LeetCode 785. Is Graph Bipartite?

Question

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.

Source: https://leetcode.com/problems/is-graph-bipartite/

Solution

A graph is bipartite means that for each node in that graph, its neighbors and itself should be in two different groups.

// BFS + coloring, Time Complexity O(|V|+|E|)
public boolean isBipartite(int[][] graph) {
    // number of nodes
    int N = graph.length;
    // node ID -> color, 1 red, -1 blue
    Map<Integer, Integer> colorMap = new HashMap<>();
    // queue for bfs
    Queue<Integer> queue = new LinkedList<>();
    for (int id = 0; id < N; id++) { // to handle forest
        if (!colorMap.containsKey(id)) {
            colorMap.put(id, 1); // group to red by default
            // BFS
            queue.add(id);
            while (!queue.isEmpty()) {
                int tid = queue.poll();
                int tcolor = colorMap.get(tid);
                for (int nid : graph[tid]) { // visit neighbors
                    if (colorMap.containsKey(nid)) {
                        int ncolor = colorMap.get(nid);
                        if (ncolor == tcolor) {
                            return false;
                        }
                    } else {
                        // color the neighbor with the opposite color
                        colorMap.put(nid, -tcolor);
                        queue.add(nid);
                    }
                }
            }
        }
    }
    return true;
}