums):

   • Iteration 1 (index 0, num = 0):

     • No flip ending (index < k)

     • (active_flips % 2) (0) == nums[0] (0), so a flip is needed

     • Mark flip start: nums[0] = 2

     • Increment active_flips to 1

     • Increment total_flips to 1

   • Iteration 2 (index 1, num = 0):

     • No flip ending (index < k)

     • (active_flips % 2) (1) != nums[1] (0), no flip needed

   • Iteration 3 (index 2, num = 0):

     • No flip ending (index < k)

     • (active_flips % 2) (1) != nums[2] (0), no flip needed

   • Iteration 4 (index 3, num = 1):

     • Flip ending: nums[index - k] (nums[0]) == 2

     • Decrement active_flips to 0

     • (active_flips % 2) (0) != nums[3] (1), no flip needed

   • Iteration 5 (index 4, num = 0):

     • No flip ending (nums[1] != 2)

     • (active_flips % 2) (0) == nums[4] (0), so a flip is needed

     • Mark flip start: nums[4] = 2

     • Increment active_flips to 1

     • Increment total_flips to 2

   • Iteration 6 (index 5, num = 1):

     • No flip ending (nums[2] != 2)

     • (active_flips % 2) (1) == nums[5] (1), so a flip is needed

     • Mark flip start: nums[5] = 2

     • Increment active_flips to 2

     • Increment total_flips to 3

   • Iteration 7 (index 6, num = 1):

     • No flip ending (nums[3] != 2)

     • (active_flips % 2) (0) != nums[6] (1), no flip needed

   • Iteration 8 (index 7, num = 0):

     • Flip ending: nums[index - k] (nums[4]) == 2

     • Decrement active_flips to 1

     • (active_flips % 2) (1) != nums[7] (0), no flip needed

3. Loop Termination:

   • The loop ends after processing all elements in nums

   • Final state: total_flips = 3, active_flips = 1

4. Visual Aid:

   Iteration Summary Table:

   Index	Element Value	Active Flips	Total Flips
   0	2	1	1
   1	0	1	1
   2	0	1	1
   3	1	0	1
   4	2	1	2
   5	2	2	3
   6	1	2	3
   7	0	1	3

5. Result Calculation/Final Steps:

   • The algorithm arrives at the final result by counting the total number of flips performed

   • The final result is the value of total_flips, which is 3

The central concept of this solution is to leverage the properties of a Binary Search Tree (BST) to efficiently convert it into a Greater Sum Tree (GST). The solution uses a reverse in-order traversal (right-root-left) to accumulate the sum of greater values for each node.

• BST Property Exploitation: In a BST, all nodes in the right subtree have greater values than the current node, and all nodes in the left subtree have smaller values.

• Reverse In-order Traversal: By traversing the tree in a right-root-left order, we process larger values first, allowing us to accumulate the sum of greater values for each node.

• Cumulative Sum Propagation: As we traverse, we maintain a running sum of all greater values, updating each node's value with this sum plus its original value.

1. Initialization:

   cumulative_sum = 0

   A variable cumulative_sum is initialized to 0. This will be used to keep track of the running sum of greater values as we traverse the tree.

2. Helper Function Definition:

   def update_node_values(node: TreeNode) -> None: nonlocal cumulative_sum

   An inner function update_node_values is defined to perform the recursive traversal and node value updates. It uses the nonlocal keyword to access and modify the cumulative_sum variable from the outer scope.

3. Right Subtree Traversal:

   if node.right: update_node_values(node.right)

   The function first recursively processes the right subtree, ensuring that all greater values are processed before the current node.

4. Node Value Update:

   cumulative_sum += node.val node.val = cumulative_sum

   After processing the right subtree, the current node's original value is added to cumulative_sum, and then the node's value is updated to be this new sum.

5. Left Subtree Traversal:

   if node.left: update_node_values(node.left)

   Finally, the function recursively processes the left subtree, which will use the updated cumulative_sum that now includes the current node's original value.

6. Traversal Initiation and Return:

   update_node_values(root) return root

   The traversal is started by calling update_node_values on the root node. After the traversal is complete, the modified root is returned.

Time Complexity:

• , where is the number of nodes in the tree. This is because each node in the tree is visited exactly once during the traversal.

Space Complexity:

• , where is the height of the tree. This space is used by the recursive call stack. In the worst case of an unbalanced tree (skewed tree), this could be , but for a balanced BST, it would be .

The space complexity is determined by the maximum depth of the recursive call stack, which corresponds to the height of the tree. Each recursive call adds a frame to the call stack, and the maximum number of simultaneous frames is equal to the height of the tree.

Input:

A visual representation of the input BST:

Step-by-Step Walkthrough:

1. Initialization:

   • Set cumulative_sum = 0

   • Call update_node_values(root) with the root node (4)

2. Main Loop (Reverse In-order Traversal):

   • Iteration 1 (Node 4):

     • Move to right child (6)

   • Iteration 2 (Node 6):

     • Move to right child (7)

   • Iteration 3 (Node 7):

     • Move to right child (8)

   • Iteration 4 (Node 8):

     • No right child, process node

     • Update cumulative_sum: 0 + 8 = 8

     • Set node value to 8

     • No left child, return to parent (7)

   • Iteration 5 (Node 7):

     • Update cumulative_sum: 8 + 7 = 15

     • Set node value to 15

     • No left child, return to parent (6)

   • Iteration 6 (Node 6):

     • Update cumulative_sum: 15 + 6 = 21

     • Set node value to 21

     • Move to left child (5)

   • Iteration 7 (Node 5):

     • No right child, process node

     • Update cumulative_sum: 21 + 5 = 26

     • Set node value to 26

     • No left child, return to parent (6), then to grandparent (4)

   • Iteration 8 (Node 4):

     • Update cumulative_sum: 26 + 4 = 30

     • Set node value to 30

     • Move to left child (1)

   • Iteration 9 (Node 1):

     • Move to right child (2)

   • Iteration 10 (Node 2):

     • Move to right child (3)

   • Iteration 11 (Node 3):

     • No right child, process node

     • Update cumulative_sum: 30 + 3 = 33

     • Set node value to 33

     • No left child, return to parent (2)

   • Iteration 12 (Node 2):

     • Update cumulative_sum: 33 + 2 = 35

     • Set node value to 35

     • No left child, return to parent (1)

   • Iteration 13 (Node 1):

     • Update cumulative_sum: 35 + 1 = 36

     • Set node value to 36

     • Move to left child (0)

   • Iteration 14 (Node 0):

     • No right child, process node

     • Update cumulative_sum: 36 + 0 = 36

     • Set node value to 36

     • No left child, return to parent (1), then to grandparent (4)

3. Loop Termination:

   • The traversal is complete when we return to the root node (4) and there are no more nodes to visit

   • Final cumulative_sum is 36

4. Visual Aid:

   Iteration Summary Table:

   Original Value	Added to Sum	New Value (Cumulative Sum)
   8	0	8
   7	8	15
   6	15	21
   5	21	26
   4	26	30
   3	30	33
   2	33	35
   1	35	36
   0	36	36

5. Result Calculation/Final Steps:

   • The algorithm has modified the BST in-place

   • Each node's value now represents the sum of its original value and all greater values in the original BST

   • The root node (originally 4) now has the value 30, which is the sum of 4 and all greater values (5, 6, 7, 8)

   • The resulting Greater Sum Tree maintains the BST structure but with updated node values

The final Greater Sum Tree looks like this:

The central concept of this solution is to convert a Binary Search Tree (BST) to a Greater Sum Tree (GST) using an iterative approach with a stack. This method simulates the reverse in-order traversal (right-root-left) without using recursion.

• Iterative Traversal: Instead of recursion, the solution uses a stack to keep track of nodes to be processed, allowing for an iterative implementation.

• Right-First Approach: By pushing all right nodes onto the stack first, the algorithm ensures that larger values are processed before smaller ones.

• Cumulative Sum Propagation: As in the recursive version, a running sum is maintained and used to update node values.

