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Check if an edge is a part of any Minimum Spanning Tree

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Given a connected undirected weighted graph in the form of a 2D array where each row is of the type [start node, end node, weight] describing an edge, and also two integers (A, B). Return if the edge formed between (A, B) is a part of any of the Minimum Spanning Tree (MST) of the graph.

Minimum Spanning Tree (MST): This is a special subgraph of the graph, such that each and every vertex is connected and the overall sum of the weights of the edges of this subgraph is as minimum as possible. A graph can have multiple minimum spanning trees.

Examples:

Input: graph = [[0 ,1, 20] , [0 , 2 , 5] , [ 0, 3, 10 ] , [ 2, 3, 10]], A = 2, B = 3
Output:  True
Explanation : 2 minimum spanning trees with can be generated which will have weight 35. The connections of the trees are 
1st: [ (0,1) , (0,3) , (0,2)] => 20 + 10 + 5 = 35
2nd: [ ( 0 , 1) , ( 0 , 2 ) , ( 2 , 3) ] => 20 + 5 + 10 = 35
As it can be seen , the edge ( 2, 3) is present in second MST.

The graph is shown in image:

Input: graph = [[0 ,1, 20] , [0 , 2 , 5] , [ 0, 3, 10 ] , [ 2, 3, 20]], A = 2, B = 3
Output: False
Explanation: Only 1 minimum spanning trees with weight 35 can be generated,
but edge (2, 3) is not included.
[(0,1) , (0,3) , (0,2)] => 20 + 10 + 5 = 35

The graph is given in the image

 

Approach : Kruskal Algorithm and Prim’s Algorithm are the two most used algorithms that can be used to find MST of any graph. In this article, the solution is based on the Kruskal algorithm. Follow the steps mentioned below to solve the problem using this approach:

  1. Find the minimum spanning tree cost of the entire graph, using the Kruskal algorithm.
  2. As the inclusion of the edge (A, B) in the MST is being checked, include this edge first in the minimum spanning tree and then include other edges subsequently.
  3. Finally check if the cost is the same for both the spanning trees including the edge(A, B) and the calculated weight of the MST.
  4. If cost is the same, then edge (A, B) is a part of some MST of the graph otherwise it is not.

Below is the implementation of the above approach: 

C++




#include <bits/stdc++.h>
using namespace std;
 
// Class to implement disjoint set union
class dsu {
public:
    unordered_map<int, int> parent;
    unordered_map<int, int> rank;
 
    // Function to find parent of a node
    int find(int x)
    {
        if (parent.count(x) == 0) {
            rank[x] = 1;
            parent[x] = x;
        }
        if (parent[x] != x) {
            parent[x] = find(parent[x]);
        }
        return parent[x];
    }
 
    // Function to perform union
    bool unite(int u, int v)
    {
        int p1 = find(u), p2 = find(v);
 
        // If do not belong to same set
        if (p1 != p2) {
            if (rank[p1] < rank[p2]) {
                parent[p1] = p2;
            }
            else if (rank[p1] > rank[p2]) {
                parent[p2] = p1;
            }
            else {
                parent[p2] = p1;
                rank[p1] += 1;
            }
            return true;
        }
 
        // Belong to same set
        else {
            return false;
        }
    }
};
 
class Solution {
public:
    // Find the MST weight
    int kruskal(bool include, vector<vector<int> >& edges,
                int a, int b)
    {
        dsu obj;
        int total = 0;
 
        // If include is True, then include edge (a, b)
        // first
        if (include) {
            for (auto edge : edges) {
                int u = edge[0], v = edge[1], wt = edge[2];
 
                // As graph is undirected so (a, b) or (b,
                // a) is same If found break the for loop
                if ((u == a && v == b)
                    || (u == b && v == a)) {
                    bool val = obj.unite(a, b);
                    total += wt;
                    break;
                }
            }
        }
 
        // Go on adding edge to the disjoint set
        for (auto edge : edges) {
            int u = edge[0], v = edge[1], wt = edge[2];
 
            // Nodes (u, v) not belong to same set include
            // it
            if (obj.unite(u, v)) {
                total += wt;
            }
        }
 
