# Is Graph Bipartite?

Nov 25, 2019

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].

char color;
bool isBipartite(vector<vector<int>>& graph) {
int size = graph.size();
vector<char> flags(size, 'w');
// w g b
color = 'b';
for(int i = 0;i < size;i++) {
if (flags[i] == 'w') {
dfs(graph, flags, i);
}
}

for(int i = 0;i < size;i++) {
for(auto j: graph[i]) {
if (flags[i] == flags[j]) return false;
}
}
return true;
}

void dfs(vector<vector<int>> &graph, vector<char> &flags, int index) {
flags[index] = color;
for(auto j: graph[index]) {
if (flags[j] == 'w') {
color = ~color;
dfs(graph, flags, j);
color = ~color;
}
}
}
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