Longest Palindromic Path in Graph

This problem requires identifying the maximum length path in a labeled graph that reads the same forwards and backwards. Use state transition dynamic programming combined with bitmasking to track visited nodes and ensure paths remain valid. Carefully handle node selection and label matching to extend palindrome candidates efficiently.

Problem Statement

Given an integer n and an undirected graph with n nodes labeled 0 to n - 1, and a 2D array edges where each edges[i] = [ui, vi] represents an edge between nodes ui and vi. Each node has an associated lowercase character from the string label, where label[i] corresponds to node i.

You may start at any node and move to any adjacent node, visiting each node at most once. Determine the length of the longest path such that the concatenated labels along the path form a palindrome.

Examples

Example 1

Input: n = 3, edges = [[0,1],[1,2]], label = "aba"

Output: 3 Exp lanation:

Example details omitted.

Example 2

Input: n = 3, edges = [[0,1],[0,2]], label = "abc"

Output: 1

Example 3

Input: n = 4, edges = [[0,2],[0,3],[3,1]], label = "bbac"

Output: 3

Constraints

• 1 <= n <= 14
• n - 1 <= edges.length <= n * (n - 1) / 2
• edges[i] == [ui, vi]
• 0 <= ui, vi <= n - 1
• ui != vi
• label.length == n
• label consists of lowercase English letters.
• There are no duplicate edges.

Solution Approach

Bitmask Dynamic Programming

Use a bitmask to track visited nodes and memoize subproblem results with state transition DP. Each state is defined by the current node pair and visited nodes, storing the maximum palindromic length achievable.

State Transition Between Node Pairs

Consider transitions from node pairs where labels match at both ends. Expand recursively or iteratively while updating DP states. Ensure each extension maintains palindrome properties and does not revisit nodes.

Maximizing Path Length

Iterate over all starting node pairs and select moves that extend palindromes. Update DP table only when the new path increases the palindrome length. Track the global maximum across all states for the final answer.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is roughly O(n^2 * 2^n) due to bitmask DP over all node pairs and visited sets. Space complexity is also O(n^2 * 2^n) to store DP states for all node pairs with all subsets of visited nodes.

What Interviewers Usually Probe

• Notice small n constraint suggests bitmask DP is feasible.
• Look for symmetry in labels to prune invalid paths early.
• Check how state transitions efficiently extend palindrome candidates without repetition.

Common Pitfalls or Variants

Common pitfalls

• Failing to track visited nodes accurately, leading to revisiting nodes.
• Not correctly handling label matching at both ends of the path.
• Overlooking state memoization, causing exponential recomputation.

Follow-up variants

• Longest palindromic path starting from a fixed node instead of any node.
• Compute the number of distinct longest palindromic paths instead of length.
• Apply the same approach on directed graphs with palindromic constraints.

FAQ

What is the best approach for Longest Palindromic Path in Graph?

Use state transition dynamic programming with bitmasking to track visited nodes and extend palindrome paths efficiently.

Can nodes be revisited in the path?

No, each node can be visited at most once to maintain valid paths and correct bitmask states.

Why is bitmasking necessary for this problem?

Bitmasking allows efficient tracking of visited nodes and avoids exponential recomputation of overlapping subproblems.

What is the time complexity of this solution?

Time complexity is O(n^2 * 2^n) since each DP state is defined by a pair of nodes and a subset of visited nodes.

How do state transitions ensure palindrome paths?

Transitions only extend paths where labels at the current node pair match, preserving palindrome properties along the path.