What is the time complexity of Largest Color Value in a Directed Graph?

The optimal solution has a time complexity of O(N+E).

What data structures are used to solve Largest Color Value in a Directed Graph?

This problem uses Hash Table, Dynamic Programming, Graph, Topological Sort, Memoization, Counting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Largest Color Value in a Directed Graph?

Insight 1: Topological Sort is crucial for processing the graph in a way that guarantees no cycles are present in the current path being considered. Using Kahn's algorithm helps detect cycles. Insight 2: Dynamic Programming can efficiently track the maximum count of each color ending at each node. We maintain a 2D array to store the maximum frequency of each color at each node.

Which companies ask LeetCode 1857. Largest Color Value in a Directed Graph in interviews?

This problem is frequently asked at Juspay.

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Largest Color Value in a Directed Graph - LeetCode 1857 Solution

Largest Color Value in a Directed Graph - Complete Solution Guide

Largest Color Value in a Directed Graph is LeetCode problem 1857, 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

There is a directed graph of n colored nodes and m edges. The nodes are numbered from 0 to n - 1 . You are given a string colors where colors[i] is a lowercase English letter representing the color of the i th node in this graph ( 0-indexed ). You are also given a 2D array edges where edges[j] = [a j , b j ] indicates that there is a directed edge from node a j to node b j . A valid path in the graph is a sequence of nodes x 1 -> x 2 -> x 3 -> ... -> x k such that there is a directed edge from x

Detailed Explanation

The problem asks us to find the largest color value of any valid path in a directed graph. The graph has 'n' nodes, each colored with a lowercase English letter. The 'color value' of a path is the maximum number of times any single color appears along that path. If the graph contains a cycle, we must return -1. The input consists of a string 'colors' indicating the color of each node, and a 2D array 'edges' representing the directed edges of the graph.

Solution Approach

The provided solution uses a combination of topological sort (Kahn's algorithm) and dynamic programming. First, it builds an adjacency list representation of the graph and computes the in-degree of each node. Then, it performs a topological sort using a queue. During the topological sort, it uses dynamic programming to keep track of the maximum color value ending at each node. If a cycle is detected (not all nodes are visited), it returns -1. Otherwise, it returns the maximum color value found during the traversal.

Step-by-Step Algorithm

Step 1: Build the adjacency list and in-degree array. Iterate through the 'edges' array and populate the adjacency list and in-degree array.
Step 2: Initialize the DP table. Create a 2D array 'dp' of size n x 26, where dp[i][c] stores the maximum number of times color 'c' appears on a path ending at node 'i'.
Step 3: Initialize the queue with nodes having an in-degree of 0. These are the starting points for topological sort.
Step 4: Perform topological sort. While the queue is not empty, remove a node 'u' from the queue. Increment the 'nodes_seen' counter.
Step 5: Update the maximum color value. Update 'max_value' with the maximum value in dp[u].
Step 6: Iterate over neighbors of 'u'. For each neighbor 'v' of 'u', update the dp[v] table based on dp[u]. dp[v][c] = max(dp[v][c], dp[u][c] + (c == color_v_idx ? 1 : 0)). This step extends the path from 'u' to 'v', updating the color counts.
Step 7: Decrement in-degree of neighbors. Decrement the in-degree of 'v'. If the in-degree of 'v' becomes 0, add 'v' to the queue.
Step 8: Cycle Detection and Return Value. After the topological sort, check if all nodes were visited (nodes_seen == n). If not, a cycle exists, and return -1. Otherwise, return 'max_value'.

Key Insights

Insight 1: Topological Sort is crucial for processing the graph in a way that guarantees no cycles are present in the current path being considered. Using Kahn's algorithm helps detect cycles.
Insight 2: Dynamic Programming can efficiently track the maximum count of each color ending at each node. We maintain a 2D array to store the maximum frequency of each color at each node.
Insight 3: The graph may contain cycles; detecting and handling cycles is essential for a correct solution. This is achieved by verifying that all nodes are visited during topological sort. If the count of visited nodes is less than the total number of nodes, a cycle exists.

Complexity Analysis

Time Complexity: O(N+E)

Space Complexity: O(N)

Topics

This problem involves: Hash Table, Dynamic Programming, Graph, Topological Sort, Memoization, Counting.

Companies

Asked at: Juspay.

Largest Color Value in a Directed Graph | LeetCode 1857

Hard

Hash Table Dynamic Programming Graph Topological Sort Memoization Counting

Time: O(N+E)Space: O(N)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int largestPathValue(String colors, int[][] edges) {
        int n = colors.length();
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        int[] inDegree = new int[n];

        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            inDegree[v] += 1;
        }

        int[][] dp = new int[n][26];
        Queue<Integer> q = new LinkedList<>();

        for (int i = 0; i < n; i++) {
            if (inDegree[i] == 0) {
                q.offer(i);
                dp[i][colors.charAt(i) - 'a'] = 1;
            }
        }

        int maxValue = 0;
        int nodesSeen = 0;

        while (!q.isEmpty()) {
            int u = q.poll();
            nodesSeen += 1;

            int maxDp = 0;
            for (int i = 0; i < 26; i++) {
                maxDp = Math.max(maxDp, dp[u][i]);
            }
            maxValue = Math.max(maxValue, maxDp);

            for (int v : adj.get(u)) {
                int colorVIdx = colors.charAt(v) - 'a';
                for (int c = 0; c < 26; c++) {
                    dp[v][c] = Math.max(dp[v][c], dp[u][c] + (c == colorVIdx ? 1 : 0));
                }

                inDegree[v] -= 1;
                if (inDegree[v] == 0) {
                    q.offer(v);
                }
            }
        }

        return nodesSeen == n ? maxValue : -1;
    }
}

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Problem Info

Description

Problem 1857: Largest Color Value in a Directed Graph - Detailed explanation and solution approaches.