        // Finally return total weight of MST
        return total;
    }
 
    // Function to find if edge (a, b) is part of any MST
    bool solve(vector<vector<int> >& edges, int a, int b)
    {
        // Sort edges according to weight in ascending order
        sort(edges.begin(), edges.end(),
             [](vector<int> a, vector<int> b) {
                 return a[2] < b[2];
             });
 
        // Not included edge (a, b)
        int overall = kruskal(false, edges, a, b);
 
        // Find mst with edge (a, b) included
        int inc = kruskal(true, edges, a, b);
 
        // Finally return True if same else False
        return inc == overall;
    }
};
int main()
{
    Solution obj;
    vector<vector<int> > graph = { { 0, 1, 20 },
                                   { 0, 2, 5 },
                                   { 0, 3, 10 },
                                   { 2, 3, 10 } };
    int A = 2, B = 3;
    bool val = obj.solve(graph, A, B);
    if (val) {
        cout << "True" << endl;
    }
    else {
        cout << "False" << endl;
    }
    return 0;
}
 
// This code is contributed by lokeshpotta20.


Java




// Java program to implement above approach
 
import java.io.*;
import java.util.*;
 
// Class to implement disjoint set union
class DSU {
    Map<Integer, Integer> parent = new HashMap<>();
    Map<Integer, Integer> rank = new HashMap<>();
 
    // Function to find parent of a node
    int find(int x)
    {
        if (!parent.containsKey(x)) {
            rank.put(x, 1);
            parent.put(x, x);
        }
        if (parent.get(x) != x) {
            parent.put(x, find(parent.get(x)));
        }
        return parent.get(x);
    }
 
    // Function to perform union
    boolean unite(int u, int v)
    {
        int p1 = find(u), p2 = find(v);
 
        // If do not belong to same set
        if (p1 != p2) {
            if (rank.get(p1) < rank.get(p2)) {
                parent.put(p1, p2);
            }
            else if (rank.get(p1) > rank.get(p2)) {
                parent.put(p2, p1);
            }
            else {
                parent.put(p2, p1);
                rank.put(p1, rank.get(p1) + 1);
            }
            return true;
        }
 
        // Belong to same set
        else {
            return false;
        }
    }
}
 
class Solution {
 
    // Find the MST weight
    int kruskal(boolean include, List<List<Integer> > edges,
                int a, int b)
    {
        DSU obj = new DSU();
        int total = 0;
        // If include is True, then include edge (a, b)
        // first
        if (include) {
            for (List<Integer> edge : edges) {
                int u = edge.get(0), v = edge.get(1),
                    wt = edge.get(2);
 
                // As graph is undirected so (a, b) or (b,
                // a) is same If found break the for loop
                if ((u == a && v == b)
                    || (u == b && v == a)) {
                    boolean val = obj.unite(a, b);
                    total += wt;
                    break;
                }
            }
        }
 
        // Go on adding edge to the disjoint set
        for (List<Integer> edge : edges) {
            int u = edge.get(0), v = edge.get(1),
                wt = edge.get(2);
 
            // Nodes (u, v) not belong to same set include
            // it
            if (obj.unite(u, v)) {
                total += wt;
            }
        }
 
        // Finally return total weight of MST
        return total;
    }
 
    // Function to find if edge (a, b) is part of any MST
    boolean solve(List<List<Integer> > edges, int a, int b)
    {
        // Sort edges according to weight in ascending order
        Collections.sort(
            edges, new Comparator<List<Integer> >() {
                @Override
                public int compare(List<Integer> a,
                                   List<Integer> b)
                {
                    return a.get(2) - b.get(2);
                }
            });
 
        // Not included edge (a, b)
        int overall = kruskal(false, edges, a, b);
 
        // Find mst with edge (a, b) included
        int inc = kruskal(true, edges, a, b);
 
        // Finally return true if same else false
        return inc == overall;
    }
}
 
class GFG {
    public static void main(String[] args)
    {
        Solution obj = new Solution();
        List<List<Integer> > graph = Arrays.asList(
            Arrays.asList(0, 1, 20), Arrays.asList(0, 2, 5),
            Arrays.asList(0, 3, 10),
            Arrays.asList(2, 3, 10));
        int A = 2, B = 3;
        boolean val = obj.solve(graph, A, B);
        if (val) {
            System.out.println("True");
        }
        else {
            System.out.println("False");
        }
    }
}
 
// This code is contributed by karthik.


