[Leetcode] Problem 886 - Possible Bipartition

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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

No.1

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

Output: true

Explanation: group1 [1,4], group2 [2,3]

No.2

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

Output: false

No.3

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

Output: false

Constraints

  • 1 <= N <= 2000
  • 0 <= dislikes.length <= 10000
  • dislikes[i].length == 2
  • 1 <= dislikes[i][j] <= N
  • dislikes[i][0] < dislikes[i][1]
  • There does not exist i != j for which dislikes[i] == dislikes[j].

Code

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private int[] color;
private List<Integer>[] graph;

public boolean possibleBipartition(int N, int[][] dislikes) {
    color = new int[N + 1];
    graph = new ArrayList[N + 1];

    for (int i = 1; i <= N; i++)
        graph[i] = new ArrayList<Integer>();
    
    for (int[] dislike : dislikes) {
        graph[dislike[0]].add(dislike[1]);
        graph[dislike[1]].add(dislike[0]);
    }

    for (int i = 1; i <= N; i++) {
        if (color[i] == 0 && !dfs(i, 1))
            return false;
    }

    return true;
}

private boolean dfs(int node, int currentColor) {
    color[node] = currentColor;

    for (int neighbor : graph[node]) {
        if (color[neighbor] == currentColor)
            return false;

        if (color[neighbor] == 0 && !dfs(neighbor, -currentColor))
            return false;
    }

    return true;
}