1. Helper Function Definition:

   def push_right_nodes(node: TreeNode) -> None: while node: stack.append(node) node = node.right

   This helper function pushes all right child nodes of a given node onto the stack. It's crucial for maintaining the right-root-left traversal order.

2. Initialization:

   stack = [] current_node = root cumulative_sum = 0

   A stack is initialized to store nodes, current_node is set to the root, and cumulative_sum is initialized to 0 to keep track of the running sum.

3. Initial Stack Population:

   push_right_nodes(current_node)

   The stack is initially populated with the root and all its right descendants.

4. Main Traversal Loop:

   while stack:

   The main loop continues as long as there are nodes in the stack to process.

5. Node Processing:

   current_node = stack.pop() cumulative_sum += current_node.val current_node.val = cumulative_sum

   For each node, we pop it from the stack, add its value to the cumulative sum, and then update its value with this new sum.

6. Left Subtree Handling:

   push_right_nodes(current_node.left)

   After processing a node, we push all right descendants of its left child onto the stack. This ensures that we process the left subtree after the current node, maintaining the reverse in-order traversal.

7. Result Return:

   return root

   After processing all nodes, the modified root is returned.

Time Complexity:

• , where is the number of nodes in the tree. Each node is visited exactly once during the traversal. Although we have a nested loop (the while loop in push_right_nodes inside the main while loop), each node is still pushed and popped from the stack exactly once.

Space Complexity:

• , where h is the height of the tree. The stack uses this space to store nodes during traversal. In the worst case of an unbalanced tree (skewed tree), this could be , but for a balanced BST, it would be .

The space complexity is determined by the maximum number of nodes that can be on the stack at any given time. This corresponds to the height of the tree because, at most, we will have a path from the root to a leaf on the stack.

Input:

A visual representation of the input BST:

Step-by-Step Walkthrough:

1. Initialization:

   • Initialize an empty stack = []

   • Set current_node = root (4)

   • Set cumulative_sum = 0

2. Initial Stack Population:

   • Call push_right_nodes(current_node)

     • Push 4 onto the stack

     • Move to right child: 6

     • Push 6 onto the stack

     • Move to right child: 7

     • Push 7 onto the stack

     • Move to right child: 8

     • Push 8 onto the stack

     • No more right children

   • Stack after pushing: [4, 6, 7, 8]

3. Main Traversal Loop:

   • Iteration 1 (Node 8):

     • Pop 8 from the stack

     • Update cumulative_sum: 0 + 8 = 8

     • Set node value to 8

     • No left child to process

   • Iteration 2 (Node 7):

     • Pop 7 from the stack

     • Update cumulative_sum: 8 + 7 = 15

     • Set node value to 15

     • No left child to process

   • Iteration 3 (Node 6):

     • Pop 6 from the stack

     • Update cumulative_sum: 15 + 6 = 21

     • Set node value to 21

     • Process left child (5):

       • Push 5 onto the stack

     • Stack after pushing: [4, 5]

   • Iteration 4 (Node 5):

     • Pop 5 from the stack

     • Update cumulative_sum: 21 + 5 = 26

     • Set node value to 26

     • No left child to process

   • Iteration 5 (Node 4):

     • Pop 4 from the stack

     • Update cumulative_sum: 26 + 4 = 30

     • Set node value to 30

     • Process left child (1):

       • Push 1 onto the stack

       • Move to right child: 2

       • Push 2 onto the stack

       • Move to right child: 3

       • Push 3 onto the stack

     • Stack after pushing: [1, 2, 3]

   • Iteration 6 (Node 3):

     • Pop 3 from the stack

     • Update cumulative_sum: 30 + 3 = 33

     • Set node value to 33

     • No left child to process

   • Iteration 7 (Node 2):

     • Pop 2 from the stack

     • Update cumulative_sum: 33 + 2 = 35

     • Set node value to 35

     • No left child to process

   • Iteration 8 (Node 1):

     • Pop 1 from the stack

     • Update cumulative_sum: 35 + 1 = 36

     • Set node value to 36

     • Process left child (0):

       • Push 0 onto the stack

     • Stack after pushing: [0]

   • Iteration 9 (Node 0):

     • Pop 0 from the stack

     • Update cumulative_sum: 36 + 0 = 36

     • Set node value to 36

     • No left child to process

4. Visual Aid:

   Iteration Summary Table:

   Original Value	Added to Sum	New Value (Cumulative Sum)
   8	0	8
   7	8	15
   6	15	21
   5	21	26
   4	26	30
   3	30	33
   2	33	35
   1	35	36
   0	36	36

5. Result Calculation/Final Steps:

   • The algorithm has modified the BST in-place

   • Each node's value now represents the sum of its original value and all greater values in the original BST

   • The root node (originally 4) now has the value 30, which is the sum of 4 and all greater values (5, 6, 7, 8)

   • The resulting Greater Sum Tree maintains the BST structure but with updated node values

The final Greater Sum Tree looks like this:

This iterative approach using a stack achieves the same result as the recursive method, converting the BST to a Greater Sum Tree by updating each node's value to be the sum of its original value and all greater values in the tree.

The central concept of this solution is to convert a Binary Search Tree (BST) to a Greater Sum Tree (GST) using the Morris Traversal algorithm. This method performs a reverse in-order traversal (right-root-left) without using recursion or an explicit stack, achieving space complexity.

• Morris Traversal: This technique creates temporary links in the tree to facilitate navigation without additional data structures.

• Threaded Binary Tree: The algorithm temporarily modifies the tree structure, creating links from successor nodes to their predecessors.

• Constant Space: By using these temporary links, the traversal is performed using only a constant amount of extra space, regardless of tree size.

1. Helper Function Definition:

   def find_successor(current_node: TreeNode) -> TreeNode: successor = current_node.right while successor.left and successor.left is not current_node: successor = successor.left return successor

   This function finds the inorder successor (leftmost node of the right subtree) of the current node in the context of reverse inorder traversal.

2. Initialization:

   cumulative_sum = 0 current_node = root

   cumulative_sum is initialized to keep track of the running sum, and current_node is set to the root to begin traversal.

3. Main Traversal Loop:

   while current_node:

   The main loop continues as long as there's a current node to process.

4. No Right Child Case:

   if not current_node.right: cumulative_sum += current_node.val current_node.val = cumulative_sum current_node = current_node.left

   If there's no right child, we process the current node and move to the left.

5. Right Child Exists Case:

   else: successor = find_successor(current_node)

   If a right child exists, we find the successor node.

6. First Visit to Successor:

   if not successor.left: successor.left = current_node current_node = current_node.right

   On the first visit, we create a temporary link from the successor to the current node and move right.

7. Second Visit to Successor:

   else: successor.left = None cumulative_sum += current_node.val current_node.val = cumulative_sum current_node = current_node.left

   On the second visit, we remove the temporary link, process the current node, and move left.

8. Result Return:

   return root

   After processing all nodes, the modified root is returned.

Time Complexity:

• , where is the number of nodes in the tree. Although each node may be visited up to three times (to create the link, to process the node, and to remove the link), this still results in linear time complexity. The total number of operations is proportional to the number of nodes.

Space Complexity:

• , constant space. This is the key advantage of Morris Traversal. It uses only a constant amount of extra space, regardless of the tree size. The temporary modifications to the tree structure do not count towards space complexity as they are made within the existing tree nodes and are reverted during the traversal.

The constant space complexity is achieved by using the existing tree structure to simulate the traversal stack, avoiding the need for additional data structures that grow with the size of the input.

