Uber·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Uber SWE interview with a tree-based string problem that looks approachable until you actually think about the constraints. The palindrome angle is a nice twist but the ancestor path traversal part is where things get tricky fast.

Questions Asked (1)

You're given a tree where each node holds a character. For a given query node, count how many ancestors (including the node itself) have the property that the substring formed by the path from the query node up to that ancestor can be rearranged into a palindrome.

Algorithms & Data Structures

Author's notes

My first instinct was to just do a DFS per query and track character frequencies as I walked up the path.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Use a bitmask to represent the parity of character counts along the root-to-node path, since a string can be rearranged into a palindrome iff at most one character has an odd count. For each query node, count how many ancestors (including itself) have a bitmask that differs from the node's bitmask by at most one bit. Preprocess the tree with DFS to compute bitmasks and use a hash map to store counts of bitmasks along the current path, enabling O(1) query per ancestor check.

Pro tip: Mention that the bitmask approach reduces the problem to counting ancestors with a bitmask that is either equal or differs by exactly one bit, and that using a hash map during DFS allows O(1) lookup per node, making the overall complexity O(N * 26) for preprocessing and O(1) per query if we store answers during DFS.

1. Understand the palindrome condition

A string can be rearranged into a palindrome if and only if at most one character has an odd frequency. Represent the parity of character counts as a bitmask of 26 bits.

2. Preprocess the tree with DFS

Perform a DFS from the root, maintaining the current bitmask (XOR of the character bit at each node). For each node, store its bitmask and also maintain a hash map counting occurrences of each bitmask along the current path from root to the current node.

3. Count valid ancestors for each node

For a node with bitmask M, the valid ancestors are those with bitmask M (even counts) or M XOR (1<<i) for some i (one odd count). Use the hash map to get counts of these bitmasks along the path, and sum them to get the answer for that node.

4. Handle queries efficiently

If queries are given offline, compute answers for all nodes during the DFS and store them in an array. If online, we can still answer each query in O(1) after preprocessing by storing the count for each node.

5. Analyze complexity and edge cases

Time complexity: O(N * 26) for preprocessing (since for each node we check up to 27 bitmasks) and O(1) per query. Space: O(N) for storing bitmasks and answers. Discuss edge cases like single-node tree, all same characters, etc.

Key Points to Mention

Palindrome rearrangement condition: at most one character with odd frequency.
Bitmask representation: 26 bits for lowercase English letters, XOR to toggle parity.
DFS with backtracking: maintain a hash map of bitmask counts along the current path.
Counting valid ancestors: check current bitmask and bitmasks with one bit flipped.
Complexity: O(N * 26) preprocessing, O(1) per query, O(N) space.
Handling queries: either precompute answers for all nodes or answer online after preprocessing.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.