Today I have done leetcode's April LeetCoding Challenge with cpp.

April LeetCoding Challenge 29

Description

Is Graph Bipartite?

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

class Solution {
public:
  bool isBipartite(vector<vector<int>>& graph) {
    int len = graph.size();
    vector<int> type(len, -1);
    int count = 0;
    while(count < len) {
      queue<int> q;
      for(int i = 0; i < len; ++i) {
        if(type[i] != -1) continue;
        q.push(i);
        type[i] = 0;
        count += 1;
        break;
      }
      while(!q.empty()) {
        int cur = q.front();
        q.pop();
        for(auto neighbor : graph[cur]) {
          if(type[neighbor] == type[cur]) return false;
          if(type[neighbor] == (type[cur] ^ 1)) continue;
          q.push(neighbor);
          type[neighbor] = type[cur] ^ 1;
          count += 1;
        }
      }
    }
    return true;
  }
};

// Accepted
// 80/80 cases passed (27 ms)
// Your runtime beats 73.07 % of cpp submissions
// Your memory usage beats 41.19 % of cpp submissions (13.7 MB)