What is the time complexity of Possible Bipartition?

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

What is the space complexity of Possible Bipartition?

The space complexity is O(n + m).

What data structures are used to solve Possible Bipartition?

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 Possible Bipartition?

Insight 1: The problem can be modeled as a graph where nodes represent people and edges represent dislikes. The goal is to check if this graph is bipartite. Insight 2: Bipartiteness can be determined by attempting to color the graph with two colors such that no adjacent nodes have the same color. If this coloring is possible, the graph is bipartite.

Which companies ask LeetCode 886. Possible Bipartition in interviews?

This problem is frequently asked at Arcesium, Coupang, Pinterest, Samsung, Waymo.

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Possible Bipartition - LeetCode 886 Solution

Possible Bipartition - Complete Solution Guide

Possible Bipartition is LeetCode problem 886, 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

We want to split a group of n people (labeled from 1 to n ) into two groups of any size . Each person may dislike some other people, and they should not go into the same group. Given the integer n and the array dislikes where dislikes[i] = [a i , b i ] indicates that the person labeled a i does not like the person labeled b i , return true if it is possible to split everyone into two groups in this way . Example 1: Input: n = 4, dislikes = [[1,2],[1,3],[2,4]] Output: true Explanation: The first

Detailed Explanation

The problem asks us to determine if a group of 'n' people can be divided into two groups such that no two people who dislike each other are in the same group. The input is 'n' (the number of people) and 'dislikes', a list of pairs, where each pair [a, b] indicates that person 'a' dislikes person 'b'. The output is 'true' if a valid division is possible and 'false' otherwise. Effectively, we're checking if the graph represented by the dislike relationships is bipartite (2-colorable).

Solution Approach

The solution uses Breadth-First Search (BFS) to color the graph with two colors (represented as 1 and -1). It iterates through each node (person) from 1 to 'n'. If a node is uncolored, it initiates a BFS starting from that node, coloring it with color 1. During the BFS, it colors each neighbor with the opposite color. If it encounters a neighbor with the same color as the current node, it means there's a conflict, and the graph is not bipartite, so it returns 'false'. If the BFS completes without any conflicts, it continues to the next uncolored node. If all nodes are processed without conflicts, it returns 'true'.

Step-by-Step Algorithm

Step 1: Create an adjacency list representation of the graph from the 'dislikes' input. adj[u] will contain a list of nodes that 'u' dislikes.
Step 2: Initialize a 'colors' array of size 'n+1' with all values set to 0. 0 represents uncolored, 1 represents one color, and -1 represents the other color.
Step 3: Iterate through each node (person) from 1 to 'n'.
Step 4: If a node 'i' is uncolored (colors[i] == 0), start a BFS from that node.
Step 5: In the BFS, color the starting node 'i' with color 1 and add it to a queue.
Step 6: While the queue is not empty, dequeue a node 'u'.
Step 7: Iterate through each neighbor 'v' of 'u' (from the adjacency list).
Step 8: If 'v' is uncolored (colors[v] == 0), color 'v' with the opposite color of 'u' (colors[v] = -colors[u]) and enqueue 'v'.
Step 9: If 'v' is already colored and has the same color as 'u' (colors[v] == colors[u]), return 'false' (because a conflict exists).
Step 10: After the BFS completes (or if a conflict is found), continue to the next uncolored node.
Step 11: If the iteration completes without returning 'false', return 'true' (the graph is bipartite).

Key Insights

Insight 1: The problem can be modeled as a graph where nodes represent people and edges represent dislikes. The goal is to check if this graph is bipartite.
Insight 2: Bipartiteness can be determined by attempting to color the graph with two colors such that no adjacent nodes have the same color. If this coloring is possible, the graph is bipartite.
Insight 3: A crucial edge case is when the 'dislikes' list is empty. In this scenario, any grouping is possible, so the answer is 'true'.

Complexity Analysis

Time Complexity: O(n + m)

Space Complexity: O(n + m)

Topics

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

Companies

Asked at: Arcesium, Coupang, Pinterest, Samsung, Waymo.

Possible Bipartition | LeetCode 886

Medium

Depth-First Search Breadth-First Search Union Find Graph

Time: O(n + m)Space: O(n + m)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Queue;

class Solution {
    public boolean possibleBipartition(int n, int[][] dislikes) {
        if (dislikes.length == 0) {
            return true;
        }

        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i <= n; i++) {
            adj.add(new ArrayList<>());
        }

        for (int[] dislike : dislikes) {
            int u = dislike[0];
            int v = dislike[1];
            adj.get(u).add(v);
            adj.get(v).add(u);
        }

        int[] colors = new int[n + 1]; // 0: uncolored, 1: group 1, -1: group 2

        for (int i = 1; i <= n; i++) {
            if (colors[i] == 0) {
                // This node belongs to a component we haven't visited yet
                Queue<Integer> queue = new LinkedList<>();
                queue.offer(i);
                colors[i] = 1; // Start coloring with color 1

                while (!queue.isEmpty()) {
                    int u = queue.poll();
                    for (int v : adj.get(u)) {
                        if (colors[v] == 0) {
                            colors[v] = -colors[u]; // Color neighbor with the opposite color
                            queue.offer(v);
                        } else if (colors[v] == colors[u]) {
                            // A neighbor has the same color, conflict!
                            return false;
                        }
                    }
                }
            }
        }

        return true;
    }
}

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Problem Info

Description

Problem 886: Possible Bipartition - Detailed explanation and solution approaches.

