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:
C++
#include<bits/stdc++.h>
using namespace std;
int minEdgeBFS(vector < int > edges[], int u,
int v, int n)
{
vector< bool > visited(n, 0);
vector< int > distance(n, 0);
queue < int > 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 ;
distance[edges[x][i]] = distance[x] + 1;
Q.push(edges[x][i]);
visited[edges[x][i]] = 1;
}
}
return distance[v];
}
void addEdge(vector < int > edges[], int u, int v)
{
edges[u].push_back(v);
edges[v].push_back(u);
}
int main()
{
int n = 9;
vector < int > edges[9];
addEdge(edges, 0, 1);
addEdge(edges, 0, 7);
addEdge(edges, 1, 7);
addEdge(edges, 1, 2);
addEdge(edges, 2, 3);
addEdge(edges, 2, 5);
addEdge(edges, 2, 8);
addEdge(edges, 3, 4);
addEdge(edges, 3, 5);
addEdge(edges, 4, 5);
addEdge(edges, 5, 6);
addEdge(edges, 6, 7);
addEdge(edges, 7, 8);
int u = 0;
int v = 5;
cout << minEdgeBFS(edges, u, v, n);
return 0;
}
|
Java
import java.util.LinkedList;
import java.util.Queue;
import java.util.Vector;
class Test
{
static int minEdgeBFS(Vector <Integer> edges[], int u,
int v, int n)
{
Vector<Boolean> visited = new Vector<Boolean>(n);
for ( int i = 0 ; i < n; i++) {
visited.addElement( false );
}
Vector<Integer> distance = new Vector<Integer>(n);
for ( int i = 0 ; i < n; i++) {
distance.addElement( 0 );
}
Queue<Integer> Q = new LinkedList<>();
distance.setElementAt( 0 , u);
Q.add(u);
visited.setElementAt( true , u);
while (!Q.isEmpty())
{
int x = Q.peek();
Q.poll();
for ( int i= 0 ; i<edges[x].size(); i++)
{
if (visited.elementAt(edges[x].get(i)))
continue ;
distance.setElementAt(distance.get(x) + 1 ,edges[x].get(i));
Q.add(edges[x].get(i));
visited.setElementAt( true ,edges[x].get(i));
}
}
return distance.get(v);
}
static void addEdge(Vector <Integer> edges[], int u, int v)
{
edges[u].add(v);
edges[v].add(u);
}
public static void main(String args[])
{
int n = 9 ;
Vector <Integer> edges[] = new Vector[ 9 ];
for ( int i = 0 ; i < edges.length; i++) {
edges[i] = new Vector<>();
}
addEdge(edges, 0 , 1 );
addEdge(edges, 0 , 7 );
addEdge(edges, 1 , 7 );
addEdge(edges, 1 , 2 );
addEdge(edges, 2 , 3 );
addEdge(edges, 2 , 5 );
addEdge(edges, 2 , 8 );
addEdge(edges, 3 , 4 );
addEdge(edges, 3 , 5 );
addEdge(edges, 4 , 5 );
addEdge(edges, 5 , 6 );
addEdge(edges, 6 , 7 );
addEdge(edges, 7 , 8 );
int u = 0 ;
int v = 5 ;
System.out.println(minEdgeBFS(edges, u, v, n));
}
}
|
Python3
import queue
def minEdgeBFS(edges, u, v, n):
visited = [ 0 ] * n
distance = [ 0 ] * n
Q = queue.Queue()
distance[u] = 0
Q.put(u)
visited[u] = True
while ( not Q.empty()):
x = Q.get()
for i in range ( len (edges[x])):
if (visited[edges[x][i]]):
continue
distance[edges[x][i]] = distance[x] + 1
Q.put(edges[x][i])
visited[edges[x][i]] = 1
return distance[v]
def addEdge(edges, u, v):
edges[u].append(v)
edges[v].append(u)
if __name__ = = '__main__' :
n = 9
