Find if there is a path of more than k length from a source
Given a graph, a source vertex in the graph and a number k, find if there is a simple path (without any cycle) starting from given source and ending at any other vertex such that the distance from source to that vertex is atleast ‘k’ length.
Example:

Weighted Undirected Graph
Input : Source s = 0, k = 58 Output : True There exists a simple path 0 -> 7 -> 1 -> 2 -> 8 -> 6 -> 5 -> 3 -> 4 Which has a total distance of 60 km which is more than 58. Input : Source s = 0, k = 62 Output : False In the above graph, the longest simple path has distance 61 (0 -> 7 -> 1-> 2 -> 3 -> 4 -> 5-> 6 -> 8, so output should be false for any input greater than 61.
We strongly recommend you to minimize your browser and try this yourself first.
One important thing to note is, simply doing BFS or DFS and picking the longest edge at every step would not work. The reason is, a shorter edge can produce longer path due to higher weight edges connected through it.
The idea is to use Backtracking. We start from given source, explore all paths from current vertex. We keep track of current distance from source. If distance becomes more than k, we return true. If a path doesn’t produces more than k distance, we backtrack.
How do we make sure that the path is simple and we don’t loop in a cycle? The idea is to keep track of current path vertices in an array. Whenever we add a vertex to path, we check if it already exists or not in current path. If it exists, we ignore the edge.
Below is implementation of above idea.
C++
// Program to find if there is a simple path with // weight more than k #include<bits/stdc++.h> using namespace std; // iPair ==> Integer Pair typedef pair< int , int > iPair; // This class represents a dipathted graph using // adjacency list representation class Graph { int V; // No. of vertices // In a weighted graph, we need to store vertex // and weight pair for every edge list< pair< int , int > > *adj; bool pathMoreThanKUtil( int src, int k, vector< bool > &path); public : Graph( int V); // Constructor // function to add an edge to graph void addEdge( int u, int v, int w); bool pathMoreThanK( int src, int k); }; // Returns true if graph has path more than k length bool Graph::pathMoreThanK( int src, int k) { // Create a path array with nothing included // in path vector< bool > path(V, false ); // Add source vertex to path path[src] = 1; return pathMoreThanKUtil(src, k, path); } // Prints shortest paths from src to all other vertices bool Graph::pathMoreThanKUtil( int src, int k, vector< bool > &path) { // If k is 0 or negative, return true; if (k <= 0) return true ; // Get all adjacent vertices of source vertex src and // recursively explore all paths from src. list<iPair>::iterator i; for (i = adj[src].begin(); i != adj[src].end(); ++i) { // Get adjacent vertex and weight of edge int v = (*i).first; int w = (*i).second; // If vertex v is already there in path, then // there is a cycle (we ignore this edge) if (path[v] == true ) continue ; // If weight of is more than k, return true if (w >= k) return true ; // Else add this vertex to path path[v] = true ; // If this adjacent can provide a path longer // than k, return true. if (pathMoreThanKUtil(v, k-w, path)) return true ; // Backtrack path[v] = false ; } // If no adjacent could produce longer path, return // false return false ; } // Allocates memory for adjacency list Graph::Graph( int V) { this ->V = V; adj = new list<iPair> [V]; } // Utility function to an edge (u, v) of weight w void Graph::addEdge( int u, int v, int w) { adj[u].push_back(make_pair(v, w)); adj[v].push_back(make_pair(u, w)); } // Driver program to test methods of graph class int main() { // create the graph given in above figure int V = 9; Graph g(V); // making above shown graph g.addEdge(0, 1, 4); g.addEdge(0, 7, 8); g.addEdge(1, 2, 8); g.addEdge(1, 7, 11); g.addEdge(2, 3, 7); g.addEdge(2, 8, 2); g.addEdge(2, 5, 4); g.addEdge(3, 4, 9); g.addEdge(3, 5, 14); g.addEdge(4, 5, 10); g.addEdge(5, 6, 2); g.addEdge(6, 7, 1); g.addEdge(6, 8, 6); g.addEdge(7, 8, 7); int src = 0; int k = 62; g.pathMoreThanK(src, k)? cout << "Yes\n" : cout << "No\n" ; k = 60; g.pathMoreThanK(src, k)? cout << "Yes\n" : cout << "No\n" ; return 0; } |