Input:

A visual representation of the input BST:

Step-by-Step Walkthrough:

1. Initialization:

   • Set cumulative_sum = 0

   • Set current_node = root (4)

2. Main Traversal Loop:

   • Processing Node 4:

     • Find successor: 5 (leftmost node of right subtree)

     • Create temporary link: 5.left = 4

     • Move to right child: 6

   • Processing Node 6:

     • Find successor: 7

     • Create temporary link: 7.left = 6

     • Move to right child: 7

   • Processing Node 7:

     • Find successor: 8

     • Create temporary link: 8.left = 7

     • Move to right child: 8

   • Processing Node 8:

     • No right child

     • Update node: 8 + 0 = 8

     • Move to left child: 7

   • Processing Node 7 (second visit):

     • Remove temporary link: 8.left = None

     • Update node: 7 + 8 = 15

     • Move to left child: 6

   • Processing Node 6 (second visit):

     • Remove temporary link: 15.left = None

     • Update node: 6 + 15 = 21

     • Move to left child: 5

   • Processing Node 5:

     • No right child

     • Update node: 5 + 21 = 26

     • Move to left child: 4

   • Processing Node 4 (second visit):

     • Remove temporary link: 26.left = None

     • Update node: 4 + 26 = 30

     • Move to left child: 1

   • Processing Node 1:

     • Find successor: 2

     • Create temporary link: 2.left = 1

     • Move to right child: 2

   • Processing Node 2:

     • Find successor: 3

     • Create temporary link: 3.left = 2

     • Move to right child: 3

   • Processing Node 3:

     • No right child

     • Update node: 3 + 30 = 33

     • Move to left child: 2

   • Processing Node 2 (second visit):

     • Remove the temporary link: 33.left = None

     • Update node: 2 + 33 = 35

     • Move to left child: 1

   • Processing Node 1 (second visit):

     • Remove the temporary link: 35.left = None

     • Update node: 1 + 35 = 36

     • Move to left child: 0

   • Processing Node 0:

     • No right child

     • Update node: 0 + 36 = 36

     • Move to left child: None (Traversal complete)

3. Visual Aid:

   Iteration Summary Table:

   Original Value	Added to Sum	New Value (Cumulative Sum)
   8	0	8
   7	8	15
   6	15	21
   5	21	26
   4	26	30
   3	30	33
   2	33	35
   1	35	36
   0	36	36

4. Result Calculation/Final Steps:

   • The algorithm has modified the BST in-place

   • Each node's value now represents the sum of its original value and all greater values in the original BST

   • The root node (originally 4) now has the value 30, which is the sum of 4 and all greater values (5, 6, 7, 8)

   • The resulting Greater Sum Tree maintains the BST structure but with updated node values

   • All temporary links created during the traversal have been removed, restoring the original tree structure

The final Greater Sum Tree looks like this:

This Morris Traversal approach achieves the same result as the recursive and stack-based methods, converting the BST to a Greater Sum Tree. It does so without using additional data structures (like a stack) or recursive calls, maintaining space complexity while still achieving time complexity.

The central concept of this solution is to leverage the properties of Binary Search Trees (BSTs) and in-order traversal to create a balanced BST. This approach consists of two main steps:

1. In-Order Traversal: Perform an in-order traversal of the original BST to get a sorted list of nodes.

2. Recursive Reconstruction: Use the sorted list to recursively construct a new balanced BST.

• BST Property Exploitation: In-order traversal of a BST yields nodes in ascending order, which is crucial for the balancing step.

• Divide and Conquer: The reconstruction phase uses a divide-and-conquer approach, selecting the median element as the root at each step to ensure balance.

1. Initialization and Helper Function Definitions:

   def balanceBST1(root: TreeNode) -> TreeNode: inorder_nodes = [] def inorder_traverse(root: TreeNode) -> None: # ... (implementation details in step 2) def build_balanced_bst(start_index: int, end_index: int) -> TreeNode | None: # ... (implementation details in step 3)

   The main function balanceBST1 is defined, along with an empty list inorder_nodes to store the sorted nodes. Two helper functions, inorder_traverse and build_balanced_bst are also defined within the main function's scope.

2. In-Order Traversal:

   def inorder_traverse(root: TreeNode) -> None: if not root: return inorder_traverse(root.left) inorder_nodes.append(root) inorder_traverse(root.right)

   This helper function performs an in-order traversal of the original BST. It recursively visits the left subtree, appends the current node to inorder_nodes, and then visits the right subtree. This results in a list of nodes sorted in ascending order.

3. Balanced BST Construction:

   def build_balanced_bst(start_index: int, end_index: int) -> TreeNode | None: if start_index > end_index: return None mid_index = start_index + (end_index - start_index) // 2 node = inorder_nodes[mid_index] node.left = build_balanced_bst(start_index, mid_index - 1) node.right = build_balanced_bst(mid_index + 1, end_index) return node

   This helper function recursively constructs a balanced BST from the sorted list of nodes. It selects the middle element as the root, recursively builds the left subtree from nodes before the middle, and the right subtree from nodes after the middle.

4. Execution and Return:

   inorder_traverse(root) return build_balanced_bst(0, len(inorder_nodes) - 1)

   Finally, the main function executes the in-order traversal on the input tree and then calls build_balanced_bst to construct and return the balanced BST.

Time Complexity:

• , where is the number of nodes in the tree. This is because:

  1. The in-order traversal visits each node once:

  2. The balanced BST construction processes each node once:

  3. The total time is the sum of these operations:

Space Complexity:

• , where is the number of nodes in the tree. This is because:

  1. The inorder_nodes list stores all n nodes:

  2. The recursion stack in both traversal and construction can go up to in the balanced case, but in the worst case (skewed tree)

  3. The total space is dominated by the largest of these:

Input:

A visual representation of the input BST:

Step-by-Step Walkthrough:

1. Initialization:

   • An empty list inorder_nodes is created to store the nodes in sorted order.

2. In-order Traversal:

   • Node 4:

     • The function starts at the root node with value 4.

     • It moves to the left child (node 3).

   • Node 3:

     • The function moves to the left child (node 2).

   • Node 2:

     • The function moves to the left child (node 1).

   • Node 1:

     • The function checks the left child, which is None.

     • Node 1 is appended to inorder_nodes.

       • inorder_nodes = [1]

     • The function checks the right child, which is None.

     • The function returns to node 2.

   • Back to Node 2:

     • Node 2 is appended to inorder_nodes.

       • inorder_nodes = [1, 2]

     • The function checks the right child, which is None.

     • The function returns to node 3.

   • Back to Node 3:

     • Node 3 is appended to inorder_nodes.

       • inorder_nodes = [1, 2, 3]

     • The function checks the right child, which is None.

     • The function returns to node 4.

   • Back to Node 4:

     • Node 4 is appended to inorder_nodes.

       • inorder_nodes = [1, 2, 3, 4]

     • The function checks the right child, which is None.

     • The in-order traversal is complete.

3. Building Balanced BST:

   • Initial Call:

     • start_index = 0, end_index = 3

     • mid_index = 1

     • Root node is set to inorder_nodes[1] (value 2)

     • Left and right subtrees are recursively constructed.

       • Left subtree: from start_index to mid_index - 1 (0 to 0)

       • Right subtree: from mid_index + 1 to end_index (2 to 3)

   • Left Subtree (0, 0):

     • start_index = 0, end_index = 0

     • mid_index = 0

     • Node with value 1 becomes the left child of 2

     • Its left and right children are set to None because the index range will be invalid

   • Right Subtree (2, 3):

     • start_index = 2, end_index = 3

     • mid_index = 2

     • Node with value 3 becomes the right child of 2

     • Left and right subtrees are recursively constructed.

       • Left subtree: from start_index to mid_index - 1 (2 to 1) - Invalid range

       • Right subtree: from mid_index + 1 to end_index (3 to 3)

   • Right Subtree of 3 (3, 3):

     • start_index = 3, end_index = 3

     • mid_index = 3

     • Node with value 4 becomes the right child of 3

     • Its left and right children are set to None because the index range will be invalid

4. Result Calculation:

   • The function returns the root of the new balanced BST (node with value 2).

   • The depth of every node in the new tree differs by at most 1, satisfying the balance condition.

A visual representation of the output Balanced BST:

The central concept of this solution is to leverage the Day-Stout-Warren (DSW) algorithm to balance the Binary Search Tree (BST). This approach consists of three main steps:

1. Vine Creation: Convert the BST into a "vine" (a right-skewed tree).

2. Perfect Tree Size Calculation: Determines the number of nodes needed for a perfect binary tree. A perfect binary tree is a binary tree in which all levels are filled except possibly for the lowest level, which is filled from left to right.

