Is Graph Bipartite?#

Problem statement#

[1]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 a bipartite graph.

Example 1#

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#

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.

Solution: Coloring the nodes by Depth First Search#

You could color the nodes in set A with one color and those in B with another color. Then two ends of every edge have different colors.

Now you can use the DFS algorithm[2] to perform the coloring on each connected component of the graph.

During the traversal, if there is an edge having the same color at two ends then return false.

Code#

#include <vector>
#include <iostream>
using namespace std;
bool isBipartite(vector<vector<int>>& graph) {
    vector<int> color(graph.size(), 0);
    for (int i = 0; i < graph.size(); i++) {
        if (color[i] != 0) continue;
        vector<int> s;
        s.push_back(i);
        color[i] = 1;
        while (!s.empty()) {
            int u = s.back();
            s.pop_back();
            for (int v : graph[u]) {
                if (color[v] == 0) {
                    color[v] = -color[u];
                    s.push_back(v);
                } else if (color[v] == color[u]) {
                    return false;
                }          
            }
        }

    }
    return true;
}
int main() {
    vector<vector<int>> graph{{1,2,3},{0,2},{0,1,3},{0,2}};
    cout << isBipartite(graph) << endl;
    graph = {{1,3},{0,2},{1,3},{0,2}};
    cout << isBipartite(graph) << endl;
}

Output:
0
1

Complexity#

Runtime: O(n), where n = graph.length.
Extra space: O(n).

Implementation tips#

This is the non-recursive implementation of DFS algorithm where you could use the stack data structure to avoid the recursion.
The stack’s methods needed in the DFS algorithm are only push and pop. There are similar ones in std::vector. They are push_back[3] and pop_back[4] which you could use well in this case.