Java
// Java Program to find if there is a simple path with // weight more than k import java.util.*; public class GFG{ static class AdjListNode { int v; int weight; AdjListNode( int _v, int _w) { v = _v; weight = _w; } int getV() { return v; } int getWeight() { return weight; } } // This class represents a dipathted graph using // adjacency list representation static class Graph { int V; // No. of vertices // In a weighted graph, we need to store vertex // and weight pair for every edge ArrayList<ArrayList<AdjListNode>> adj; // Allocates memory for adjacency list Graph( int V) { this .V = V; adj = new ArrayList<ArrayList<AdjListNode>>(V); for ( int i = 0 ; i < V; i++) { adj.add( new ArrayList<AdjListNode>()); } } // Utility function to an edge (u, v) of weight w void addEdge( int u, int v, int weight) { AdjListNode node1 = new AdjListNode(v, weight); adj.get(u).add(node1); // Add v to u's list AdjListNode node2 = new AdjListNode(u, weight); adj.get(v).add(node2); // Add u to v's list } // Returns true if graph has path more than k length boolean pathMoreThanK( int src, int k) { // Create a path array with nothing included // in path boolean path[] = new boolean [V]; Arrays.fill(path, false ); // Add source vertex to path path[src] = true ; return pathMoreThanKUtil(src, k, path); } // Prints shortest paths from src to all other vertices boolean pathMoreThanKUtil( int src, int k, boolean [] path) { // If k is 0 or negative, return true; if (k <= 0 ) return true ; // Get all adjacent vertices of source vertex src and // recursively explore all paths from src. ArrayList<AdjListNode> it = adj.get(src); int index = 0 ; for ( int i = 0 ; i < adj.get(src).size(); i++) { AdjListNode vertex = adj.get(src).get(i); // Get adjacent vertex and weight of edge int v = vertex.v; int w = vertex.weight; // increase theindex index++; // If vertex v is already there in path, then // there is a cycle (we ignore this edge) if (path[v] == true ) continue ; // If weight of is more than k, return true if (w >= k) return true ; // Else add this vertex to path path[v] = true ; // If this adjacent can provide a path longer // than k, return true. if (pathMoreThanKUtil(v, k-w, path)) return true ; // Backtrack path[v] = false ; } // If no adjacent could produce longer path, return // false return false ; } } // Driver program to test methods of graph class public static void main(String[] args) { // create the graph given in above figure int V = 9 ; Graph g = new Graph(V); // making above shown graph g.addEdge( 0 , 1 , 4 ); g.addEdge( 0 , 7 , 8 ); g.addEdge( 1 , 2 , 8 ); g.addEdge( 1 , 7 , 11 ); g.addEdge( 2 , 3 , 7 ); g.addEdge( 2 , 8 , 2 ); g.addEdge( 2 , 5 , 4 ); g.addEdge( 3 , 4 , 9 ); g.addEdge( 3 , 5 , 14 ); g.addEdge( 4 , 5 , 10 ); g.addEdge( 5 , 6 , 2 ); g.addEdge( 6 , 7 , 1 ); g.addEdge( 6 , 8 , 6 ); g.addEdge( 7 , 8 , 7 ); int src = 0 ; int k = 62 ; if (g.pathMoreThanK(src, k)) System.out.println( "YES" ); else System.out.println( "NO" ); k = 60 ; if (g.pathMoreThanK(src, k)) System.out.println( "YES" ); else System.out.println( "NO" ); } } // This code is contributed by adityapande88. |
Python3
# Program to find if there is a simple path with # weight more than k # This class represents a dipathted graph using # adjacency list representation class Graph: # Allocates memory for adjacency list def __init__( self , V): self .V = V self .adj = [[] for i in range (V)] # Returns true if graph has path more than k length def pathMoreThanK( self ,src, k): # Create a path array with nothing included # in path path = [ False ] * self .V # Add source vertex to path path[src] = 1 return self .pathMoreThanKUtil(src, k, path) # Prints shortest paths from src to all other vertices def pathMoreThanKUtil( self ,src, k, path): # If k is 0 or negative, return true if (k < = 0 ): return True # Get all adjacent vertices of source vertex src and # recursively explore all paths from src. i = 0 while i ! = len ( self .adj[src]): # Get adjacent vertex and weight of edge v = self .adj[src][i][ 0 ] w = self .adj[src][i][ 1 ] i + = 1 # If vertex v is already there in path, then # there is a cycle (we ignore this edge) if (path[v] = = True ): continue # If weight of is more than k, return true if (w > = k): return True # Else add this