3. Tree Balancing: Perform a series of left rotations to balance the tree.

• In-Place Transformation: The DSW algorithm achieves balance through a series of tree rotations, without requiring extra space for node storage.

• Guaranteed Balance: It ensures a balanced tree with a height of .

• Space Efficiency: Unlike methods that require auxiliary space, this approach uses only extra space.

1. Helper Function Definitions:

   def right_rotate(parent: TreeNode, node: TreeNode) -> None: left_child = node.left node.left = left_child.right left_child.right = node parent.right = left_child

   The right_rotate function performs a right rotation on the given node:

   • It saves the left child of the node.

   • It sets the node's left child to be the right child of its former left child.

   • It makes the node the right child of its former left child.

   • It updates the parent to point to the new root of this subtree (the former left child).

   def left_rotate(parent: TreeNode, node: TreeNode) -> None: right_child = node.right node.right = right_child.left right_child.left = node parent.right = right_child

   The left_rotate function performs a left rotation on the given node:

   • It saves the right child of the node.

   • It sets the node's right child to be the left child of its former right child.

   • It makes the node the left child of its former right child.

   • It updates the parent to point to the new root of this subtree (the former right child).

   def compress_vine(vine_root: TreeNode, rotations: int) -> None: current_node = vine_root for _ in range(rotations): child = current_node.right left_rotate(current_node, child) current_node = current_node.right

   The compress_vine function performs a series of left rotations to balance part of the vine:

   • It starts from the vine_root and performs the specified number of rotations.

   • For each rotation, it performs a left rotation on the current node's right child.

   • After each rotation, it moves to the new right child to continue the process.

2. Initialization:

   dummy_root = TreeNode() dummy_root.right = root current_node = dummy_root

   A dummy root is created to handle edge cases and simplify the vine creation process. The original root becomes the right child of this dummy node.

3. Vine Creation (Step 1):

   while current_node.right: if current_node.right.left: right_rotate(current_node, current_node.right) else: current_node = current_node.right

   This loop converts the BST into a right-skewed tree (vine) by performing right rotations whenever a left child is encountered.

4. Node Counting and Perfect Tree Size Calculation (Step 2):

   node_count = 0 current_node = dummy_root.right while current_node: node_count += 1 current_node = current_node.right perfect_tree_nodes = 2 ** math.floor(math.log2(node_count + 1)) - 1

   This section counts the total number of nodes and calculates the size of the largest perfect binary tree that can be formed with these nodes. The perfect tree size is determined by finding the largest power of 2 less than or equal to the node count. This is expressed as .

5. Tree Balancing (Step 3):

   compress_vine(dummy_root, node_count - perfect_tree_nodes) remaining_nodes = perfect_tree_nodes while remaining_nodes > 1: remaining_nodes //= 2 compress_vine(dummy_root, remaining_nodes)

   The balancing process involves an initial compression to remove excess nodes, followed by a series of compressions that transform the vine into a balanced tree. Here we perform the initial compression and then iteratively reduce the number of remaining nodes by half, performing compressions at each step.

6. Result Return:

   return dummy_root.right

   The function returns the right child of the dummy root, which is the root of the balanced BST.

Time Complexity:

• , where is the number of nodes in the tree. This is because:

  1. Vine creation visits each node at most once:

  2. Node counting is a linear traversal:

  3. Tree balancing performs a constant number of operations per node:

  4. Despite multiple passes, each node is processed a constant number of times, resulting in linear time complexity.

Space Complexity:

• , which is constant space. This is because:

  1. The algorithm uses only a fixed number of additional variables (e.g., dummy_root, current_node)

  2. All operations are performed in-place, without requiring additional data structures that grow with input size

  3. The recursion depth (or iteration count) in helper functions is independent of the tree size

This space efficiency is a key advantage of the DSW algorithm, especially for large trees where extra space might be prohibitive.

Input:

A visual representation of the input BST:

Step-by-Step Walkthrough:

1. Initialization:

   • A dummy root is created and connected to the tree root.

   • The current_node is set to the dummy root.

2. Step 1: Convert BST to vine (right-leaning linked list)

   • We start with current_node as the dummy root.

   • We continue this process while current_node.right exists.

   • If current_node.right.left exists, we perform a right rotation.

   • Otherwise, we move current_node to current_node.right.

   • Right Rotation 1:

     • current_node is dummy, current_node.right is 4, and current_node.right.left is 3.

     • We perform a right rotation:

       1. Set left_child = node.left (left_child = 3)

       2. Set node.left = left_child.right (4.left = None)

       3. Set left_child.right = node (3.right = 4)

       4. Set parent.right = left_child (dummy.right = 3)

     • After rotation:

       

   • Right Rotation 2:

     • current_node is still dummy, current_node.right is now 3, and current_node.right.left is 2.

     • We perform another right rotation:

       1. Set left_child = node.left (left_child = 2)

       2. Set node.left = left_child.right (3.left = None)

       3. Set left_child.right = node (2.right = 3)

       4. Set parent.right = left_child (dummy.right = 2)

     • After rotation:

       

   • Right Rotation 3:

     • current_node is still dummy, current_node.right is now 2, and current_node.right.left is 1.

     • We perform the final right rotation:

       1. Set left_child = node.left (left_child = 1)

       2. Set node.left = left_child.right (2.left = None)

       3. Set left_child.right = node (1.right = 2)

       4. Set parent.right = left_child (dummy.right = 1)

     • After rotation:

       

   • Vine creation complete:

     • The BST is now a right-leaning linked list (vine)

     • Vine nodes: [1, 2, 3, 4]

3. Step 2: Count nodes and calculate perfect tree size

   • We traverse the vine to count the nodes:

     • Start with current_node = dummy_root.right

     • While current_node exists, increment node_count and move to current_node.right

   • Total node count: 4

   • Calculate perfect tree nodes:

4. Step 3: Balance the tree through a series of left rotations

   • We perform compressions to balance the tree.

   • Initial Compression (1 rotation):

     • Number of rotations = node_count - perfect_tree_nodes = 4 - 3 = 1

     • We perform 1 left rotation:

       1. Set right_child = node.right (right_child = 2)

       2. Set node.right = right_child.left (1.right = None)

       3. Set right_child.left = node (2.left = 1)

       4. Set parent.right = right_child (dummy.right = 2)

     • After compression (Left rotation):

       

   • Remaining Compression (1 rotation):

     • We calculate remaining_nodes = perfect_tree_nodes = 3

     • While remaining_nodes > 1, we divide it by 2: 3 // 2 = 1

     • We perform one more left rotation:

       1. Set right_child = node.right (right_child = 3)

       2. Set node.right = right_child.left (2.right = None)

       3. Set right_child.left = node (3.left = 2)

       4. Set parent.right = right_child (dummy.right = 3)

     • After compression (Left rotation):

       

5. Compression Summary:

   • Initial compressions: 1

   • Compression rounds: [1]

6. Final Balanced Tree:

   • The tree is now balanced, with a height difference of at most 1 between any two subtrees.

The central concept of this solution is to leverage the unique property of a star graph to efficiently identify the center node. In a star graph, the center node is connected to all other nodes, which means it appears on every edge.

• Minimum Edge Requirement: We only need to examine two edges to find the center node.

• Common Node Principle: The center node will be the only node that appears in both of these edges.

1. Function Definition and Edge Selection:

   def findCenter1(edges: List[List[int]]) -> int: reference_edge, comparison_edge = edges[0], edges[1]

   The function takes a list of edges as input. It immediately selects the first two edges from this list, assigning them to reference_edge and comparison_edge. This selection is arbitrary but sufficient due to the star graph's properties.

2. Center Node Identification:

   return (reference_edge[0] if reference_edge[0] in comparison_edge else reference_edge[1])

   This single line performs the core logic of the function:

   • It checks if the first node of the reference_edge is present in the comparison_edge.

   • If true, this node must be the center (as it's in both edges) and is returned.

   • If false, the second node of the reference_edge must be the center (since one of the two must be the center, and we've eliminated the first).

