14. Is Graph Bipartite.cpp

/*
Is Graph Bipartite?
===================

Given an undirected graph, return true if and only if it is bipartite.

Recall that a graph is bipartite if we can split its 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: graph = [[1,3],[0,2],[1,3],[0,2]]
Output: true
Explanation: We can divide the vertices into two groups: {0, 2} and {1, 3}.

Example 2:
Input: graph = [[1,2,3],[0,2],[0,1,3],[0,2]]
Output: false
Explanation: We cannot find a way to divide the set of nodes into two independent subsets.

Constraints:
1 <= graph.length <= 100
0 <= graph[i].length < 100
0 <= graph[i][j] <= graph.length - 1
graph[i][j] != i
All the values of graph[i] are unique.
The graph is guaranteed to be undirected. 
*/

class Solution
{
public:
  bool bfs(int curr, vector<vector<int>> &graph, vector<int> &color)
  {
    queue<int> pending;
    pending.push(curr);
    color[curr] = 1;

    while (pending.size())
    {
      auto curr = pending.front();
      pending.pop();
      for (auto &i : graph[curr])
      {
        if (color[i] != -1 && color[i] != 1 - color[curr])
          return false;
        else if (color[i] == -1)
        {
          color[i] = 1 - color[curr];
          pending.push(i);
        }
      }
    }
    return true;
  }

  bool isBipartite(vector<vector<int>> &graph)
  {
    int n = graph.size();
    vector<int> color(n, -1);

    for (int i = 0; i < n; ++i)
    {
      if (color[i] == -1)
      {
        if (!bfs(i, graph, color))
          return false;
      };
    }

    return true;
  }
};