Minimum number of edges between two vertices of a Graph

You are given an 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];
}

Implementation:

3

Time Complexity: O(V + E), for performing Breadth-First Search.
Auxiliary Space: O(V), as V vertices can be present in the queue in the worst case.

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