Is Graph Bipartite?

Problem

There is an undirected graph with n nodes, where each node is numbered between 0 and n - 1. You are given a 2D array graph, where graph[u] is an array of nodes that node u is adjacent to. More formally, for each v in graph[u], there is an undirected edge between node u and node v. The graph has the following properties:

• There are no self-edges (graph[u] does not contain u).
• There are no parallel edges (graph[u] does not contain duplicate values).
• If v is in graph[u], then u is in graph[v] (the graph is undirected).
• The graph may not be connected, meaning there may be two nodes u and v such that there is no path between them.

A graph is bipartite if the nodes can be partitioned into two independent sets A and B such that every edge in the graph connects a node in set A and a node in set B.

Return true if and only if it is bipartite.

Example 1:

Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
Output: false
Explanation: There is no way to partition the nodes into two independent sets such that every edge connects a node in one and a node in the other.

Example 2:

Input: graph = [[1,3],[0,2],[1,3],[0,2]]
Output: true
Explanation: We can partition the nodes into two sets: {0, 2} and {1, 3}.

Constraints:

• graph.length == n
• 1 <= n <= 100
• 0 <= graph[u].length < n
• 0 <= graph[u][i] <= n - 1
• graph[u] does not contain u.
• All the values of graph[u] are unique.
• If graph[u] contains v, then graph[v] contains u.

Solution

With union-find

class Solution {
    int[] root;
    int[] rank;

    public boolean isBipartite(int[][] graph) {
        root = new int[graph.length];
        rank = new int[graph.length];

        for (var i = 0; i < graph.length; i++) {
            root[i] = i;
            rank[i] = 1;
        }

        for (var i = 0; i < graph.length; i++) {
            var node = graph[i];
            for (var adj : node) {
                if (find(i) == find(adj)) {
                    return false;
                }

                union(node[0], adj);
            }
        }

        return true;
    }

    int find(int x) {
        if (root[x] != x) {
            root[x] = find(root[x]);
        }
        return root[x];
    }

    void union(int x, int y) {
        var rootX = find(x);
        var rootY = find(y);
        if (rank[rootX] > rank[rootY]) {
            rootX ^= rootY;
            rootY ^= rootX;
            rootX ^= rootY;
        }
        rank[rootY] += rank[rootX];
        root[rootX] = rootY;
    }
}