Time Complexity:

• , where is the number of nodes in the graph. This is because the function performs a fixed number of operations regardless of the input size. It only accesses two edges and performs a single comparison.

Space Complexity:

• . The function uses only a constant amount of extra space to store the two selected edges and perform the comparison, regardless of the input size.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • The function receives the edges list as input.

   • reference_edge is set to [1, 2] (the first edge in the list).

   • comparison_edge is set to [5, 1] (the second edge in the list).

2. Center Node Identification:

   • The function checks if the first node of reference_edge (1) is in comparison_edge [5, 1].

   • Condition: 1 in [5, 1] evaluates to True.

   • Since the condition is true, the center node is identified as 1.

3. Visual Aids: A decision summary table is created:

   Reference Edge	Comparison Edge	Condition	Center Node
   [1, 2]	[5, 1]	1 in [5, 1]	1

4. Result Calculation/Final Steps:

   • The function determines that 1 is the center node.

   • This is correct because node 1 appears in all edges of the input, confirming it's connected to all other nodes.

The central concept of this solution is to leverage a greedy approach to maximize the total importance of roads. The key insight is that assigning higher values to cities with more connections will result in the maximum total importance.

• Connection Count: The solution starts by counting how many connections each city has.

• Value Assignment: Cities are then assigned values from 1 to n based on their connection count, with more connected cities receiving higher values.

• Greedy Strategy: This greedy strategy ensures that the cities contributing to more roads get higher values, thus maximizing the overall importance.

1. Initialization and Connection Counting:

   city_connections = [0] * n for city1, city2 in roads: city_connections[city1] += 1 city_connections[city2] += 1

   This step initializes a list city_connections to keep track of how many roads are connected to each city. It then iterates through the roads list, incrementing the count for both cities connected by each road.

2. Initialization of Result Variables:

   total_importance = 0 city_value = 1

   total_importance will store the final result, while city_value is used to assign increasing values to cities.

3. Sorting and Importance Calculation:

   for connections in sorted(city_connections): total_importance += city_value * connections city_value += 1

   This is the core of the algorithm. It sorts the city_connections list, effectively ordering cities from least to most connected. Then, it iterates through this sorted list, calculating the contribution to the total importance for each city and incrementing the city_value for the next iteration.

4. Result Return:

   return total_importance

   The final calculated total_importance is returned as the result.

Time Complexity:

• , where is the number of cities and is the number of roads.

  • Counting connections:

  • Sorting the connection counts:

  • Final calculation loop:

  The overall complexity is dominated by the sorting step, hence .

Space Complexity:

• , where is the number of cities.

  • The city_connections list uses space.

  • The sorting operation in Python (Timsort) uses space in the worst case.

  • All other variables use constant space.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • We start with n = 5 cities and a list of roads connecting them.

   • Initialize city_connections = [0, 0, 0, 0, 0], a list to count connections for each city.

2. Counting Connections:

   • Iterate through each road in roads:

     • Road 1: 0 <-> 1 → Update: city_connections = [1, 1, 0, 0, 0]

     • Road 2: 1 <-> 2 → Update: city_connections = [1, 2, 1, 0, 0]

     • Road 3: 2 <-> 3 → Update: city_connections = [1, 2, 2, 1, 0]

     • Road 4: 0 <-> 2 → Update: city_connections = [2, 2, 3, 1, 0]

     • Road 5: 1 <-> 3 → Update: city_connections = [2, 3, 3, 2, 0]

     • Road 6: 2 <-> 4 → Update: city_connections = [2, 3, 4, 2, 1]

3. Sorting Connections:

   • Sort city_connections in ascending order: [1, 2, 2, 3, 4]

   • Here is how the cities are sorted based on their connection counts:

   City #	Connections
   4	1
   0/3	2
   3/0	2
   1	3
   2	4

4. Calculating Importance:

   • Initialize total_importance = 0 and city_value = 1

   • Iterate through sorted connections:

     • Iteration 1:

       • We start with city 4 (1 connection) and assign it value 1.

       • 1 * 1 = 1, total_importance = 1, incremented city_value for next iteration: 2

     • Iteration 2:

       • The next city is 0/3 (2 connections) and assigned value 2.

       • 2 * 2 = 4, total_importance = 5 (1 + 4), incremented city_value: 3

     • Iteration 3:

       • The next city is 3/0 (2 connections) and assigned value 3.

       • 3 * 2 = 6, total_importance = 11 (5 + 6), incremented city_value: 4

     • Iteration 4:

       • The next city is 1 (3 connections) and assigned value 4.

       • 4 * 3 = 12, total_importance = 23 (11 + 12), incremented city_value: 5

     • Iteration 5:

       • The final city is 2 (4 connections) and assigned value 5.

       • 5 * 4 = 20, total_importance = 43 (23 + 20)

5. Visual Aid:

   Iteration Summary Table:

   City #	Connections	Assigned Value	Importance Contribution	Total Importance
   4	1	1	1	1
   0/3	2	2	4	5
   3/0	2	3	6	11
   1	3	4	12	23
   2	4	5	20	43

6. Result Calculation:

   • The final total_importance is 43, which is the sum of all importance contributions.

   • This value represents the maximum total importance of all roads after optimally assigning values to cities.

The central concept of this solution is to leverage a depth-first search (DFS) approach to propagate ancestor information through a Directed Acyclic Graph (DAG). This method efficiently builds a complete ancestor list for each node by traversing the graph structure.

• Graph Representation: The solution uses an adjacency list to represent the DAG, which allows for efficient lookup of a node's direct children.

• Ancestor Propagation: Instead of finding ancestors for each node independently, the algorithm propagates each node as an ancestor to all of its descendants.

• Recursive Traversal: A depth-first search is implemented recursively to ensure all descendants of a node receive the ancestor information.

1. Initialization:

   direct_children = defaultdict(list) ancestors: List[List[int]] = [[] for _ in range(n)]

   The code initializes two key data structures:

   • direct_children: A defaultdict to store the adjacency list representation of the graph.

   • ancestors: A list of lists to store the ancestors for each node.

2. Building the Adjacency List:

   for parent, child in edges: direct_children[parent].append(child)

   This loop processes the input edges to construct the adjacency list. For each edge, it adds the child node to the list of direct children for the parent node.

3. Defining the Propagation Function:

   def propagate_ancestor(current_ancestor: int, current_node: int): for child in direct_children[current_node]: if (not ancestors[child] or ancestors[child][-1] != current_ancestor): ancestors[child].append(current_ancestor) propagate_ancestor(current_ancestor, child)

   This helper function recursively propagates ancestry information:

   • It iterates through all direct children of the current node.

   • For each child, it checks if the current ancestor needs to be added (avoiding duplicates).

     • If the child has no ancestors or the last ancestor is not the current ancestor, it adds the current ancestor.

       • The last check works because we iterate in ascending order, ensuring the current ancestor is always the latest one.

   • If needed, it adds the ancestor and recursively calls itself for the child node.

4. Initiating Ancestor Propagation:

   for node in range(n): propagate_ancestor(node, node)

   This loop initiates the propagation process for each node in the graph. Each node is considered as its own ancestor to start the propagation.

5. Result Return:

   return ancestors

   The function returns the ancestors list, which now contains the complete list of ancestors for each node.

Time Complexity:

• , where is the number of nodes and is the number of edges. This is because:

  • The outer loop runs times, once for each node.

  • For each iteration, the propagate_ancestor function potentially traverses all descendants of the node, which in the worst case could be all nodes and edges in the graph .

Space Complexity:

• , where is the number of nodes and is the number of edges. This is because:

  • The direct_children adjacency list takes space.

  • The ancestors list takes space in the worst case, where each node could potentially have all other nodes as ancestors.

  • The recursion stack in the worst case could go up to depth.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Initialize direct_children as an empty defaultdict(list).

   • Create ancestors as a list of eight empty lists: [[], [], [], [], [], [], [], []].

