# Is Graph Bipartite? Problem

## Description

LeetCode Problem 785.

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:

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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:

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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 containu.
- All the values of graph[u] are unique.
- If graph[u] contains v, then graph[v] contains u.

## Sample C++ Code

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class Solution {
public:
bool isBipartite(vector<vector<int>>& graph) {
int n = graph.size();
vector<int> colors(n, 0);
queue<int> q;
for (int i = 0; i < n; i++) {
if (colors[i]) continue;
colors[i] = 1;
q.push(i);
while (!q.empty()) {
int temp = q.front();
for (auto neighbor : graph[temp]) {
// Color neighbor with opposite color
if (!colors[neighbor]) {
colors[neighbor] = -colors[temp];
q.push(neighbor);
}
// If the neighbor has the same color - can't bipartite.
else if (colors[neighbor] == colors[temp])
return false;
}
q.pop();
}
}
return true;
}
};