Is Graph Bipartite?

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2020-05-27

785. Is Graph Bipartite?

Question:

Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split it’s set of nodes into two independent subsets A and B such that every edge in the graph has one node in A and another node in B.

The graph is given in the following form: graph[i] is a list of indexes j for which the edge between nodes i and j exists. Each node is an integer between 0 and graph.length - 1. There are no self edges or parallel edges: graph[i] does not contain i, and it doesn’t contain any element twice.

Example 1:

Example 1:
Input: [[1,3], [0,2], [1,3], [0,2]]
Output: true
Explanation: 
The graph looks like this:
0----1
|    |
|    |
3----2
We can divide the vertices into two groups: {0, 2} and {1, 3}.

Example 2:

Example 2:
Input: [[1,2,3], [0,2], [0,1,3], [0,2]]
Output: false
Explanation: 
The graph looks like this:
0----1
| \  |
|  \ |
3----2
We cannot find a way to divide the set of nodes into two independent subsets.

Note:

graph will have length in range [1, 100].
graph[i] will contain integers in range [0, graph.length - 1].
graph[i] will not contain i or duplicate values.
The graph is undirected: if any element j is in graph[i], then i will be in graph[j].

Solution:

Think as a graph, if we can color all the node with two colors, then the graph is Bipartite. The neighbours node should have a different color than yours. Then using BFS or DFS to check or assign the color to the node next to it.

Using BFS:

class Solution {
    public boolean isBipartite(int[][] graph) {
        if (graph == null || graph.length == 0) return true;

        int[] colors = new int[graph.length];
        Queue<Integer> que = new LinkedList<>();
        
        for (int i = 0; i < colors.length; i++) {
            if (colors[i] != 0) continue;
            
            colors[i] = 1;
            que.offer(i);
            
            while (!que.isEmpty()) {
                int curr = que.poll();
                
                for (int next : graph[curr]) {
                    if (colors[curr] == colors[next]) return false;
                    if (colors[next] == 0) {
                        colors[next] = -colors[curr];
                        que.add(next);
                    }
                }
            }
        }
        return true;
    }
}

Using DFS:

class Solution {
    public boolean isBipartite(int[][] graph) {
        if (graph == null || graph.length == 0) return true;

        int[] colors = new int[graph.length];
        for (int i = 0; i < colors.length; i++) {
            if (colors[i] == 0 && !dfs(i, 1, graph, colors)) return false;
        }
        
        return true;
    }
    
    private boolean dfs(int i, int targetcolor, int[][] graph, int[] colors){
        if (colors[i] != 0) return colors[i] == targetcolor;
        
        colors[i] = targetcolor;
        
        if (graph[i] == null) return true;
        
        for (int next: graph[i]) {
            if (!dfs(next, -targetcolor, graph, colors)) return false;
        }
        
        return true;
    }
}

Time complexity: O(V+E)
Space complexity: O(V)

785. Is Graph Bipartite?

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