Problem

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:

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:

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

DFS

class Solution {
    public boolean isBipartite(int[][] graph) {
        int[] colors = new int[graph.length];
        
        for (int i = 0; i < graph.length; i++) {
            if (colors[i] == 0 && !dfs(graph, colors, i, 1)) return false;
        }
        
        return true;
    }
    
    private boolean dfs(int[][] graph, int[] colors, int node, int color) {
        if (colors[node] != 0) return colors[node] == color;
        else colors[node] = color;
        
        for (int neighbor: graph[node]) {
            if (!dfs(graph, colors, neighbor, -color)) return false;
        }
        
        return true;
    }
}

BFS

class Solution {
    public boolean isBipartite(int[][] graph) {
        int[] colors = new int[graph.length];
        
        for (int i = 0; i < graph.length; i++) {
            if (graph[i].length != 0 && colors[i] == 0) {
                colors[i] = 1;
                Queue<Integer> queue = new LinkedList<>();
                queue.offer(i);
                while (!queue.isEmpty()) {
                    int node = queue.poll();
                    for (int neighbor: graph[node]) {
                        if (colors[neighbor] == 0) {
                            colors[neighbor] = -colors[node];
                            queue.offer(neighbor);
                        } else {
                            if (colors[neighbor] == colors[node]) return false;
                        }
                    }
                }
            }
        }
        
        return true;
    }
}

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