edges = [[] for i in range (n)]
addEdge(edges, 0 , 1 )
addEdge(edges, 0 , 7 )
addEdge(edges, 1 , 7 )
addEdge(edges, 1 , 2 )
addEdge(edges, 2 , 3 )
addEdge(edges, 2 , 5 )
addEdge(edges, 2 , 8 )
addEdge(edges, 3 , 4 )
addEdge(edges, 3 , 5 )
addEdge(edges, 4 , 5 )
addEdge(edges, 5 , 6 )
addEdge(edges, 6 , 7 )
addEdge(edges, 7 , 8 )
u = 0
v = 5
print (minEdgeBFS(edges, u, v, n))
|
C#
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
static int minEdgeBFS(ArrayList []edges, int u,
int v, int n)
{
ArrayList visited = new ArrayList();
for ( int i = 0; i < n; i++)
{
visited.Add( false );
}
ArrayList distance = new ArrayList();
for ( int i = 0; i < n; i++)
{
distance.Add(0);
}
Queue Q = new Queue();
distance[u] = 0;
Q.Enqueue(u);
visited[u] = true ;
while (Q.Count != 0)
{
int x = ( int )Q.Dequeue();
for ( int i = 0; i < edges[x].Count; i++)
{
if (( bool )visited[( int )edges[x][i]])
continue ;
distance[( int )edges[x][i]] = ( int )distance[x] + 1;
Q.Enqueue(( int )edges[x][i]);
visited[( int )edges[x][i]] = true ;
}
}
return ( int )distance[v];
}
static void addEdge(ArrayList []edges,
int u, int v)
{
edges[u].Add(v);
edges[v].Add(u);
}
public static void Main( string []args)
{
int n = 9;
ArrayList []edges = new ArrayList[9];
for ( int i = 0; i < 9; i++)
{
edges[i] = new ArrayList();
}
addEdge(edges, 0, 1);
addEdge(edges, 0, 7);
addEdge(edges, 1, 7);
addEdge(edges, 1, 2);
addEdge(edges, 2, 3);
addEdge(edges, 2, 5);
addEdge(edges, 2, 8);
addEdge(edges, 3, 4);
addEdge(edges, 3, 5);
addEdge(edges, 4, 5);
addEdge(edges, 5, 6);
addEdge(edges, 6, 7);
addEdge(edges, 7, 8);
int u = 0;
int v = 5;
Console.Write(minEdgeBFS(edges, u, v, n));
}
}
|
Javascript
<script>
function minEdgeBFS(edges,u,v,n)
{
let visited = [];
for (let i = 0; i < n; i++) {
visited.push( false );
}
let distance = [];
for (let i = 0; i < n; i++) {
distance.push(0);
}
let Q = [];
distance[u] = 0;
Q.push(u);
visited[u] = true ;
while (Q.length!=0)
{
let x = Q.shift();
for (let i=0; i<edges[x].length; i++)
{
if (visited[edges[x][i]])
continue ;
distance[edges[x][i]] = distance[x] + 1;
Q.push(edges[x][i]);
visited[edges[x][i]]= true ;
}
}
return distance[v];
}
function addEdge(edges,u,v)
{
edges[u].push(v);
edges[v].push(u);
}
let n = 9;
let edges = new Array(9);
for (let i = 0; i < edges.length; i++) {
edges[i] = [];
}
addEdge(edges, 0, 1);
addEdge(edges, 0, 7);
addEdge(edges, 1, 7);
addEdge(edges, 1, 2);
addEdge(edges, 2, 3);
addEdge(edges, 2, 5);
addEdge(edges, 2, 8);
addEdge(edges, 3, 4);
addEdge(edges, 3, 5);
addEdge(edges, 4, 5);
addEdge(edges, 5, 6);
addEdge(edges, 6, 7);
addEdge(edges, 7, 8);
let u = 0;
let v = 5;
document.write(minEdgeBFS(edges, u, v, n));
</script>
|
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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Last Updated :
15 Sep, 2023
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