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

The optimal solution has a time complexity of O(n).

What data structures are used to solve Longest Path With Different Adjacent Characters?

This problem uses Array, String, Tree, Depth-First Search, Graph, Topological Sort. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Longest Path With Different Adjacent Characters?

Insight 1: The longest path doesn't necessarily have to pass through the root node (node 0). Insight 2: We can use Depth-First Search (DFS) to traverse the tree and compute the longest path from each node.

Which companies ask LeetCode 2246. Longest Path With Different Adjacent Characters in interviews?

This problem is frequently asked at Hudson River Trading, Target.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Longest Path With Different Adjacent Characters - LeetCode 2246 Solution

Longest Path With Different Adjacent Characters - Complete Solution Guide

Longest Path With Different Adjacent Characters is LeetCode problem 2246, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a tree (i.e. a connected, undirected graph that has no cycles) rooted at node 0 consisting of n nodes numbered from 0 to n - 1 . The tree is represented by a 0-indexed array parent of size n , where parent[i] is the parent of node i . Since node 0 is the root, parent[0] == -1 . You are also given a string s of length n , where s[i] is the character assigned to node i . Return the length of the longest path in the tree such that no pair of adjacent nodes on the path have the same ch

Detailed Explanation

The problem asks us to find the longest path in a tree where no two adjacent nodes have the same character. We are given a tree represented by a `parent` array, where `parent[i]` is the parent of node `i`, and a string `s` where `s[i]` is the character of node `i`. The goal is to return the length of the longest valid path.

Solution Approach

The solution uses Depth-First Search (DFS) to traverse the tree. For each node, we recursively find the longest path from each of its children. While traversing, we maintain two variables, `longest_1` and `longest_2`, which represent the lengths of the two longest paths from the children of the current node that have different characters than the current node. We update the global `max_len` variable with the maximum path length found so far. The length of the path passing through the current node is `1 + longest_1 + longest_2`.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the tree from the `parent` array.
Step 2: Initialize `max_len` to 0. This variable will store the length of the longest valid path found so far.
Step 3: Define a recursive DFS function that takes a node as input.
Step 4: Inside the DFS function, initialize `longest_1` and `longest_2` to 0. These variables will store the lengths of the two longest paths from the children of the current node.
Step 5: Iterate through the children of the current node.
Step 6: For each child, recursively call the DFS function to find the longest path from that child.
Step 7: If the character of the current node is different from the character of the child, update `longest_1` and `longest_2` accordingly. If the path length from the child is greater than `longest_1`, update `longest_2` to the previous value of `longest_1` and update `longest_1` to the path length from the child. Otherwise, if the path length from the child is greater than `longest_2`, update `longest_2` to the path length from the child.
Step 8: Update `max_len` with the maximum of its current value and `1 + longest_1 + longest_2`. The 1 accounts for the current node in the path.
Step 9: Return `1 + longest_1` to the parent node. This represents the length of the longest path from the current node to one of its descendants.
Step 10: Call the DFS function starting from the root node (node 0).
Step 11: Return the value of `max_len`.

Key Insights

Insight 1: The longest path doesn't necessarily have to pass through the root node (node 0).
Insight 2: We can use Depth-First Search (DFS) to traverse the tree and compute the longest path from each node.
Insight 3: For each node, we need to consider the longest two paths from its children that have different characters from the node itself. This is because these two paths can be combined to form a longer path passing through the current node.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Tree, Depth-First Search, Graph, Topological Sort.

Companies

Asked at: Hudson River Trading, Target.

Longest Path With Different Adjacent Characters | LeetCode 2246

Hard

Array String Tree Depth-First Search Graph Topological Sort

Time: O(n)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    private int max_len;

    public int longestPath(int[] parent, String s) {
        int n = parent.length;
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        for (int i = 1; i < n; i++) {
            adj.get(parent[i]).add(i);
        }

        this.max_len = 0;

        dfs(0, adj, s);
        return this.max_len;
    }

    private int dfs(int node, List<List<Integer>> adj, String s) {
        int longest_1 = 0;
        int longest_2 = 0;

        for (int child : adj.get(node)) {
            int path_len_from_child = dfs(child, adj, s);

            if (s.charAt(node) != s.charAt(child)) {
                if (path_len_from_child > longest_1) {
                    longest_2 = longest_1;
                    longest_1 = path_len_from_child;
                } else if (path_len_from_child > longest_2) {
                    longest_2 = path_len_from_child;
                }
            }
        }

        this.max_len = Math.max(this.max_len, 1 + longest_1 + longest_2);

        return 1 + longest_1;
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2246: Longest Path With Different Adjacent Characters - Detailed explanation and solution approaches.