2. Building the Adjacency List:

   • Iterate through each edge in edges:

     • For [0, 3]: Add 3 to direct_children[0]

     • For [0, 4]: Add 4 to direct_children[0]

     • For [1, 3]: Add 3 to direct_children[1]

     • For [2, 4]: Add 4 to direct_children[2]

     • For [2, 7]: Add 7 to direct_children[2]

     • For [3, 5]: Add 5 to direct_children[3]

     • For [3, 6]: Add 6 to direct_children[3]

     • For [3, 7]: Add 7 to direct_children[3]

     • For [4, 6]: Add 6 to direct_children[4]

   Final direct_children: {0: [3, 4], 1: [3], 2: [4, 7], 3: [5, 6, 7], 4: [6]}

3. Main Loop: Propagating Ancestors:

   • Iterate through each node from 0 to 7:

   • Node 0:

     • Call propagate_ancestor(0, 0)

     • Propagate to children 3 and 4:

       • For child 3: Add 0 to ancestors[3], then recursively propagate

         • Add 0 to ancestors[5], ancestors[6], and ancestors[7]

       • For child 4: Add 0 to ancestors[4], then recursively propagate

         • Add 0 to ancestors[6] (already present, so skip)

   • Node 1:

     • Call propagate_ancestor(1, 1)

     • Propagate to child 3:

       • Add 1 to ancestors[3], then recursively propagate

         • Add 1 to ancestors[5], ancestors[6], and ancestors[7]

   • Node 2:

     • Call propagate_ancestor(2, 2)

     • Propagate to children 4 and 7:

       • For child 4: Add 2 to ancestors[4], then recursively propagate

         • Add 2 to ancestors[6]

       • For child 7: Add 2 to ancestors[7]

   • Node 3:

     • Call propagate_ancestor(3, 3)

     • Propagate to children 5, 6, and 7:

       • Add 3 to ancestors[5], ancestors[6], and ancestors[7]

   • Node 4:

     • Call propagate_ancestor(4, 4)

     • Propagate to child 6:

       • Add 4 to ancestors[6]

   • Nodes 5, 6, and 7:

     • Call propagate_ancestor for each, but they have no children, so no further propagation occurs

4. Visual Aid: Iteration Summary

   Node	Ancestors	Direct Children
   0	[]	[3, 4]
   1	[]	[3]
   2	[]	[4, 7]
   3	[0, 1]	[5, 6, 7]
   4	[0, 2]	[6]
   5	[0, 1, 3]	[]
   6	[0, 1, 2, 3, 4]	[]
   7	[0, 1, 2, 3]	[]

5. Result Calculation/Final Steps:

   • The ancestors list now contains the final result: [[], [], [], [0, 1], [0, 2], [0, 1, 3], [0, 1, 2, 3, 4], [0, 1, 2, 3]]

   • This list represents the ancestors of each node in ascending order:

     • Node 0: No ancestors

     • Node 1: No ancestors

     • Node 2: No ancestors

     • Node 3: Ancestors are 0 and 1

     • Node 4: Ancestors are 0 and 2

     • Node 5: Ancestors are 0, 1, and 3

     • Node 6: Ancestors are 0, 1, 2, 3, and 4

     • Node 7: Ancestors are 0, 1, 2, and 3

   The function returns this ancestors list as the final result.

The central concept of this solution is to leverage a depth-first search (DFS) approach combined with memoization to efficiently compute and store the ancestors of each node in a Directed Acyclic Graph (DAG). This method builds a complete ancestor list for each node by traversing the graph structure upwards and caching results to avoid redundant computations.

• Reverse Graph Representation: The solution uses a reverse adjacency list (direct_parents) to represent the DAG, allowing for efficient upward traversal from a node to its parents.

• Recursive Memoization: A depth-first search is implemented recursively with memoization, ensuring that each node's ancestors are computed only once and then cached for future use.

• Set Operations: The use of sets for intermediate computations allows for efficient union operations and automatic elimination of duplicates.

1. Initialization:

   direct_parents = defaultdict(list) ancestors = [[] for _ in range(n)]

   The code initializes two key data structures:

   • direct_parents: A defaultdict to store the reverse adjacency list of the graph.

   • ancestors: A list of lists to store the final sorted ancestors for each node.

2. Building the Reverse Adjacency List:

   for parent, child in edges: direct_parents[child].append(parent)

   This loop processes the input edges to construct the reverse adjacency list. For each edge, it adds the parent node to the list of direct parents for the child node.

3. Defining the Recursive Ancestor-Finding Function:

   def find_all_ancestors(node: int) -> set: if ancestors[node]: return set(ancestors[node]) result = set() for parent in direct_parents[node]: result.update(find_all_ancestors(parent)) result.update(direct_parents[node]) ancestors[node] = list(result) return result

   This helper function recursively finds and memoizes all ancestors of a given node:

   • It first checks if the ancestors for the node are already computed (memoization).

   • If not, it recursively computes the ancestors of each parent and updates the result set.

   • It then adds the direct parents to the result set.

   • Finally, it caches the result in the ancestors list and returns the set.

4. Initiating Ancestor Computation:

   for node in range(n): if not ancestors[node]: find_all_ancestors(node)

   This loop initiates the ancestor computation process for each node in the graph that hasn't been processed yet.

5. Sorting the Ancestors:

   for node in range(n): ancestors[node].sort()

   This step sorts the ancestors for each node in ascending order, as required by the problem statement.

6. Result Return:

   return ancestors

   The function returns the ancestors list, which now contains the complete, sorted list of ancestors for each node.

Time Complexity:

• , where is the number of nodes and is the number of edges. This is because:

  • In the worst case, each node might need to traverse all edges to find its ancestors.

  • The sorting step for each node takes time, but this is dominated by the traversal in sparse graphs.

  • The memoization ensures that each node's ancestors are computed only once.

Space Complexity:

• , where is the number of nodes and is the number of edges. This is because:

  • The direct_parents adjacency list takes space.

  • The ancestors list takes space in the worst case, where each node could potentially have all other nodes as ancestors.

  • The recursion stack in the worst case could go up to depth.

  • The set used for intermediate computations can take up to space for each recursive call.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Initialize direct_parents as an empty defaultdict(list).

   • Create ancestors as a list of eight empty lists: [[], [], [], [], [], [], [], []].

2. Building the Reverse Adjacency List:

   • Iterate through each edge in edges:

     • For [0, 3]: Add 0 to direct_parents[3]

     • For [0, 4]: Add 0 to direct_parents[4]

     • For [1, 3]: Add 1 to direct_parents[3]

     • For [2, 4]: Add 2 to direct_parents[4]

     • For [2, 7]: Add 2 to direct_parents[7]

     • For [3, 5]: Add 3 to direct_parents[5]

     • For [3, 6]: Add 3 to direct_parents[6]

     • For [3, 7]: Add 3 to direct_parents[7]

     • For [4, 6]: Add 4 to direct_parents[6]

   Final direct_parents: {3: [0, 1], 4: [0, 2], 7: [2, 3], 5: [3], 6: [3, 4]}

3. Main Loop: Computing Ancestors:

   • Iterate through each node from 0 to 7:

   • Node 0:

     • Call find_all_ancestors(0)

     • No direct parents, return empty set

     • ancestors[0] remains []

   • Node 1:

     • Call find_all_ancestors(1)

     • No direct parents, return empty set

     • ancestors[1] remains []

   • Node 2:

     • Call find_all_ancestors(2)

     • No direct parents, return empty set

     • ancestors[2] remains []

   • Node 3:

     • Call find_all_ancestors(3)

     • Direct parents: [0, 1]

     • Recursively call for parent 0 (returns an empty set)

     • Recursively call for parent 1 (returns an empty set)

     • Add direct parents [0, 1] to result

     • Set ancestors[3] = [0, 1]

   • Node 4:

     • Call find_all_ancestors(4)

     • Direct parents: [0, 2]

     • Recursively call for parent 0 (returns an empty set)

     • Recursively call for parent 2 (returns an empty set)

     • Add direct parents [0, 2] to result

     • Set ancestors[4] = [0, 2]

   • Node 5:

     • Call find_all_ancestors(5)

     • Direct parents: [3]

