Check if an edge is a part of any Minimum Spanning Tree
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 = 35The 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:
- Find the minimum spanning tree cost of the entire graph, using the Kruskal algorithm.
- 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.
- 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.
- 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" ); } |
True
Time Complexity: O(E * logV). where E is the number of edges and V is the number of vertices.
Auxiliary Space: O(V)
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