Python3




# Python program to implement above approach
 
# Class to implement disjoint set union
class dsu:
    def __init__(self):
        self.parent = {}
        self.rank = {}
 
    # Function to find parent of a node
    def find(self, x):
        if (x not in self.parent):
            self.rank[x] = 1
            self.parent[x] = x
        if (self.parent[x] != x):
            self.parent[x] = \
                self.find(self.parent[x])
        return (self.parent[x])
 
    # Function to perform union
    def union(self, u, v):
        p1 = self.find(u)
        p2 = self.find(v)
 
        # If do not belong to same set
        if (p1 != p2):
            if (self.rank[p1]
                    < self.rank[p2]):
                self.parent[p1] = p2
 
            elif(self.rank[p1]
                 > self.rank[p1]):
                self.parent[p2] = p1
            else:
                self.parent[p2] = p1
                self.rank[p1] += 1
            return (True)
 
        # Belong to same set
        else:
            return False
 
 
class Solution:
 
    # Find the MST weight
    def kruskal(self, include, edges, a, b):
        obj = dsu()
        total = 0
 
        # If include is True , then include
        # edge (a,b) first
        if (include):
            for (u, v, wt) in edges:
 
                # As graph is undirected so
                # (a,b) or (b,a) is same
                # If found break the for loop
                if (u, v) == (a, b) or \
                        (b, a) == (u, v):
                    val = obj.union(a, b)
                    total += wt
                    break
 
        # Go on adding edge to the disjoint set
        for (u, v, wt) in edges:
 
            # Nodes (u,v) not belong to
            # same set include it
            if (obj.union(u, v)):
                total += wt
 
        # Finally return total weight of MST
        return (total)
 
    # Function to find if edge (a, b)
    # is part of any MST
    def solve(self, edges, a, b):
 
        # Sort edges according to weight
        # in ascending order
        edges.sort(key=lambda it: it[2])
 
        # Not included edge (a,b)
        overall = self.kruskal(False,
                               edges, a, b)
 
        # Find mst with edge (a,b) included
        inc = self.kruskal(True,
                           edges, a, b)
 
        # Finally return True if same
        # else False
        if (inc == overall):
            return (True)
        else:
            return (False)
 
# Driver code
if __name__ == "__main__":
    obj = Solution()
    graph = [[0, 1, 20], [0, 2, 5],
             [0, 3, 10], [2, 3, 10]]
    A, B = 2, 3
    val = obj.solve(graph, A, B)
    if (val):
        print("True")
    else:
        print("False")


C#




// C# program to implement above approach
using System;
using System.Collections.Generic;
using System.Linq;
 
// Class to implement disjoint set union
public class dsu {
  Dictionary<int, int> parent
    = new Dictionary<int, int>();
  Dictionary<int, int> rank = new Dictionary<int, int>();
 
  // Function to find parent of a node
  public int Find(int x)
  {
    if (!parent.ContainsKey(x)) {
      rank[x] = 1;
      parent[x] = x;
    }
    if (parent[x] != x) {
      parent[x] = Find(parent[x]);
    }
    return parent[x];
  }
 
  // Function to perform union
  public bool Unite(int u, int v)
  {
    int p1 = Find(u), p2 = Find(v);
 
    // If do not belong to same set
    if (p1 != p2) {
      if (rank[p1] < rank[p2]) {
        parent[p1] = p2;
      }
      else if (rank[p1] > rank[p2]) {
        parent[p2] = p1;
      }
      else {
        parent[p2] = p1;
        rank[p1] += 1;
      }
      return true;
    }
 
    // Belong to same set
    else {
      return false;
    }
  }
}
 
public class Solution {
  // Find the MST weight
  int Kruskal(bool include, List<List<int> > edges, int x,
              int y)
  {
    dsu obj = new dsu();
    int total = 0;
 