     • Recursively call for parent 3 (returns {0, 1})

     • Add direct parent 3 to result

     • Set ancestors[5] = [0, 1, 3]

   • Node 6:

     • Call find_all_ancestors(6)

     • Direct parents: [3, 4]

     • Recursively call for parent 3 (returns {0, 1})

     • Recursively call for parent 4 (returns {0, 2})

     • Add direct parents [3, 4] to result

     • Set ancestors[6] = [0, 1, 2, 3, 4]

   • Node 7:

     • Call find_all_ancestors(7)

     • Direct parents: [2, 3]

     • Recursively call for parent 2 (returns an empty set)

     • Recursively call for parent 3 (returns {0, 1})

     • Add direct parents [2, 3] to result

     • Set ancestors[7] = [0, 1, 2, 3]

4. Sorting Ancestors:

   • For each node, sort its list of ancestors:

     • Node 0: [] (no change)

     • Node 1: [] (no change)

     • Node 2: [] (no change)

     • Node 3: [0, 1] (already sorted)

     • Node 4: [0, 2] (already sorted)

     • Node 5: [0, 1, 3] (already sorted)

     • Node 6: [0, 1, 2, 3, 4] (already sorted)

     • Node 7: [0, 1, 2, 3] (already sorted)

5. Visual Aid: Iteration Summary

   Node	Ancestors	Direct Parents
   0	[]	[]
   1	[]	[]
   2	[]	[]
   3	[0, 1]	[0, 1]
   4	[0, 2]	[0, 2]
   5	[0, 1, 3]	[3]
   6	[0, 1, 2, 3, 4]	[3, 4]
   7	[0, 1, 2, 3]	[2, 3]

6. Result Calculation/Final Steps:

   • The ancestors list now contains the final result: [[], [], [], [0, 1], [0, 2], [0, 1, 3], [0, 1, 2, 3, 4], [0, 1, 2, 3]]

   • This list represents the sorted ancestors of each node:

     • Node 0: No ancestors

     • Node 1: No ancestors

     • Node 2: No ancestors

     • Node 3: Ancestors are 0 and 1

     • Node 4: Ancestors are 0 and 2

     • Node 5: Ancestors are 0, 1, and 3

     • Node 6: Ancestors are 0, 1, 2, 3, and 4

     • Node 7: Ancestors are 0, 1, 2, and 3

   The function returns this ancestors list as the final result.

The central concept of this solution is to combine a breadth-first search (BFS) approach with topological sorting to efficiently compute the ancestors of each node in a Directed Acyclic Graph (DAG). This method processes nodes in an order that ensures all ancestors are computed before they're needed, allowing for a single pass through the graph.

• Graph Representation: The solution uses an adjacency list (direct_children) to represent the DAG, facilitating efficient traversal of child nodes.

• Topological Sorting: By using in-degree counting and a queue-based approach, the algorithm processes nodes in topological order, ensuring that a node is only processed after all its ancestors have been computed.

• Set Operations: The use of sets for storing ancestors allows for efficient update operations and automatic deduplication.

1. Initialization:

   direct_children = defaultdict(list) ancestors = [set() for _ in range(n)] in_degree = [0] * n

   The code initializes three key data structures:

   • direct_children: A defaultdict to store the adjacency list of the graph.

   • ancestors: A list of sets to store the ancestors for each node.

   • in_degree: An array to keep track of the number of incoming edges for each node.

2. Building the Graph and Initial Ancestor Sets:

   for parent, child in edges: direct_children[parent].append(child) ancestors[child].add(parent) in_degree[child] += 1

   This loop processes the input edges to:

   • Construct the adjacency list (direct_children).

   • Initialize the direct ancestors for each node.

   • Calculate the in-degree for each node.

3. Initializing the Queue for Topological Sort:

   queue = deque(node for node in range(n) if in_degree[node] == 0)

   This step identifies all nodes with no incoming edges (in-degree of 0) and adds them to the queue. These nodes serve as the starting points for the topological traversal.

4. Processing Nodes in Topological Order:

   while queue: current_node = queue.popleft() for child_node in direct_children[current_node]: ancestors[child_node].update(ancestors[current_node]) in_degree[child_node] -= 1 if in_degree[child_node] == 0: queue.append(child_node)

   This is the core BFS loop that processes nodes in topological order:

   • It pops a node from the queue and processes all its children.

   • For each child, it updates its ancestor set by adding all ancestors of the current node.

   • It decrements the in-degree of the child node.

   • If a child's in-degree becomes 0, it's added to the queue for processing.

5. Result Preparation and Return:

   return [sorted(node_ancestors) for node_ancestors in ancestors]

   This final step converts the sets of ancestors to sorted lists, as required by the problem statement, and returns the result.

Time Complexity:

• in the worst case, where is the number of nodes and is the number of edges. This is because:

  • Building the initial graph structure takes time.

  • In the worst case (e.g., a chain structure), each node might accumulate all previous nodes as ancestors, leading to total ancestors across all nodes.

  • Sorting these large ancestor lists takes time per node, resulting in for all nodes.

  • The BFS traversal itself processes each edge once, taking time.

Space Complexity:

• , which simplifies to , where is the number of nodes and is the number of edges. This is because:

  • The ancestors list of sets can take up to space in the worst case, where each node has all other nodes as ancestors.

  • The direct_children adjacency list takes space.

  • The in_degree array and queue take space.

It's worth noting that while this worst-case time complexity appears higher than the previous approaches, this algorithm often performs better in practice for many graph structures, especially sparse graphs, due to its non-recursive nature and efficient set operations.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Initialize direct_children as an empty defaultdict(list).

   • Create ancestors as a list of eight empty sets: [set(), set(), set(), set(), set(), set(), set(), set()].

   • Initialize in_degree as a list of 8 zeros: [0, 0, 0, 0, 0, 0, 0, 0].

2. Building Graph Data Structures:

   • Iterate through each edge in edges:

     • For [0, 3]: Add 3 to direct_children[0], add 0 to ancestors[3], increment in_degree[3]

     • For [0, 4]: Add 4 to direct_children[0], add 0 to ancestors[4], increment in_degree[4]

     • For [1, 3]: Add 3 to direct_children[1], add 1 to ancestors[3], increment in_degree[3]

     • For [2, 4]: Add 4 to direct_children[2], add 2 to ancestors[4], increment in_degree[4]

     • For [2, 7]: Add 7 to direct_children[2], add 2 to ancestors[7], increment in_degree[7]

     • For [3, 5]: Add 5 to direct_children[3], add 3 to ancestors[5], increment in_degree[5]

     • For [3, 6]: Add 6 to direct_children[3], add 3 to ancestors[6], increment in_degree[6]

     • For [3, 7]: Add 7 to direct_children[3], add 3 to ancestors[7], increment in_degree[7]

     • For [4, 6]: Add 6 to direct_children[4], add 4 to ancestors[6], increment in_degree[6]

   Final state:

   • direct_children: {0: [3, 4], 1: [3], 2: [4, 7], 3: [5, 6, 7], 4: [6]}

   • ancestors: [set(), set(), set(), {0, 1}, {0, 2}, {3}, {3, 4}, {2, 3}]

   • in_degree: [0, 0, 0, 2, 2, 1, 2, 2]

3. Initializing Queue:

   • Add nodes with in-degree 0 to the queue: [0, 1, 2]

4. Processing Nodes in Topological Order:

   • Process Node 0:

     • Current queue: [1, 2]

     • Children: [3, 4]

     • For child 3:

       • Update ancestors[3] to {0, 1}

       • Decrement in_degree[3] to 1

     • For child 4:

       • Update ancestors[4] to {0, 2}

       • Decrement in_degree[4] to 1

   • Process Node 1:

     • Current queue: [2]

     • Children: [3]

     • For child 3:

       • Update ancestors[3] to {0, 1}

       • Decrement in_degree[3] to 0

       • Add 3 to queue: [2, 3]

   • Process Node 2:

     • Current queue: [3]

     • Children: [4, 7]

     • For child 4:

       • Update ancestors[4] to {0, 2}

       • Decrement in_degree[4] to 0

       • Add 4 to queue: [3, 4]

     • For child 7:

       • Update ancestors[7] to {2, 3}

       • Decrement in_degree[7] to 1

   • Process Node 3:

     • Current queue: [4]

     • Children: [5, 6, 7]

     • For child 5:

       • Update ancestors[5] to {0, 1, 3}

       • Decrement in_degree[5] to 0

       • Add 5 to queue: [4, 5]

     • For child 6:

       • Update ancestors[6] to {0, 1, 3, 4}

       • Decrement in_degree[6] to 1

     • For child 7:

       • Update ancestors[7] to {0, 1, 2, 3}

       • Decrement in_degree[7] to 0

       • Add 7 to queue: [4, 5, 7]

   • Process Node 4:

     • Current queue: [5, 7]

     • Children: [6]

     • For child 6:

       • Update ancestors[6] to {0, 1, 2, 3, 4}

       • Decrement in_degree[6] to 0

       • Add 6 to queue: [5, 7, 6]

   • Process Node 5:

     • Current queue: [7, 6]

     • No children, hence we do nothing

   • Process Node 7:

     • Current queue: [6]

     • No children, hence we do nothing

   • Process Node 6:

     • Current queue: []

     • No children, hence we do nothing

5. Sorting Ancestors:

   • For each node, convert its set of ancestors to a sorted list:

     • Node 0: [] (empty set to empty list)

     • Node 1: [] (empty set to empty list)

     • Node 2: [] (empty set to empty list)

     • Node 3: [0, 1]

     • Node 4: [0, 2]

     • Node 5: [0, 1, 3]

     • Node 6: [0, 1, 2, 3, 4]

     • Node 7: [0, 1, 2, 3]

6. Visual Aid: Iteration Summary

   Processed Node	Ancestors	Direct Children	Final In-Degree
   0	set()	[3, 4]	0
   1	set()	[3]	0
   2	set()	[4, 7]	0
   3	{0, 1}	[5, 6, 7]	0
   4	{0, 2}	[6]	0
   5	{0, 1, 3}	[]	0
   7	{0, 1, 2, 3}	[]	0
   6	{0, 1, 2, 3, 4}	[]	0

7. Result Calculation/Final Steps:

   • The ancestors list now contains the final result: [[], [], [], [0, 1], [0, 2], [0, 1, 3], [0, 1, 2, 3, 4], [0, 1, 2, 3]]

   • This list represents the sorted ancestors of each node:

     • Node 0: No ancestors

     • Node 1: No ancestors

     • Node 2: No ancestors

     • Node 3: Ancestors are 0 and 1

     • Node 4: Ancestors are 0 and 2

     • Node 5: Ancestors are 0, 1, and 3

     • Node 6: Ancestors are 0, 1, 2, 3, and 4

     • Node 7: Ancestors are 0, 1, 2, and 3

   The function returns this ancestors list as the final result.

The central concept of this solution is to leverage the Union-Find data structure to efficiently track connected components for both Alice and Bob separately. The solution processes edges in two passes, prioritizing shared (type 3) edges to potentially maximize removable edges.

• Dual Union-Find Structures: The solution uses two separate Union-Find structures, one for Alice and one for Bob, to track their respective traversable paths independently.

• Two-Pass Edge Processing: Edges are processed in two passes - first type 3 (shared) edges, then type 1 ( Alice-only) and type 2 (Bob-only) edges. This prioritization helps maximize the number of removable edges.

• Essential Edge Tracking: The solution keeps count of "essential" edges that must be kept for full traversal, which is used to calculate the maximum removable edges.

1. Initialization:

   alice_uf = UnionFind(size=n, offset=1) bob_uf = UnionFind(size=n, offset=1) essential_edges = 0

   Two UnionFind structures are initialized, one for Alice and one for Bob. The offset=1 is used because node numbering starts from 1. essential_edges keeps track of edges that must be kept.

2. First Pass - Processing Type 3 Edges:

   for edge_type, u, v in edges: if edge_type == 3: essential_edges += alice_uf.union(u, v) | bob_uf.union(u, v) if alice_uf.is_single_component(): return len(edges) - essential_edges

   This loop processes all type 3 edges first. The union operation is performed for both Alice and Bob, and the result is OR'ed to increment essential_edges if either union was successful. If Alice's graph becomes fully connected, we can immediately return the result as Bob's graph will also be fully connected.

3. Second Pass - Processing Type 1 and Type 2 Edges:

   for edge_type, u, v in edges: if edge_type == 1: if alice_uf.union(u, v): essential_edges += 1 elif edge_type == 2: if bob_uf.union(u, v): essential_edges += 1 if alice_uf.is_single_component() and bob_uf.is_single_component(): return len(edges) - essential_edges

   This loop processes type 1 (Alice-only) and type 2 (Bob-only) edges. It performs the union operation and increments essential_edges if the union was successful. After each edge, it checks if both graphs are fully connected.

4. Result Calculation/Return:

   return -1 # Full traversal is not possible for both Alice and Bob

   If the function reaches this point, it means full traversal is not possible for both Alice and Bob, so it returns -1.

Time Complexity:

• , where is the number of nodes, is the number of edges, and is the inverse Ackermann function.

• This is because each Union-Find operation (union and find) takes amortized time, and we perform at most such operations (once for each edge in each pass).

• The inverse Ackermann function grows extremely slowly and is effectively constant for all practical values of .

Space Complexity:

• , where is the number of nodes.

• This is due to the two UnionFind data structures, each storing parent and rank information for nodes.

• The space used is independent of the number of edges, making it efficient for dense graphs.

Input:

Step-by-Step Walkthrough:

1. Initialization:

   • Initialize two UnionFind data structures: alice_uf and bob_uf, both with size 4 and offset 1.

   • Set essential_edges to 0.

2. First Pass: Processing Type 3 Edges

   • Iteration 1:

     • Process edge [3, 1, 2]

     • For both Alice and Bob:

       • Union nodes 1 and 2

       • This is a new connection, so increment essential_edges to 1

     • Check if Alice's graph is fully connected (it's not)

   • Iteration 2:

     • Process edge [3, 2, 3]

     • For both Alice and Bob:

       • Union nodes 2 and 3

       • This is a new connection, so increment essential_edges to 2

     • Check if Alice's graph is fully connected (it's not)

3. Second Pass: Processing Type 1 and Type 2 Edges

   • Iteration 3:

     • Process edge [1, 1, 3]

     • For Alice only:

       • Attempt to union nodes 1 and 3

       • They are already connected, so no change to essential_edges

     • Check if both graphs are fully connected (they're not)

   • Iteration 4:

     • Process edge [1, 2, 4]

     • For Alice only:

       • Union nodes 2 and 4

       • This is a new connection, so increment essential_edges to 3

     • Check if both graphs are fully connected (Alice's is, Bob's isn't)

   • Iteration 5:

     • Process edge [1, 1, 2]

     • For Alice only:

       • Attempt to union nodes 1 and 2

       • They are already connected, so no change to essential_edges

     • Check if both graphs are fully connected (Alice's is, Bob's isn't)

   • Iteration 6:

     • Process edge [2, 3, 4]

     • For Bob only:

       • Union nodes 3 and 4

       • This is a new connection, so increment essential_edges to 4

     • Check if both graphs are fully connected (they are)

4. Visual Aid:

   Here's a table summarizing the edge processing:

   Edge #	Type	U	V	Alice Union	Bob Union	Essential Edges
   1	3	1	2	True	True	1
   2	3	2	3	True	True	2
   3	1	1	3	False	-	2
   4	1	2	4	True	-	3
   5	1	1	2	False	-	3
   6	2	3	4	-	True	4

5. Result Calculation:

   • Total number of edges: 6

   • Number of essential edges: 4

   • Maximum number of edges that can be removed: 6 - 4 = 2

The algorithm determines that two edges can be removed while still allowing both Alice and Bob to fully traverse the graph. These removable edges are the ones that didn't result in new connections: [1, 1, 3] and [1, 1, 2].