Topics

Hash Table Dynamic Programming Graph Topological Sort Memoization Counting

Companies

Juspay

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1857

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

8 steps

1
Step 1: Build the adjacency list and in-degree array. Iterate through the 'edges' array and populate the adjacency list and in-degree array.
2
Step 2: Initialize the DP table. Create a 2D array 'dp' of size n x 26, where dp[i][c] stores the maximum number of times color 'c' appears on a path ending at node 'i'.
3
Step 3: Initialize the queue with nodes having an in-degree of 0. These are the starting points for topological sort.
4
Step 4: Perform topological sort. While the queue is not empty, remove a node 'u' from the queue. Increment the 'nodes_seen' counter.
5
Step 5: Update the maximum color value. Update 'max_value' with the maximum value in dp[u].
6
Step 6: Iterate over neighbors of 'u'. For each neighbor 'v' of 'u', update the dp[v] table based on dp[u]. dp[v][c] = max(dp[v][c], dp[u][c] + (c == color_v_idx ? 1 : 0)). This step extends the path from 'u' to 'v', updating the color counts.
7
Step 7: Decrement in-degree of neighbors. Decrement the in-degree of 'v'. If the in-degree of 'v' becomes 0, add 'v' to the queue.
8
Step 8: Cycle Detection and Return Value. After the topological sort, check if all nodes were visited (nodes_seen == n). If not, a cycle exists, and return -1. Otherwise, return 'max_value'.

💡

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+E)

💾

Space Complexity

O(N)

📝

Why?

The time complexity is dominated by the topological sort and DP updates. Each node and edge are processed at most once. The space complexity arises from storing the graph (adjacency list), in-degrees, the DP table and the queue. The DP table contributes O(N) space (O(26*N) reduces to O(N)), as well as the other auxilary data structures. Therefore overall it is O(N+E). However, if we assume the number of edges are bounded by N, we say that the space complexity is O(N).

📖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

Depth First Search (DFS) with Memoization

Instead of topological sort, we can use DFS to explore the graph. To avoid cycles, we can use a 'visited' array and a 'recursion stack' array. We also use memoization (like the 'dp' array in the main solution) to store the maximum color values for each node.

Time: O(N + E)Space: O(N)

Trade-offs: DFS can be more intuitive for some, but it requires careful handling of recursion stack to detect cycles. The topological sort approach is generally considered more robust and easier to reason about in terms of cycle detection.

✨

Key Insights

1Insight 1: Topological Sort is crucial for processing the graph in a way that guarantees no cycles are present in the current path being considered. Using Kahn's algorithm helps detect cycles.
2Insight 2: Dynamic Programming can efficiently track the maximum count of each color ending at each node. We maintain a 2D array to store the maximum frequency of each color at each node.
3Insight 3: The graph may contain cycles; detecting and handling cycles is essential for a correct solution. This is achieved by verifying that all nodes are visited during topological sort. If the count of visited nodes is less than the total number of nodes, a cycle exists.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Graph

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

Edge Cases

Edge case 1: Empty graph (n=0). The code correctly handles this case because the loops won't execute, and the function will return 0 if there were no edges.
Edge case 2: Graph with a single node and no edges. The code correctly initializes dp for this node.
Edge case 3: Graph with cycles. The code correctly detects cycles by checking if nodes_seen == n. If not, it means a cycle exists, and the function returns -1.
Edge case 4: Graph with no edges. The code calculates the 'max_value' correctly based on the independent nodes' colors.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to detect cycles. If cycles are not detected, the algorithm might get stuck in an infinite loop or produce incorrect results.

✗Common mistake 2: Incorrectly updating the DP table. The 'dp' table must be updated carefully to reflect the maximum color values along the path. It is important to add the appropriate value only when the current color matches the color being tracked.

✗Common mistake 3: Not considering the case where no nodes has in-degree 0. For example, if all nodes are in a cycle, then the queue would never be initialized, and thus would cause an incorrect result.

✗Common mistake 4: Integer overflow with large path lengths. Although not explicitly addressed in the constraints or example, the problem could lead to integer overflow if a path has extremely high repetition of the same color. Using larger integer types might be required for specific edge cases.

✅

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: Dependency resolution in software projects. The directed graph represents dependencies between software modules. Finding the 'largest color value' can be analogous to finding the longest chain of dependencies of a certain type (e.g., dependencies that require a specific library version).

🏦Real-world scenario 2: Task scheduling. The directed graph represents tasks and their dependencies. The color can represent a resource type required by the task. Finding the largest color value corresponds to finding the longest sequence of tasks that require the same resource.

💡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

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

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: July 28, 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

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

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Largest Color Value in a Directed Graph"?

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

🤖 Need help with this problem?