Start by treating the tree as a directed graph with parent-child relationships. Use a DFS from the root to calculate the maximum path length while ensuring no two adjacent nodes share the same character. Combine the longest branches at each node to track the global maximum path length efficiently.
Problem Statement
You are given a tree rooted at node 0 represented by an array parent of size n, where parent[i] is the parent of node i and parent[0] == -1. Each node i has a character s[i] assigned from a lowercase English string s.
Return the length of the longest path in this tree such that no two adjacent nodes along the path have the same character. For example, traversing from parent to child nodes while avoiding repeated characters maximizes the path length.
Examples
Example 1
Input: parent = [-1,0,0,1,1,2], s = "abacbe"
Output: 3
The longest path where each two adjacent nodes have different characters in the tree is the path: 0 -> 1 -> 3. The length of this path is 3, so 3 is returned. It can be proven that there is no longer path that satisfies the conditions.
Example 2
Input: parent = [-1,0,0,0], s = "aabc"
Output: 3
The longest path where each two adjacent nodes have different characters is the path: 2 -> 0 -> 3. The length of this path is 3, so 3 is returned.
Constraints
- n == parent.length == s.length
- 1 <= n <= 105
- 0 = 1
- parent[0] == -1
- parent represents a valid tree.
- s consists of only lowercase English letters.
Solution Approach
DFS From Root
Perform a depth-first search starting at node 0, recursively computing the longest path in each subtree. At each node, track the two longest child paths that satisfy the adjacent-character constraint to combine into a potential longer path through this node.
Combine Branches Carefully
At every node, consider only child paths whose end character differs from the current node. Maintain the top two lengths to calculate the longest path passing through the current node. Update a global maximum whenever a longer valid path is found.
Leverage Tree Structure Efficiently
Construct the adjacency list from the parent array to traverse children efficiently. Avoid revisiting nodes by leveraging the tree's inherent acyclic structure, which guarantees each node has a single path back to the root, ensuring O(n) time complexity.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
Time complexity is O(n) because each node is visited once during DFS. Space complexity is O(n) for the adjacency list and recursion stack. The approach scales linearly with tree size, avoiding redundant traversal of nodes or subtrees.
What Interviewers Usually Probe
- Focus on graph indegree plus topological ordering.
- Watch for combining two longest child paths at each node.
- Verify edge cases where characters repeat consecutively along branches.
Common Pitfalls or Variants
Common pitfalls
- Counting paths with repeated adjacent characters.
- Ignoring second-longest child when combining paths through a node.
- Mismanaging recursion stack leading to excessive memory use.
Follow-up variants
- Find the longest path where adjacent nodes differ by at least k characters in a string.
- Compute the longest path in a k-ary tree with unique node labels.
- Determine the maximum path length avoiding a specified subset of characters.
How GhostInterview Helps
- Guides DFS implementation while tracking character constraints at each node.
- Highlights pattern of combining top two child paths to maximize global path length.
- Provides step-by-step example validation like parent = [-1,0,0,1,1,2] with s = 'abacbe'.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the main strategy for Longest Path With Different Adjacent Characters?
Use DFS from the root to track longest paths per subtree, combining branches while ensuring adjacent characters differ.
Why is topological reasoning important in this problem?
It ensures nodes are processed from leaves to root correctly, capturing the longest path contributions from all children.
How do I handle adjacent nodes with the same character?
Exclude paths where the child character matches the current node when calculating maximum path lengths.
What is the time complexity for this tree path problem?
O(n) because each node is visited once in DFS and adjacency list construction is linear.
Can GhostInterview help me debug path calculations for example inputs?
Yes, it walks through sample trees step by step, verifying branch combinations and maximum path updates efficiently.
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