How do you solve Longest Path With Different Adjacent Characters on LeetCode?

Longest Path With Different Adjacent Characters (LeetCode #2246) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The longest path can be derived from the longest two distinct character paths from any node.

What is the time complexity of Longest Path With Different Adjacent Characters?

The optimal solution for Longest Path With Different Adjacent Characters runs in O(n) time and uses O(n) space. This complexity is due to the single traversal of all nodes in the tree, which is efficient for the given constraints.

What is the brute force solution for Longest Path With Different Adjacent Characters?

The brute force approach for Longest Path With Different Adjacent Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves exploring all possible paths in the tree and checking the characters of adjacent nodes. This method is straightforward but inefficient, as it doesn't leverage the tree structure effectively. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Longest Path With Different Adjacent Characters?

Clearly explain your thought process and how you are structuring your solution. Consider edge cases, such as trees with only one node or all nodes having the same character. Practice DFS problems to become comfortable with recursive tree traversal.

What are common mistakes when solving Longest Path With Different Adjacent Characters?

Not considering the character constraints when calculating path lengths. Failing to properly manage the state during recursive DFS calls.

#2246

Longest Path With Different Adjacent Characters

Hard

ArrayStringTreeDepth-First SearchGraph TheoryTopological SortDepth-First SearchTree Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses Depth-First Search (DFS) to traverse the tree while keeping track of the longest paths from each node. By only considering paths where adjacent characters differ, we efficiently compute the maximum path length.

⚙️

Algorithm

3 steps

1Step 1: Build the tree from the parent array.
2Step 2: Perform a DFS starting from the root node, calculating the longest valid path for each node.
3Step 3: For each node, keep track of the two longest paths from its children that have different characters.

solution.py26 lines

1# Full working Python code
2from collections import defaultdict
3
4def longestPath(parent, s):
5    n = len(s)
6    graph = defaultdict(list)
7    for i in range(1, n):
8        graph[parent[i]].append(i)
9    max_length = 0
10
11    def dfs(node):
12        nonlocal max_length
13        longest1, longest2 = 0, 0
14        for child in graph[node]:
15            child_length = dfs(child)
16            if s[node] != s[child]:
17                if child_length > longest1:
18                    longest2 = longest1
19                    longest1 = child_length
20                elif child_length > longest2:
21                    longest2 = child_length
22        max_length = max(max_length, longest1 + longest2 + 1)
23        return longest1 + 1
24
25    dfs(0)
26    return max_length

ℹ

Complexity note: This complexity is due to the single traversal of all nodes in the tree, which is efficient for the given constraints.

1The longest path can be derived from the longest two distinct character paths from any node.
2DFS is a powerful technique for tree traversal, allowing us to explore all paths efficiently.

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