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Minimum edges to be removed from given undirected graph to remove any existing path between nodes A and B

  • Difficulty Level : Hard
  • Last Updated : 14 Dec, 2021

Given two integers N and M denoting the number of vertices and edges in the graph and array edges[][] of size M, denoting an edge between edges[i][0] and edges[i][1], the task is to find the minimum edges directly connected with node B that must be removed such that there exist no path between vertex A and B.

Examples:

Input: N = 4, A = 3, B = 2, edges[][] = {{3, 1}, {3, 4}, {1, 2}, {4, 2}}
Output: 2
Explanation: The edges at index 2 and 3 i.e., {1, 2} and {4, 2} must be removed as they both are in the path from vertex A to vertex B.

Input: N = 6, A = 1, B = 6, edges[][] = {{1, 2}, {1, 6}, {2, 6}, {1, 4}, {4, 6}, {4, 3}, {2, 4}}
Output: 3

Approach: The given problem can be solved using a Depth-first search algorithm. It can be observed that all the edges associated with the ending vertex B and exist in any path from starting node A and ending at node B must be removed. Hence, perform a dfs starting from node A and maintain all the visited vertices from it. Follow the steps below to solve the problem:

  • Create an adjacency matrix g[][] which stores the edges between two nodes.
  • Initialize an array v[], to mark the node which can be reached from node A.
  • Create a variable cnt, which stores the count of nodes needed to be removed. Initially, cnt = 0.
  • Iterate through all the nodes and if it is reachable from A and is directly connected with B, increment the value of cnt.
  • The value stored in cnt is the required answer.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function for Depth first Search
void dfs(int s, vector<vector<int> > g,
         vector<int>& v)
{
    for (auto i : g[s]) {
 
        // If current vertex is
        // not visited yet
        if (!v[i]) {
            v[i] = 1;
 
            // Recursive call for
            // dfs function
            dfs(i, g, v);
        }
    }
}
 
// Function to find the out the minimum
// number of edges that must be removed
int deleteEdges(int n, int m, int a, int b,
                vector<vector<int> > edges)
{
    // Creating Adjacency Matrix
    vector<vector<int> > g(n, vector<int>());
    for (int i = 0; i < m; i++) {
        g[edges[i][0] - 1].push_back(edges[i][1] - 1);
        g[edges[i][1] - 1].push_back(edges[i][0] - 1);
    }
 
    // Vector for marking visited
    vector<int> v(n, 0);
    v[a - 1] = 1;
 
    // Calling dfs function
    dfs(a - 1, g, v);
 
    // Stores the final count
    int cnt = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If current node can not
        // be visited from node A
        if (v[i] == 0)
            continue;
        for (int j = 0; j < g[i].size(); j++) {
 
            // If a node between current
            // node and node b exist
            if (g[i][j] == b - 1) {
                cnt++;
            }
        }
    }
 
    // Return Answer
    return cnt;
}
 
// Driver Code
int main()
{
    int N = 6;
    int M = 7;
    int A = 1;
    int B = 6;
    vector<vector<int> > edges{
        { 1, 2 }, { 5, 2 }, { 2, 4 },
        { 2, 3 }, { 3, 6 }, { 4, 6 }, { 5, 6 }
    };
 
    cout << deleteEdges(N, M, A, B, edges);
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function for Depth first Search
static void dfs(int s, Vector<Integer> [] g,
         int[] v)
{
    for (int i : g[s]) {
 
        // If current vertex is
        // not visited yet
        if (v[i] == 0) {
            v[i] = 1;
 
            // Recursive call for
            // dfs function
            dfs(i, g, v);
        }
    }
}
 
// Function to find the out the minimum
// number of edges that must be removed
static int deleteEdges(int n, int m, int a, int b,
                int[][] edges)
{
   
    // Creating Adjacency Matrix
    Vector<Integer> []g = new Vector[n];
    for (int i = 0; i < g.length; i++)
        g[i] = new Vector<Integer>();
    for (int i = 0; i < m; i++) {
        g[edges[i][0] - 1].add(edges[i][1] - 1);
        g[edges[i][1] - 1].add(edges[i][0] - 1);
    }
 
    // Vector for marking visited
    int []v = new int[n];
    v[a - 1] = 1;
 
    // Calling dfs function
    dfs(a - 1, g, v);
 
    // Stores the final count
    int cnt = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If current node can not
        // be visited from node A
        if (v[i] == 0)
            continue;
        for (int j = 0; j < g[i].size(); j++) {
 
            // If a node between current
            // node and node b exist
            if (g[i].get(j) == b - 1) {
                cnt++;
            }
        }
    }
 
    // Return Answer
    return cnt;
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 6;
    int M = 7;
    int A = 1;
    int B = 6;
    int[][] edges ={
        { 1, 2 }, { 5, 2 }, { 2, 4 },
        { 2, 3 }, { 3, 6 }, { 4, 6 }, { 5, 6 }
    };
 
    System.out.print(deleteEdges(N, M, A, B, edges));
 
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python program for the above approach
 
