How do you solve Longest Palindromic Path in Graph on LeetCode?

Longest Palindromic Path in Graph (LeetCode #3615) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * 2^n) time complexity and O(2^n) space complexity. Key insight: Utilizing bitmasking allows efficient tracking of visited nodes.

What is the time complexity of Longest Palindromic Path in Graph?

The optimal solution for Longest Palindromic Path in Graph runs in O(n * 2^n) time and uses O(2^n) space. The complexity arises from the bitmasking approach, where each subset of nodes is processed.

What is the brute force solution for Longest Palindromic Path in Graph?

The brute force approach for Longest Palindromic Path in Graph is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible paths in the graph and check if the concatenated labels form a palindrome. This approach is straightforward but inefficient due to the exponential number of paths. The optimal Optimal Solution approach improves this to O(n * 2^n).

What are the interview tips for Longest Palindromic Path in Graph?

Explain your thought process clearly. Discuss edge cases and how they affect your solution. Practice similar problems to build confidence.

What are common mistakes when solving Longest Palindromic Path in Graph?

Overlooking the need to track character counts for odd/even occurrences. Not considering all possible paths from each node.

#3615

Longest Palindromic Path in Graph

Hard

StringDynamic ProgrammingBit ManipulationGraph TheoryBitmaskBitmaskingDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * 2^n)
Space	O(1)	O(2^n)

💡

Intuition

Time O(n * 2^n)Space O(2^n)

Use bitmask dynamic programming to track visited nodes and character counts. This allows efficient checking of potential palindrome formations.

⚙️

Algorithm

3 steps

1Step 1: Initialize a DP table with bitmasks representing visited nodes and their character counts.
2Step 2: Iterate through each node, updating the DP table for each valid path.
3Step 3: Calculate the maximum palindrome length based on character counts from the DP table.

solution.py23 lines

1def longestPalindromicPath(n, edges, label):
2    from collections import defaultdict
3    graph = defaultdict(list)
4    for u, v in edges:
5        graph[u].append(v)
6        graph[v].append(u)
7    dp = {}  # (mask, odd_count) -> max_length
8    def dfs(mask, odd_count):
9        if (mask, odd_count) in dp:
10            return dp[(mask, odd_count)]
11        max_len = 0
12        for i in range(n):
13            if mask & (1 << i) == 0:
14                new_mask = mask | (1 << i)
15                new_odd_count = odd_count
16                if label[i] in odd_count:
17                    new_odd_count.remove(label[i])
18                else:
19                    new_odd_count.add(label[i])
20                max_len = max(max_len, dfs(new_mask, new_odd_count) + 1)
21        dp[(mask, odd_count)] = max_len
22        return max_len
23    return dfs(0, set())

ℹ

Complexity note: The complexity arises from the bitmasking approach, where each subset of nodes is processed.

1Utilizing bitmasking allows efficient tracking of visited nodes.
2Character counts are crucial for palindrome formation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.