    // If include is True, then include edge (a, b)
    // first
    if (include) {
      foreach(var edge in edges)
      {
        int u = edge[0], v = edge[1], wt = edge[2];
 
        // As graph is undirected so (a, b) or (b,
        // a) is same If found break the for loop
        if ((u == x && v == y)
            || (u == y && v == x)) {
          if (obj.Unite(x, y)) {
            total += wt;
            break;
          }
        }
      }
    }
 
    // Go on adding edge to the disjoint set
    foreach(var edge in edges)
    {
      int u = edge[0], v = edge[1], wt = edge[2];
 
      // Nodes (u, v) not belong to same set include
      // it
      if (obj.Unite(u, v)) {
        total += wt;
      }
    }
 
    // Finally return total weight of MST
    return total;
  }
 
  // Function to find if edge (a, b) is part of any MST
  public bool Solve(List<List<int> > edges, int x, int y)
  {
    // Sort edges according to weight in ascending order
    edges.Sort((a, b) => a[2] - b[2]);
 
    // Not included edge (a, b)
    int overall = Kruskal(false, edges, x, y);
 
    // Find mst with edge (a, b) included
    int inc = Kruskal(true, edges, x, y);
 
    // Finally return true if same else false
    return inc == overall;
  }
}
 
public class GFG {
 
  static public void Main()
  {
 
    // Code
    Solution obj = new Solution();
    List<List<int> > graph = new List<List<int> >{
      new List<int>{ 0, 1, 20 },
      new List<int>{ 0, 2, 5 },
      new List<int>{ 0, 3, 10 },
      new List<int>{ 2, 3, 10 }
    };
    int A = 2, B = 3;
    bool val = obj.Solve(graph, A, B);
    if (val) {
      Console.WriteLine("True");
    }
    else {
      Console.WriteLine("False");
    }
  }
}
 
// This code is contributed by sankar.


Javascript




// Class to implement disjoint set union
class DSU {
    constructor() {
        this.parent = {};
        this.rank = {};
    }
 // Function to find parent of a node
    find(x) {
        if (!(x in this.parent)) {
            this.rank[x] = 1;
            this.parent[x] = x;
        }
        if (this.parent[x] !== x) {
            this.parent[x] = this.find(this.parent[x]);
        }
        return this.parent[x];
    }
 // Function to perform union
    union(u, v) {
        let p1 = this.find(u);
        let p2 = this.find(v);
        if (p1 !== p2) {
        // If do not belong to same set
            if (this.rank[p1] < this.rank[p2]) {
                this.parent[p1] = p2;
            } else if (this.rank[p1] > this.rank[p1]) {
                this.parent[p2] = p1;
            } else {
                this.parent[p2] = p1;
                this.rank[p1] += 1;
            }
            return true;
        } else {
            return false;
        }
    }
}
 
class Solution {
 // Find the MST weight
    kruskal(include, edges, a, b) {
        let obj = new DSU();
        let total = 0;
        if (include) {
            for (let i = 0; i < edges.length; i++) {
                let [u, v, wt] = edges[i];
                if ((u === a && v === b) || (b === a && u === v)) {
                    let val = obj.union(a, b);
                    total += wt;
                    break;
                }
            }
        }
        for (let i = 0; i < edges.length; i++) {
            let [u, v, wt] = edges[i];
            if (obj.union(u, v)) {
                total += wt;
            }
        }
        return total;
    }
 
    solve(edges, a, b) {
     // As graph is undirected so (a, b) or (b,
                // a) is same If found break the for loop
        edges.sort((a, b) => a[2] - b[2]);
        let overall = this.kruskal(false, edges, a, b);
        let inc = this.kruskal(true, edges, a, b);
        return inc === overall;
    }
}
 
let obj = new Solution();
let graph = [[0, 1, 20], [0, 2, 5], [0, 3, 10], [2, 3, 10]];
let A = 2, B = 3;
let val = obj.solve(graph, A, B);
if (val) {
    console.log("True");
} else {
 // Finally return True if same else False
    console.log("False");
}


Output

True

Time Complexity: O(E * logV). where E is the number of edges and V is the number of vertices.
Auxiliary Space: O(V)



Last Updated : 02 Mar, 2023
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