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

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].

class Solution {
    // Color -> 0 not colored YET
    // Color -> 1 (one of the colors)
    // Color -> -1 (The other color)
    // colors array works as color storage as well as visited array
    int[] colors;
    List<Integer>[] graph;

    public boolean possibleBipartition(int N, int[][] dislikes) {
        // Creating adjacency list graph
        graph = new ArrayList[N];
        for (int i = 0; i < N; i++)
            graph[i] = new ArrayList<>();
        for (int[] pair : dislikes) {
            graph[pair[0] - 1].add(pair[1] - 1);
            graph[pair[1] - 1].add(pair[0] - 1);
        }
        colors = new int[N];
        for (int i = 0; i < N; i++)
            if (colors[i] == 0 && !paint(i, 1))
                return false;
        return true;
    }

    boolean paint(int u, int color) {
        // "color" -> the color we want this person ("u") to be
        if (colors[u] != 0)
            return colors[u] == color;
        colors[u] = color;
        for (int neighbor : graph[u])
            // If we cannot paint the neighbor with opposite color
            if (!paint(neighbor, -color))
                return false;
        return true;
    }
}