Topics

Array String Tree Depth-First Search Graph Topological Sort

Companies

Hudson River TradingTarget

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2246

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

11 steps

1
Step 1: Build the adjacency list representation of the tree from the `parent` array.
2
Step 2: Initialize `max_len` to 0. This variable will store the length of the longest valid path found so far.
3
Step 3: Define a recursive DFS function that takes a node as input.
4
Step 4: Inside the DFS function, initialize `longest_1` and `longest_2` to 0. These variables will store the lengths of the two longest paths from the children of the current node.
5
Step 5: Iterate through the children of the current node.
6
Step 6: For each child, recursively call the DFS function to find the longest path from that child.
7
Step 7: If the character of the current node is different from the character of the child, update `longest_1` and `longest_2` accordingly. If the path length from the child is greater than `longest_1`, update `longest_2` to the previous value of `longest_1` and update `longest_1` to the path length from the child. Otherwise, if the path length from the child is greater than `longest_2`, update `longest_2` to the path length from the child.
8
Step 8: Update `max_len` with the maximum of its current value and `1 + longest_1 + longest_2`. The 1 accounts for the current node in the path.
9
Step 9: Return `1 + longest_1` to the parent node. This represents the length of the longest path from the current node to one of its descendants.
10
Step 10: Call the DFS function starting from the root node (node 0).
11
Step 11: Return the value of `max_len`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is O(n) because the DFS algorithm visits each node in the tree exactly once. Building the adjacency list also takes O(n) time. The space complexity is O(n) because in the worst case, the adjacency list might store all possible edges, and the call stack for DFS could reach a depth proportional to n (for a skewed tree).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative Depth-First Search

An iterative DFS can be implemented using a stack. This could potentially reduce the space complexity by avoiding the implicit call stack overhead of recursive DFS, although the overall space complexity would still likely be O(n) due to the adjacency list.

Time: O(n)Space: O(n)

Trade-offs: Iterative DFS can avoid potential stack overflow issues with very deep trees and might be slightly faster in some cases, but the implementation can be more complex than the recursive approach.

✨

Key Insights

1Insight 1: The longest path doesn't necessarily have to pass through the root node (node 0).
2Insight 2: We can use Depth-First Search (DFS) to traverse the tree and compute the longest path from each node.
3Insight 3: For each node, we need to consider the longest two paths from its children that have different characters from the node itself. This is because these two paths can be combined to form a longer path passing through the current node.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Tree

Hierarchical structure with nodes. Like a family tree!

Graph

Nodes connected by edges. Like a social network or city map!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Tree with only one node. The solution correctly returns 1 in this case because the DFS still runs on the root node and the max_len will be updated with 1.
Edge case 2: A tree where all nodes have the same character. The solution correctly handles this case because the `if` condition `s[node] != s[child]` prevents the inclusion of paths with adjacent nodes having the same character.
Edge case 3: Skewed Tree. The solution works correctly and handles the worst-case space complexity of O(n) for the call stack during DFS.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider the case where the longest path does not pass through the root node. The solution must correctly explore all possible paths in the tree.

✗Common mistake 2: Incorrectly updating `longest_1` and `longest_2`, leading to inaccurate path length calculations. It's crucial to ensure that only paths with different adjacent characters are considered and that the two longest paths are correctly maintained.

✗Common mistake 3: Not handling the edge case of a single-node tree. The program should return 1 in this case.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network routing where nodes represent routers and characters represent network protocols. The longest path with different adjacent protocols could represent the maximum number of compatible hops.

🏦Real-world scenario 2: Phylogenetic tree analysis where nodes represent species and characters represent genetic traits. The longest path with different adjacent traits could indicate evolutionary divergence.

🎮Real-world scenario 3: Organizational chart analysis where nodes represent employees and characters represent departments. Finding the longest path with different adjacent departments could help identify potential communication bottlenecks or cross-departmental collaboration opportunities.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 26, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 14

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Longest Path With Different Adjacent Characters"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?