Topics

Depth-First Search Breadth-First Search Union Find Graph

Companies

ArcesiumCoupangPinterestSamsungWaymo

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:886

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: Create an adjacency list representation of the graph from the 'dislikes' input. adj[u] will contain a list of nodes that 'u' dislikes.
2
Step 2: Initialize a 'colors' array of size 'n+1' with all values set to 0. 0 represents uncolored, 1 represents one color, and -1 represents the other color.
3
Step 3: Iterate through each node (person) from 1 to 'n'.
4
Step 4: If a node 'i' is uncolored (colors[i] == 0), start a BFS from that node.
5
Step 5: In the BFS, color the starting node 'i' with color 1 and add it to a queue.
6
Step 6: While the queue is not empty, dequeue a node 'u'.
7
Step 7: Iterate through each neighbor 'v' of 'u' (from the adjacency list).
8
Step 8: If 'v' is uncolored (colors[v] == 0), color 'v' with the opposite color of 'u' (colors[v] = -colors[u]) and enqueue 'v'.
9
Step 9: If 'v' is already colored and has the same color as 'u' (colors[v] == colors[u]), return 'false' (because a conflict exists).
10
Step 10: After the BFS completes (or if a conflict is found), continue to the next uncolored node.
11
Step 11: If the iteration completes without returning 'false', return 'true' (the graph is bipartite).

💡

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 + m)

💾

Space Complexity

O(n + m)

📝

Why?

The time complexity is determined by the BFS traversal, which explores all nodes and edges once. The space complexity is dominated by the adjacency list and the colors array, which store information about the graph structure and coloring state.

📖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, DFS can also be used to color the graph. The approach is similar, but instead of using a queue, a recursive function is used to explore each node and its neighbors.

Time: O(n + m)Space: O(n + m) in the worst case (due to recursion depth), but can be O(n) on average for balanced graphs.

Trade-offs: DFS might use slightly less space on average due to not needing to explicitly store a queue, but it can have a higher space complexity in the worst case due to the recursion stack. BFS may be preferred if you anticipate the graph being very deep, to avoid potential stack overflow issues.

Union Find (Disjoint Set Union)

This approach involves iterating through each dislike pair (u, v) and merging 'u' with all the enemies of 'v', and 'v' with all the enemies of 'u'. The key idea is that if 'u' and 'v' dislike each other, they cannot be in the same group. So, all of 'u's enemies should belong to 'v's group and vice-versa. If 'u' and 'v' are already in the same group, it means that the graph cannot be bipartite, so the result is 'false'.

Time: O(m * α(n)), where α(n) is the inverse Ackermann function, which grows very slowly and can be considered almost constant. Hence effectively O(m) Space: O(n)

Trade-offs: Union Find can be more efficient in practice compared to BFS or DFS, especially if the dislike pairs form a pattern suitable for quick merging. It can be a bit less intuitive to implement compared to BFS or DFS.

✨

Key Insights

1Insight 1: The problem can be modeled as a graph where nodes represent people and edges represent dislikes. The goal is to check if this graph is bipartite.
2Insight 2: Bipartiteness can be determined by attempting to color the graph with two colors such that no adjacent nodes have the same color. If this coloring is possible, the graph is bipartite.
3Insight 3: A crucial edge case is when the 'dislikes' list is empty. In this scenario, any grouping is possible, so the answer is 'true'.

📚

Key Terms Explained

For Students

Graph

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

Edge Cases

Edge case 1: Empty 'dislikes' list. The solution correctly handles this by returning 'true' because if there are no dislikes, any division is valid.
Edge case 2: A single connected component versus multiple disconnected components. The outer loop in the solution ensures that each disconnected component of the graph is processed, so it handles both cases correctly.
Edge case 3: A person disliking themselves (though the problem states ai < bi, it's good to consider). If a person disliked themselves, the algorithm would detect a conflict immediately when coloring their neighbor (which is themselves).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling disconnected components. Failing to iterate through all nodes to find uncolored nodes can lead to incorrect results if the graph is not fully connected.

✗Common mistake 2: Incorrectly implementing the coloring logic. Forgetting to color the neighbors with the opposite color or failing to check for conflicts between adjacent nodes can cause the algorithm to fail.

✗Common mistake 3: Not accounting for an empty dislikes list. It is important to handle the initial condition when dislikes is empty

✅

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: Resource allocation. Consider assigning programmers to projects where some programmers dislike each other and shouldn't work on the same project. This problem becomes a bipartite graph problem.

🏦Real-world scenario 2: Conflict resolution. Identifying if two conflicting groups can be separated into two distinct factions where internal conflicts are resolved.

🎮Real-world scenario 3: Job scheduling. Assigning tasks to workers such that certain tasks cannot be performed by the same worker due to skill conflicts or resource constraints.

💡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

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

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary 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: November 14, 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 13

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

@InterviewProJul 6

❓

Frequently Asked Questions

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 "Possible Bipartition"?

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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