785

785. Is Graph Bipartite? (Medium)

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.

Companies:
Amazon, LinkedIn, Google, ByteDance

Related Topics:
Depth-First Search, Breadth-First Search, Union Find, Graph

Solution 1. DFS

// OJ: https://leetcode.com/problems/is-graph-bipartite/
// Author: github.com/lzl124631x
// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
    bool isBipartite(vector<vector<int>>& G) {
        int N = G.size();
        vector<int> id(N); // 0 unseen, 1 and -1 are different colors
        function<bool(int, int)> dfs = [&](int u, int color) {
            if (id[u]) return id[u] == color;
            id[u] = color;
            for (int v : G[u]) {
                if (!dfs(v, -color)) return false;
            }
            return true;
        };
        for (int i = 0; i < N; ++i) {
            if (!id[i] && !dfs(i, 1)) return false;
        }
        return true;
    }
};

Solution 2. BFS

// OJ: https://leetcode.com/problems/is-graph-bipartite/
// Author: github.com/lzl124631x
// Time: O(V + E)
// Space: O(V + E)
class Solution {
public:
    bool isBipartite(vector<vector<int>>& G) {
        int N = G.size();
        vector<int> id(N); // 0 unseen, 1 and -1 are different colors
        queue<int> q;
        for (int i = 0; i < N; ++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;
    }
};