The palindrome condition clicked pretty fast for me, at most one character with odd frequency, so you can encode a path as a 26-bit XOR mask.
Use an iterative post-order traversal to avoid stack overflow, and for each node, maintain a bitmask of character parities along the path from the root. Count valid pairs by checking if the XOR of two path masks has at most one bit set, using a hash map to track frequencies of masks in the subtree.
Pro tip: Mention that the bitmask approach reduces the palindrome condition to a simple XOR check, and that using DSU on tree (small-to-large merging) can achieve O(N log N) time, which is crucial for large N.
Confirm that a path between two nodes is the unique simple path, and that rearranging characters into a palindrome requires at most one character with odd frequency. Define the parity mask: a 26-bit integer where each bit indicates odd count of a character.
Implement an iterative post-order DFS using an explicit stack to compute the parity mask from root to each node without recursion, avoiding call stack overflow.
For each node, count pairs with nodes in its subtree where the XOR of their path masks has at most one bit set. Use a hash map to store mask frequencies and merge subtrees using small-to-large merging (DSU on tree) to keep time O(N log N).
Consider pairs where one node is an ancestor of the other (including the root), and ensure the algorithm counts each pair exactly once. Also handle N=1 and large N gracefully.
Explain that the time complexity is O(26 * N log N) due to bitmask checks and merging, and space complexity is O(N) for the hash map and stack. Emphasize that this avoids recursion depth issues.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.