Soyo

May Leetcode Challenge - Day 27

Possible Bipartition

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

Build graph, use dfs

class Solution {
    func possibleBipartition(_ N: Int, _ dislikes: [[Int]]) -> Bool {
        var graph = [[Int]](repeating: [Int](repeating: 0, count: N + 1), count: N + 1)
        for d in dislikes{
            graph[d[0]][d[1]] = 1
            graph[d[1]][d[0]] = 1
        }
        var colors = [Int](repeating: 0, count: N + 1)
        for i in 1...N {
            if colors[i] == 0 && !dfs(&graph, i, 1, &colors) { return false }
        }
        return true;
    }
    
    func dfs(_ graph: inout [[Int]], _ cIndex: Int, _ cColor: Int, _ colors: inout [Int]) -> Bool {
        colors[cIndex] = cColor
        for i in 0..<colors.count {
            if graph[cIndex][i] == 1 {
                if colors[i] == cColor {
                    return false    
                }
                if colors[i] == 0 && !dfs(&graph, i, -cColor, &colors) {
                    return false   
                }
            }
        }
        return true
    }
}

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