What is the time complexity of Is Graph Bipartite??

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

What data structures are used to solve Is Graph Bipartite??

This problem uses Depth-First Search, Breadth-First Search, Union Find, Graph. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Is Graph Bipartite??

Insight 1: Bipartiteness can be determined by assigning colors to nodes such that no adjacent nodes have the same color. We only need two colors. Insight 2: If a graph is not connected, we need to check each connected component independently.

Which companies ask LeetCode 785. Is Graph Bipartite? in interviews?

This problem is frequently asked at Pinterest, Samsung.

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Is Graph Bipartite? - LeetCode 785 Solution

Is Graph Bipartite? - Complete Solution Guide

Is Graph Bipartite? is LeetCode problem 785, 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

There is an undirected graph with n nodes, where each node is numbered between 0 and n - 1 . You are given a 2D array graph , where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u] , there is an undirected edge between node u and node v . The graph has the following properties: There are no self-edges ( graph[u] does not contain u ). There are no parallel edges ( graph[u] does not contain duplicate values). If v is in graph[u] , then u is in graph[v

Detailed Explanation

The problem asks us to determine if a given undirected graph is bipartite. A graph is bipartite if its nodes can be divided into two disjoint sets A and B such that every edge connects a node in A to a node in B. The input is a 2D array 'graph' representing the adjacency list of the graph. `graph[u]` contains a list of nodes that node `u` is connected to. The graph may not be connected, meaning there may be multiple disconnected components. The output should be `true` if the graph is bipartite and `false` otherwise. The nodes are numbered from 0 to n-1.

Solution Approach

The provided solutions use Breadth-First Search (BFS) to color the nodes of the graph. We iterate through each node in the graph. If a node hasn't been colored yet, we start a BFS traversal from that node. We assign a color (1 or -1) to the starting node and then alternate colors for each neighbor. If we encounter a neighbor that has already been colored and has the same color as the current node, it means the graph is not bipartite. We repeat this process for each uncolored node in the graph until we have checked all connected components.

Step-by-Step Algorithm

Step 1: Initialize a 'colors' array of size n, filled with 0. 0 means uncolored, 1 and -1 represent the two colors.
Step 2: Iterate through each node 'i' from 0 to n-1.
Step 3: If colors[i] is 0 (uncolored), it means we need to start a BFS traversal from this node.
Step 4: Create a queue and add the starting node 'i' to it. Assign colors[i] = 1 (or any arbitrary color).
Step 5: While the queue is not empty:
Step 6: Dequeue a node 'u' from the queue.
Step 7: Iterate through each neighbor 'v' of node 'u' in graph[u].
Step 8: If colors[v] is 0 (uncolored), assign colors[v] = -colors[u] (opposite color of 'u') and enqueue 'v'.
Step 9: If colors[v] is equal to colors[u], it means there's a conflict, and the graph is not bipartite. Return false.
Step 10: After the BFS traversal completes, if no conflict was found, continue to the next uncolored node.
Step 11: If all nodes have been processed without finding any conflicts, return true.

Key Insights

Insight 1: Bipartiteness can be determined by assigning colors to nodes such that no adjacent nodes have the same color. We only need two colors.
Insight 2: If a graph is not connected, we need to check each connected component independently.
Insight 3: A graph with a cycle of odd length is not bipartite.

Complexity Analysis

Time Complexity: O(V+E)

Space Complexity: O(V)

Topics

This problem involves: Depth-First Search, Breadth-First Search, Union Find, Graph.

Companies

Asked at: Pinterest, Samsung.

Is Graph Bipartite? | LeetCode 785

Medium

Depth-First Search Breadth-First Search Union Find Graph

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

Solution Code

Available in:

import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;

class Solution {
    public boolean isBipartite(int[][] graph) {
        int n = graph.length;
        int[] colors = new int[n]; // 0: uncolored, 1: color A, -1: color B
        Arrays.fill(colors, 0);

        for (int i = 0; i < n; i++) {
            if (colors[i] == 0) { // If this node hasn't been colored yet, start a traversal
                Queue<Integer> queue = new LinkedList<>();
                queue.offer(i);
                colors[i] = 1; // Start with color 1

                while (!queue.isEmpty()) {
                    int node = queue.poll();
                    for (int neighbor : graph[node]) {
                        if (colors[neighbor] == 0) {
                            // If the neighbor is uncolored, assign the opposite color
                            colors[neighbor] = -colors[node];
                            queue.offer(neighbor);
                        } else if (colors[neighbor] == colors[node]) {
                            // If the neighbor has the same color, conflict found
                            return false;
                        }
                    }
                }
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 785: Is Graph Bipartite? - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Union Find Graph

