Formatted question description: https://leetcode.ca/all/785.html

785. Is Graph Bipartite? (Medium)

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].

Related Topics:
Depth-first Search, Breadth-first Search, Graph

Solution 1. DFS

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

// Time: O(V + E)
// Space: O(V + E)
class Solution {
    bool dfs(vector<vector<int>> &G, vector<int> &id, int u, int prev = 1) {
        if (id[u]) return id[u] != prev;
        id[u] = -prev;
        for (int v : G[u]) {
            if (!dfs(G, id, v, id[u])) return false;
        }
        return true;
    }
public:
    bool isBipartite(vector<vector<int>>& G) {
        vector<int> id(G.size());
        for (int i = 0; i < G.size(); ++i) {
            if (id[i]) continue;
            if (!dfs(G, id, i)) return false;
        }
        return true;
    }
};

Solution 2. BFS

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

// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
    bool isBipartite(vector<vector<int>>& G) {
        vector<int> id(G.size());
        queue<int> q;
        for (int i = 0; i < G.size(); ++i) {
            if (id[i]) continue;
            q.push(i);
            id[i] = 1;
            while (q.size()) {
                int u = q.front();
                q.pop();
                for (int v : G[u]) {
                    if (id[v]) {
                        if (id[v] != -id[u]) return false;
                        continue;
                    }
                    id[v] = -id[u];
                    q.push(v);
                }
            }
        }
        return true;
    }
};

Java

class Solution {
    public boolean isBipartite(int[][] graph) {
        int nodes = graph.length;
        int[] colors = new int[nodes];
        Arrays.fill(colors, -1);
        for (int i = 0; i < nodes; i++) {
            if (colors[i] == -1) {
                colors[i] = 0;
                Stack<Integer> stack = new Stack<Integer>();
                stack.push(i);
                while (!stack.isEmpty()) {
                    int start = stack.pop();
                    int color = colors[start];
                    int[] ends = graph[start];
                    for (int end : ends) {
                        if (colors[end] == -1) {
                            colors[end] = 1 - color;
                            stack.push(end);
                        } else if (colors[start] == colors[end])
                            return false;
                    }
                }
            }
        }
        return true;
    }
}

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