# Function for Depth first Search
def dfs(s, g, v):
    for i in g[s]:
 
        # If current vertex is
        # not visited yet
        if not v[i]:
            v[i] = 1
 
            # Recursive call for
            # dfs function
            dfs(i, g, v)
 
# Function to find the out the minimum
# number of edges that must be removed
def deleteEdges(n, m, a, b, edges):
 
    # Creating Adjacency Matrix
    g = [0] * m
    for i in range(len(g)):
        g[i] = []
 
    for i in range(m):
        g[edges[i][0] - 1].append(edges[i][1] - 1)
        g[edges[i][1] - 1].append(edges[i][0] - 1)
 
    # Vector for marking visited
    v = [0] * n
    v[a - 1] = 1
 
    # Calling dfs function
    dfs(a - 1, g, v)
 
    # Stores the final count
    cnt = 0
 
    for i in range(n):
 
        # If current node can not
        # be visited from node A
        if (v[i] == 0):
            continue
 
        for j in range(len(g[i])):
 
            # If a node between current
            # node and node b exist
            if (g[i][j] == b - 1):
                cnt += 1
 
    # Return Answer
    return cnt
 
# Driver Code
N = 6
M = 7
A = 1
B = 6
edges = [[1, 2], [5, 2], [2, 4],
         [2, 3], [3, 6], [4, 6],
         [5, 6]]
 
print(deleteEdges(N, M, A, B, edges))
 
# This code is contributed by gfgking

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
public class GFG{
 
// Function for Depth first Search
static void dfs(int s, List<int> [] g,
         int[] v)
{
    foreach (int i in g[s]) {
 
        // If current vertex is
        // not visited yet
        if (v[i] == 0) {
            v[i] = 1;
 
            // Recursive call for
            // dfs function
            dfs(i, g, v);
        }
    }
}
 
// Function to find the out the minimum
// number of edges that must be removed
static int deleteEdges(int n, int m, int a, int b,
                int[,] edges)
{
   
    // Creating Adjacency Matrix
    List<int> []g = new List<int>[n];
    for (int i = 0; i < g.Length; i++)
        g[i] = new List<int>();
    for (int i = 0; i < m; i++) {
        g[edges[i,0] - 1].Add(edges[i,1] - 1);
        g[edges[i,1] - 1].Add(edges[i,0] - 1);
    }
 
    // List for marking visited
    int []v = new int[n];
    v[a - 1] = 1;
 
    // Calling dfs function
    dfs(a - 1, g, v);
 
    // Stores the readonly count
    int cnt = 0;
 
    for (int i = 0; i < n; i++) {
 
        // If current node can not
        // be visited from node A
        if (v[i] == 0)
            continue;
        for (int j = 0; j < g[i].Count; j++) {
 
            // If a node between current
            // node and node b exist
            if (g[i][j] == b - 1) {
                cnt++;
            }
        }
    }
 
    // Return Answer
    return cnt;
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 6;
    int M = 7;
    int A = 1;
    int B = 6;
    int[,] edges ={
        { 1, 2 }, { 5, 2 }, { 2, 4 },
        { 2, 3 }, { 3, 6 }, { 4, 6 }, { 5, 6 }
    };
 
    Console.Write(deleteEdges(N, M, A, B, edges));
}
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function for Depth first Search
function dfs(s, g, v)
{
    for(let i of g[s])
    {
         
        // If current vertex is
        // not visited yet
        if (!v[i])
        {
            v[i] = 1;
 
            // Recursive call for
            // dfs function
            dfs(i, g, v);
        }
    }
}
 
// Function to find the out the minimum
// number of edges that must be removed
function deleteEdges(n, m, a, b, edges)
{
     
    // Creating Adjacency Matrix
    let g = new Array(m);
    for(let i = 0; i < g.length; i++)
    {
        g[i] = [];
    }
 
    for(let i = 0; i < m; i++)
    {
        g[edges[i][0] - 1].push(edges[i][1] - 1);
        g[edges[i][1] - 1].push(edges[i][0] - 1);
    }
 
    // Vector for marking visited
    let v = new Array(n).fill(0)
    v[a - 1] = 1;
 
    // Calling dfs function
    dfs(a - 1, g, v);
 
    // Stores the final count
    let cnt = 0;
 
    for(let i = 0; i < n; i++)
    {
         
        // If current node can not
        // be visited from node A
        if (v[i] == 0)
            continue;
             
        for(let j = 0; j < g[i].length; j++)
        {
             
            // If a node between current
            // node and node b exist
            if (g[i][j] == b - 1)
            {
                cnt++;
            }
        }
    }
     
    // Return Answer
    return cnt;
}
 
// Driver Code
let N = 6;
let M = 7;
let A = 1;
let B = 6;
let edges = [ [ 1, 2 ], [ 5, 2 ], [ 2, 4 ],
              [ 2, 3 ], [ 3, 6 ], [ 4, 6 ],
              [ 5, 6 ] ];
 
document.write(deleteEdges(N, M, A, B, edges));
 
// This code is contributed by Potta Lokesh
 
</script>
Output
3

Time Complexity: O(N)
Auxiliary Space: O(N)


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