# Is Graph Bipartite?

## Problem

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.

## Solution

With union-find

```
class Solution {
int[] root;
int[] rank;
public boolean isBipartite(int[][] graph) {
root = new int[graph.length];
rank = new int[graph.length];
for (var i = 0; i < graph.length; i++) {
root[i] = i;
rank[i] = 1;
}
for (var i = 0; i < graph.length; i++) {
var node = graph[i];
for (var adj : node) {
if (find(i) == find(adj)) {
return false;
}
union(node[0], adj);
}
}
return true;
}
int find(int x) {
if (root[x] != x) {
root[x] = find(root[x]);
}
return root[x];
}
void union(int x, int y) {
var rootX = find(x);
var rootY = find(y);
if (rank[rootX] > rank[rootY]) {
rootX ^= rootY;
rootY ^= rootX;
rootX ^= rootY;
}
rank[rootY] += rank[rootX];
root[rootX] = rootY;
}
}
```