What is the time complexity of Node With Highest Edge Score?

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

What data structures are used to solve Node With Highest Edge Score?

This problem uses Hash Table, Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Node With Highest Edge Score?

Insight 1: The problem requires calculating the sum of node labels pointing to each node, suggesting the use of an array or hash map to store these sums efficiently. Insight 2: Since we need to return the node with the smallest index in case of ties, iterating through the calculated scores sequentially and updating the result will ensure the correct output.

Which companies ask LeetCode 2374. Node With Highest Edge Score in interviews?

This problem is frequently asked at Juspay.

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Node With Highest Edge Score - LeetCode 2374 Solution

Node With Highest Edge Score - Complete Solution Guide

Node With Highest Edge Score is LeetCode problem 2374, a Medium 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 directed graph with n nodes labeled from 0 to n - 1 , where each node has exactly one outgoing edge. The graph is represented by a given 0-indexed integer array edges of length n , where edges[i] indicates that there is a directed edge from node i to node edges[i] . The edge score of a node i is defined as the sum of the labels of all the nodes that have an edge pointing to i . Return the node with the highest edge score . If multiple nodes have the same edge score , return the n

Detailed Explanation

The problem asks us to find the node with the highest 'edge score' in a directed graph. The graph is represented by an array `edges`, where `edges[i]` indicates that there is a directed edge from node `i` to node `edges[i]`. The 'edge score' of a node `i` is defined as the sum of all the node labels that point to `i`. The goal is to return the node with the highest edge score. If multiple nodes have the same highest edge score, return the node with the smallest index.

Solution Approach

The solution iterates through the `edges` array. For each element `edges[i]`, it adds `i` (the node label) to the `scores` array at index `edges[i]` (the destination node). This effectively accumulates the edge scores for each node. After calculating all the scores, the solution iterates through the `scores` array to find the node with the maximum score. If a node has a higher score than the current maximum, it updates the `max_score` and `result_node`. If multiple nodes have the same maximum score, the node encountered earlier in the iteration (i.e., with the smaller index) will be selected, satisfying the problem's requirement.

Step-by-Step Algorithm

Step 1: Initialize an array `scores` of size `n` (number of nodes) to store the edge scores of each node. All elements are initialized to 0.
Step 2: Iterate through the `edges` array. For each index `i` and value `dest = edges[i]`, add `i` to `scores[dest]`. This accumulates the edge score for each node.
Step 3: Initialize `max_score` to a small value (e.g., -1) and `result_node` to -1.
Step 4: Iterate through the `scores` array. For each index `i` and score `score = scores[i]`, check if `score` is greater than `max_score`. If it is, update `max_score` to `score` and `result_node` to `i`.
Step 5: Return `result_node`.

Key Insights

Insight 1: The problem requires calculating the sum of node labels pointing to each node, suggesting the use of an array or hash map to store these sums efficiently.
Insight 2: Since we need to return the node with the smallest index in case of ties, iterating through the calculated scores sequentially and updating the result will ensure the correct output.
Insight 3: The node labels can be large (up to 10^5), so using a data type that can hold large sums like `long` is important to avoid potential overflow errors.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Hash Table, Graph.

Companies

Asked at: Juspay.

Node With Highest Edge Score | LeetCode 2374

Medium

Hash Table Graph

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

Solution Code

Available in:

class Solution {
    public int edgeScore(int[] edges) {
        int n = edges.length;
        long[] scores = new long[n];

        for (int i = 0; i < n; i++) {
            scores[edges[i]] += i;
        }
        
        long max_score = -1;
        int result_node = -1;
        
        for (int i = 0; i < n; i++) {
            if (scores[i] > max_score) {
                max_score = scores[i];
                result_node = i;
            }
        }
        
        return result_node;
    }
}

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Problem Info

Description

Problem 2374: Node With Highest Edge Score - Detailed explanation and solution approaches.

Topics

Hash Table Graph

Companies

Juspay

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2374

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

5 steps

1
Step 1: Initialize an array `scores` of size `n` (number of nodes) to store the edge scores of each node. All elements are initialized to 0.
2
Step 2: Iterate through the `edges` array. For each index `i` and value `dest = edges[i]`, add `i` to `scores[dest]`. This accumulates the edge score for each node.
3
Step 3: Initialize `max_score` to a small value (e.g., -1) and `result_node` to -1.
4
Step 4: Iterate through the `scores` array. For each index `i` and score `score = scores[i]`, check if `score` is greater than `max_score`. If it is, update `max_score` to `score` and `result_node` to `i`.
5
Step 5: Return `result_node`.

💡

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 linear because the code iterates through the input arrays a fixed number of times. The space complexity is linear because the `scores` array scales linearly with the number of nodes in the graph. There are no nested loops or recursive calls that would increase the complexity.

📖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

Using a HashMap

Instead of using an array, we can use a HashMap to store the edge scores. The key would be the node, and the value would be its edge score. This approach is useful if the node labels are not consecutive or if the number of nodes is very large but the number of edges is relatively small.

Time: O(n), where n is the number of edges. Assuming HashMap operations take O(1) on average.Space: O(n), where n is the number of nodes with incoming edges. In the worst case, every node has an incoming edge, so the HashMap would store n entries.

Trade-offs: The HashMap approach may be slightly less efficient in terms of constant factors due to the overhead of HashMap operations. However, it can be more flexible if the node labels are not consecutive integers.

✨

Key Insights

1Insight 1: The problem requires calculating the sum of node labels pointing to each node, suggesting the use of an array or hash map to store these sums efficiently.
2Insight 2: Since we need to return the node with the smallest index in case of ties, iterating through the calculated scores sequentially and updating the result will ensure the correct output.
3Insight 3: The node labels can be large (up to 10^5), so using a data type that can hold large sums like `long` is important to avoid potential overflow errors.

📚

Key Terms Explained

For Students

Hash Table

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

Graph

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

Edge Cases

Edge case 1: All nodes point to the same node. The algorithm correctly calculates the edge score for the target node and returns it. Example: edges = [0, 0, 0]. The algorithm returns 0.
Edge case 2: The graph has only two nodes. The algorithm correctly calculates the edge scores for both nodes and returns the node with the higher score or the node with the smaller index if the scores are equal. Example: edges = [1, 0].
Edge case 3: Empty Graph: Although the constraints specify n >= 2, let's consider if n = 0 or n = 1. The code would likely crash or give incorrect result, though not relevant given constraints.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using an integer data type to store the edge scores, which can lead to integer overflow if the node labels are large. The solutions correctly use `long` in Java and C++ or Python which handles large integers, avoiding this overflow issue.

✗Common mistake 2: Not considering the requirement to return the node with the smallest index in case of ties. This can be easily avoided by updating the result only when a strictly higher score is found.

✅

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: Analyzing website traffic. Imagine nodes represent web pages and edges represent links between pages. The edge score can represent the number of incoming links to a page, indicating its popularity or importance. Finding the node with the highest edge score corresponds to finding the most popular page.

🏦Real-world scenario 2: Social network analysis. Nodes can represent users and edges can represent followers. The edge score can represent the number of followers a user has. Finding the node with the highest edge score corresponds to finding the most influential user.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 5, 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

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

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I represent a graph in code?

Use adjacency list (HashMap or array of lists) for sparse graphs - it's space-efficient O(V+E). Use adjacency matrix for dense graphs or when you need O(1) edge lookup. For weighted graphs, store (neighbor, weight) pairs in adjacency lists.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Node With Highest Edge Score"?

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

🤖 Need help with this problem?