Prefix sum with a hash map is the clean solution here.
Transform the problem by treating 0s as -1 and 1s as +1, then find the longest subarray with sum 0. Use a hash map to store the first occurrence of each prefix sum, and for each index, check if the current prefix sum has been seen before; if so, update the maximum length. This yields an O(n) time and O(n) space solution.
Pro tip: Start by discussing the brute-force O(n^2) approach to show understanding, then optimize to O(n) using prefix sums and a hash map. Emphasize that the key insight is converting 0s to -1, which transforms the problem into finding the longest subarray with sum zero.
Confirm the problem: given a binary array, find the maximum length of a contiguous subarray with equal number of 0s and 1s. Ask about edge cases like empty array or no such subarray.
Mention that a naive solution checks all subarrays, counting 0s and 1s, which takes O(n^2) time. This shows you can think of a baseline.
Explain the transformation: replace 0 with -1. Then the problem reduces to finding the longest subarray with sum 0. Use a hash map to store the first index where each prefix sum occurs.
Initialize a hash map with {0: -1} to handle subarrays starting at index 0. Iterate through the array, maintain a running sum, and for each index, if the sum is in the map, update max length; else, store the sum with the current index.
State that time complexity is O(n) and space is O(n). Discuss edge cases: all 0s or all 1s (result 0), and arrays where the entire array has equal counts.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
The 'return the path' part is what got me.
Use a post-order DFS that returns the longest strictly increasing downward path starting at each node, and track the global maximum path. At each node, compare its value with children: if child value > node value, extend the child's path; otherwise start a new path of length 1. Store the actual node values in the returned path to reconstruct the final answer.
Pro tip: Clarify whether the path must be strictly increasing and whether it can go through a node with only one child; also mention that storing paths as lists simplifies reconstruction but may use O(n) space in the worst case, so you can optimize by storing parent pointers if needed.
Confirm that the path must be strictly increasing (parent value < child value) and that it can start and end at any nodes. Ask if the tree can be empty or have duplicate values.
Design a DFS function that returns the longest strictly increasing path starting from the current node as a list of values. The path includes the current node and extends to at most one child (since it's a downward path).
Recursively get paths from left and right children. If a child's value is greater than the current node's value, the path can be extended by prepending the current node to that child's path. Choose the longer valid extension (or start a new path of length 1 if none).
Compare the length of the path starting at the current node with the global maximum length. If it's longer, update the global maximum and store the actual path.
After the DFS completes, return the stored global maximum path. Discuss time and space complexity: O(n) time and O(h) space for recursion, plus O(n) for storing paths in the worst case.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.