vertex to path path[v] = True # If this adjacent can provide a path longer # than k, return true. if ( self .pathMoreThanKUtil(v, k - w, path)): return True # Backtrack path[v] = False # If no adjacent could produce longer path, return # false return False # Utility function to an edge (u, v) of weight w def addEdge( self ,u, v, w): self .adj[u].append([v, w]) self .adj[v].append([u, w]) # Driver program to test methods of graph class if __name__ = = '__main__' : # create the graph given in above figure V = 9 g = Graph(V) # making above shown graph g.addEdge( 0 , 1 , 4 ) g.addEdge( 0 , 7 , 8 ) g.addEdge( 1 , 2 , 8 ) g.addEdge( 1 , 7 , 11 ) g.addEdge( 2 , 3 , 7 ) g.addEdge( 2 , 8 , 2 ) g.addEdge( 2 , 5 , 4 ) g.addEdge( 3 , 4 , 9 ) g.addEdge( 3 , 5 , 14 ) g.addEdge( 4 , 5 , 10 ) g.addEdge( 5 , 6 , 2 ) g.addEdge( 6 , 7 , 1 ) g.addEdge( 6 , 8 , 6 ) g.addEdge( 7 , 8 , 7 ) src = 0 k = 62 if g.pathMoreThanK(src, k): print ( "Yes" ) else : print ( "No" ) k = 60 if g.pathMoreThanK(src, k): print ( "Yes" ) else : print ( "No" ) |
C#
// C# Program to find if there is a simple path with // weight more than k using System; using System.Collections.Generic; class Program { // Driver program to test methods of graph class static void Main( string [] args) { // create the graph given in above figure int V = 9; Graph g = new Graph(V); // making above shown graph g.addEdge(0, 1, 4); g.addEdge(0, 7, 8); g.addEdge(1, 2, 8); g.addEdge(1, 7, 11); g.addEdge(2, 3, 7); g.addEdge(2, 8, 2); g.addEdge(2, 5, 4); g.addEdge(3, 4, 9); g.addEdge(3, 5, 14); g.addEdge(4, 5, 10); g.addEdge(5, 6, 2); g.addEdge(6, 7, 1); g.addEdge(6, 8, 6); g.addEdge(7, 8, 7); int src = 0; int k = 62; if (g.pathMoreThanK(src, k)) Console.WriteLine( "YES" ); else Console.WriteLine( "NO" ); k = 60; if (g.pathMoreThanK(src, k)) Console.WriteLine( "YES" ); else Console.WriteLine( "NO" ); } } class AdjListNode { public int v; public int weight; public AdjListNode( int _v, int _w) { v = _v; weight = _w; } public int getV() { return v; } public int getWeight() { return weight; } } // This class represents a dipathted graph using // adjacency list representation class Graph { public int V; // No. of vertices // In a weighted graph, we need to store vertex // and weight pair for every edge public List<List<AdjListNode> > adj = new List<List<AdjListNode> >(); // Allocates memory for adjacency list public Graph( int V) { this .V = V; for ( int i = 0; i < V; i++) adj.Add( new List<AdjListNode>()); } // Utility function to an edge (u, v) of weight w public void addEdge( int u, int v, int weight) { AdjListNode node1 = new AdjListNode(v, weight); adj[u].Add(node1); // Add v to u's list AdjListNode node2 = new AdjListNode(u, weight); adj[v].Add(node2); // Add u to v's list } // Prints shortest paths from src to all other vertices bool pathMoreThanKUtil( int src, int k, bool [] path) { // If k is 0 or negative, return true; if (k <= 0) return true ; // Get all adjacent vertices of source vertex src // and recursively explore all paths from src. List<AdjListNode> it = adj[src]; for ( int i = 0; i < adj[src].Count; i++) { AdjListNode vertex = adj[src][i]; // Get adjacent vertex and weight of edge int v = vertex.v; int w = vertex.weight; // If vertex v is already there in path, then // there is a cycle (we ignore this edge) if (path[v] == true ) continue ; // If weight of is more than k, return true if (w >= k) return true ; // Else add this vertex to path path[v] = true ; // If this adjacent can provide a path longer // than k, return true. if (pathMoreThanKUtil(v, k - w, path)) return true ; // Backtrack path[v] = false ; } // If no adjacent could produce longer path, return // false return false ; } // Returns true if graph has path more than k length public bool pathMoreThanK( int src, int k) { // Create a path array with nothing included // in path bool [] path = new bool [V]; for ( int i = 0; i < V; i++) path[i] = false ; // Add source vertex to path path[src] = true ; return pathMoreThanKUtil(src, k, path); } } // This code is contributed by Tapesh (tapeshdua420) |
Javascript