Companies

PinterestSamsung

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:785

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: Initialize a 'colors' array of size n, filled with 0. 0 means uncolored, 1 and -1 represent the two colors.
2
Step 2: Iterate through each node 'i' from 0 to n-1.
3
Step 3: If colors[i] is 0 (uncolored), it means we need to start a BFS traversal from this node.
4
Step 4: Create a queue and add the starting node 'i' to it. Assign colors[i] = 1 (or any arbitrary color).
5
Step 5: While the queue is not empty:
6
Step 6: Dequeue a node 'u' from the queue.
7
Step 7: Iterate through each neighbor 'v' of node 'u' in graph[u].
8
Step 8: If colors[v] is 0 (uncolored), assign colors[v] = -colors[u] (opposite color of 'u') and enqueue 'v'.
9
Step 9: If colors[v] is equal to colors[u], it means there's a conflict, and the graph is not bipartite. Return false.
10
Step 10: After the BFS traversal completes, if no conflict was found, continue to the next uncolored node.
11
Step 11: If all nodes have been processed without finding any conflicts, return true.

💡

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(V+E)

💾

Space Complexity

O(V)

📝

Why?

The BFS traversal explores each node and its neighbors. In an adjacency list representation, we iterate through the neighbors of each node, resulting in O(E) operations. The 'colors' array takes O(V) space. The queue in BFS can hold up to V nodes in the worst-case scenario.

📖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)

Instead of BFS, we can use DFS to traverse the graph and color the nodes. The logic for assigning colors and detecting conflicts remains the same.

Time: O(V+E)Space: O(V) in the worst case, due to the recursion stack.

Trade-offs: DFS and BFS have similar performance characteristics in this case. DFS might be slightly more memory-efficient in some scenarios due to the smaller stack size compared to the queue size in BFS, but in worst case could be larger.

Union Find

Using the Union Find data structure, we can iterate through all edges (u,v) and union u and v's 'enemy' (initially set to null). If we ever encounter u and v are already in the same set, we return false.

Time: O(E * α(V)) where α(V) is the inverse Ackermann function, effectively constant time per operation.Space: O(V)

Trade-offs: Union find can be faster when it's highly optimized, but in practice it is often slower than BFS or DFS in this particular problem.

✨

Key Insights

1Insight 1: Bipartiteness can be determined by assigning colors to nodes such that no adjacent nodes have the same color. We only need two colors.
2Insight 2: If a graph is not connected, we need to check each connected component independently.
3Insight 3: A graph with a cycle of odd length is not bipartite.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: Empty graph (n=0). The loops will not execute, and it will correctly return `true` (an empty graph is considered bipartite).
Edge case 2: Disconnected graph. The outer loop ensures that each connected component is checked independently. If any component is not bipartite, the function returns `false`.
Edge case 3: Single node graph. The inner loop will never execute and will return `true`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling disconnected graphs. Failing to iterate through all nodes to check each connected component.

✗Common mistake 2: Assuming that the graph is always connected. This can lead to incorrect results if there are multiple disconnected components.

✗Common mistake 3: Incorrectly assigning colors. Forgetting to alternate colors between adjacent nodes, or not checking for conflicts when assigning colors.

✅

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: Job scheduling where tasks need to be assigned to two different processors such that conflicting tasks are assigned to different processors.

🏦Real-world scenario 2: Network routing where nodes in a network need to be divided into two groups to avoid conflicts in communication.

🎮Real-world scenario 3: Compiler design: determining whether a program's variables can be assigned to a fixed number of registers without conflicts. This is graph coloring applied to an interference graph.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 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

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

@DevExpert92Jul 13

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

@AlgoMasterJul 6

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

@InterviewProJun 29

❓

Frequently Asked Questions

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Is Graph Bipartite?"?

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

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