Possible Bipartition

May 27, 2020

Introduction

Given a set of N people (numbered 1, 2, …, N), we would like to split everyone into two groups of any size.

Each person may dislike some other people, and they should not go into the same group.

Formally, if dislikes[i] = [a, b], it means it is not allowed to put the people numbered a and b into the same group.

Return true if and only 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: group1 [1,4], group2 [2,3]

Example 2:

Input: N = 3, dislikes = [[1,2],[1,3],[2,3]]
Output: false

Example 3:

Input: N = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
Output: false

Note:

1 <= N <= 2000
0 <= dislikes.length <= 10000
1 <= dislikes[i][j] <= N
dislikes[i][0] < dislikes[i][1]
There does not exist i != j for which dislikes[i] == dislikes[j].

Solution

Let’s use BFS search for each person through his dislikes to check if we have conflict.

func possibleBipartition(N int, dislikes [][]int) bool {
    if N == 0 {
        return false
    }
    dis := make([][]int, N)
    for _, pair := range dislikes {        
        a, b := pair[0]-1, pair[1]-1
        dis[a] = append(dis[a], b)
        dis[b] = append(dis[b], a)
    }
    color := make([]int, N)
    visited := make([]bool, N)
    for a := 0; a < N; a++ {
        if visited[a] {
            continue
        }
        color[a] = 1
        q := []int { a }
        for len(q) > 0 {
            a = q[len(q)-1]
            q = q[:len(q)-1]
            if visited[a] {
                continue
            }
            visited[a] = true
            for _, b := range dis[a] {
                if color[a] == color[b] {
                    return false
                }
                color[b] = color[a] * -1
                if !visited[b] {
                    q = append(q, b)
                }
            }

        }
    }     
    return true
}