You are given a undirected graph G(V, E) with N vertices and M edges. We need to find the minimum number of edges between a given pair of vertices (u, v).

Examples: 

Input : For given graph G. Find minimum number
        of edges between (1, 5).

Output : 2
Explanation: (1, 2) and (2, 5) are the only
edges resulting into shortest path between 1
and 5.

The idea is to perform BFS from one of given input vertex(u). At the time of BFS maintain an array of distance[n] and initialize it to zero for all vertices. Now, suppose during BFS, vertex x is popped from queue and we are pushing all adjacent non-visited vertices(i) back into queue at the same time we should update distance[i] = distance[x] + 1;. 
Finally, distance[v] gives the minimum number of edges between u and v. 

Algorithm: 

int minEdgeBFS(int u, int v, int n)
{
    // visited[n] for keeping track of visited
    // node in BFS
    bool visited[n] = {0};

    // Initialize distances as 0
    int distance[n] = {0};
 
    // queue to do BFS.
    queue  Q;
    distance[u] = 0;
    
    Q.push(u);
    visited[u] = true;
    while (!Q.empty())
    {
        int x = Q.front();
        Q.pop();
 
        for (int i=0; i<edges[x].size(); i++)
        {
            if (visited[edges[x][i]])
                continue;

            // update distance for i
            distance[edges[x][i]] = distance[x] + 1;
            Q.push(edges[x][i]);
            visited[edges[x][i]] = 1;
        }
    }
    return distance[v];
}

Output: 

3

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.