// JavaScript Program to find if there is a simple path with // weight more than k class AdjListNode { constructor(v, weight) { this .v = v; this .weight = weight; } } // This class represents a dipathted graph using // adjacency list representation class Graph { // Allocates memory for adjacency list constructor(V) { this .V = V; // No. of vertices // In a weighted graph, we need to store vertex // and weight pair for every edge this .adj = []; for ( var i = 0; i < V; i++) { this .adj[i] = []; } } // Utility function to an edge (u, v) of weight w addEdge(u, v, weight) { var node1 = new AdjListNode(v, weight); this .adj[u].push(node1); // Add v to u's list var node2 = new AdjListNode(u, weight); this .adj[v].push(node2); // Add u to v's list } // Prints shortest paths from src to all other vertices pathMoreThanKUtil(src, k, path) { // If k is 0 or negative, return true; if (k <= 0) return true ; // Get all adjacent vertices of source vertex src and // recursively explore all paths from src. var it = this .adj[src]; for ( var i = 0; i < this .adj[src].length; i++) { var vertex = this .adj[src][i]; // Get adjacent vertex and weight of edge var v = vertex.v; var w = vertex.weight; // If vertex v is already there in path, then // there is a cycle (we ignore this edge) if (path[v] == true ) continue ; // If weight of is more than k, return true if (w >= k) return true ; // Else add this vertex to path path[v] = true ; // If this adjacent can provide a path longer // than k, return true. if ( this .pathMoreThanKUtil(v, k - w, path)) return true ; // Backtrack path[v] = false ; } // If no adjacent could produce longer path, return // false return false ; } // Returns true if graph has path more than k length pathMoreThanK(src, k) { // Create a path array with nothing included // in path var path = new Array(V); for ( var i = 0; i < V; i++) { path[i] = false ; } // Add source vertex to path path[src] = true ; return this .pathMoreThanKUtil(src, k, path); } } // Driver program to test methods of graph class // create the graph given in above figure var V = 9; var g = new Graph(V); // making above shown graph g.addEdge(0, 1, 4); g.addEdge(0, 7, 8); g.addEdge(1, 2, 8); g.addEdge(1, 7, 11); g.addEdge(2, 3, 7); g.addEdge(2, 8, 2); g.addEdge(2, 5, 4); g.addEdge(3, 4, 9); g.addEdge(3, 5, 14); g.addEdge(4, 5, 10); g.addEdge(5, 6, 2); g.addEdge(6, 7, 1); g.addEdge(6, 8, 6); g.addEdge(7, 8, 7); var src = 0; var k = 62; if (g.pathMoreThanK(src, k)) { console.log( "YES" ); } else { console.log( "NO" ); } k = 60 if (g.pathMoreThanK(src, k)) { console.log( "YES" ); } else { console.log( "NO" ); } // This code is contributed by Tapesh(tapeshdua420) |
Output:
No Yes
Exercise:
Modify the above solution to find weight of longest path from a given source.
Time Complexity: The time complexity of this algorithm is O(V+E) where V is the number of vertices and E is the number of edges in the graph. This is because the algorithm performs a depth-first search through the graph, visiting each vertex and checking if there is a path with a weight greater than k.
Auxiliary Space: O(V)
Explanation:
From the source node, we one-by-one visit all the paths and check if the total weight is greater than k for each path. So, the worst case will be when the number of possible paths is maximum. This is the case when every node is connected to every other node.
Beginning from the source node we have n-1 adjacent nodes. The time needed for a path to connect any two nodes is 2. One for joining the source and the next adjacent vertex. One for breaking the connection between the source and the old adjacent vertex.
After selecting a node out of n-1 adjacent nodes, we are left with n-2 adjacent nodes (as the source node is already included in the path) and so on at every step of selecting a node our problem reduces by 1 node.
We can write this in the form of a recurrence relation as: F(n) = n*(2+F(n-1)) This expands to: 2n + 2n*(n-1) + 2n*(n-1)*(n-2) + ……. + 2n(n-1)(n-2)(n-3)…..1 As n times 2n(n-1)(n-2)(n-3)….1 is greater than the given expression so we can safely say time complexity is: n*2*n! Here in the question the first node is defined so time complexity becomes F(n-1) = 2(n-1)*(n-1)! = 2*n*(n-1)! – 2*1*(n-1)! = 2*n!-2*(n-1)! = O(n!)
This article is contributed by Shivam Gupta. The explanation for time complexity is contributed by Pranav Nambiar. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Please Login to comment...