#   # 785. Is Graph Bipartite?

## Graph Coloring by Depth First Search

### Problem statement

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`.

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 to color 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 note

• 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`, which are `push_back` and `pop_back` which you could use well.

### References

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