Given a Graph consisting of N vertices and M weighted edges and an array edges[][], with each row representing the two vertices connected by the edge and the weight of the edge, the task is to find the path with the least sum of weights from a given source vertex src to a given destination vertex dst, made up of K intermediate vertices. If no such path exists, then print -1.
Examples:
Input: N = 3, M = 3, src = 0, dst = 2, K = 1, edges[] = {{0, 1, 100}, {1, 2, 100}, {0, 2, 500}}
Output: 200
Explanation: The path 0 -> 1 -> 2 is the least weighted sum (= 100 + 100 = 200) path connecting src (= 0) and dst(= 2) with exactly K (= 1) intermediate vertex.Input: N = 3, M = 3, src = 0, dst = 2, K = 0, edges[] = { { 0, 1, 100 }, { 1, 2, 100 }, { 0, 2, 500 } }
Output: 500
Explanation: The direct edge 0 -> 2 with weight 500 is the required path.
Approach: The given problem can be solved using Priority Queue and perform BFS. Follow the steps below to solve this problem:
- Initialize a priority queue to store the tuples {cost to reach this vertex, vertex, number of stops}.
- Push {0, src, k+1} as the first starting point.
- Pop-out the top element of priority queue. If all stops are exhausted, then repeat this step.
- If the destination is reached, then print the cost to reach the current vertex.
- Otherwise, find the neighbor of this vertex which required the smallest cost to reach that vertex. Push it into the priority queue.
- Repeat from step 2.
- If no path is found after performing the above steps, print -1.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find the minimum cost path// from the source vertex to destination// vertex via K intermediate verticesint leastWeightedSumPath(int n, vector<vector<int> >& edges, int src, int dst, int K){ // Initialize the adjacency list unordered_map<int, vector<pair<int, int> > > graph; // Generate the adjacency list for (vector<int>& edge : edges) { graph[edge[0]].push_back( make_pair(edge[1], edge[2])); } // Initialize the minimum priority queue priority_queue<vector<int>, vector<vector<int> >, greater<vector<int> > > pq; // Stores the minimum cost to // travel between vertices via // K intermediate nodes vector<vector<int> > costs(n, vector<int>( K + 2, INT_MAX)); costs[src][K + 1] = 0; // Push the starting vertex, // cost to reach and the number // of remaining vertices pq.push({ 0, src, K + 1 }); while (!pq.empty()) { // Pop the top element // of the stack auto top = pq.top(); pq.pop(); // If destination is reached if (top[1] == dst) // Return the cost return top[0]; // If all stops are exhausted if (top[2] == 0) continue; // Find the neighbour with minimum cost for (auto neighbor : graph[top[1]]) { // Pruning if (costs[neighbor.first][top[2] - 1] < neighbor.second + top[0]) { continue; } // Update cost costs[neighbor.first][top[2] - 1] = neighbor.second + top[0]; // Update priority queue pq.push({ neighbor.second + top[0], neighbor.first, top[2] - 1 }); } } // If no path exists return -1;}// Driver Codeint main(){ int n = 3, src = 0, dst = 2, k = 1; vector<vector<int> > edges = { { 0, 1, 100 }, { 1, 2, 100 }, { 0, 2, 500 } }; // Function Call to find the path // from src to dist via k nodes // having least sum of weights cout << leastWeightedSumPath(n, edges, src, dst, k); return 0;} |
Java
// Java code to implement the approachimport java.util.ArrayList;import java.util.Collections;import java.util.List;import java.util.PriorityQueue;class GFG { // Function to find the minimum cost path // from the source vertex to destination // vertex via K intermediate vertices static int LeastWeightedSumPath(int n, List<List<Integer> > edges, int src, int dst, int K) { // Initialize the adjacency list var graph = new ArrayList<List<int[]> >( Collections.nCopies(n, null)); // Generate the adjacency list for (int i = 0; i < edges.size(); i++) { var edge = edges.get(i); if (graph.get(edge.get(0)) == null) { graph.set(edge.get(0), new ArrayList<int[]>()); } graph.get(edge.get(0)) .add( new int[] { edge.get(1), edge.get(2) }); } // Initialize the minimum priority queue var pq = new PriorityQueue<int[]>( (i1, i2) -> i1[0] - i2[0]); // Stores the minimum cost to // travel between vertices via // K intermediate nodes var costs = new int[n][K + 2]; for (int i = 0; i < n; i++) { for (int j = 0; j < K + 2; j++) { costs[i][j] = Integer.MAX_VALUE; } } costs[src][K + 1] = 0; // Push the starting vertex, // cost to reach and the number // of remaining vertices pq.offer(new int[] { 0, src, K + 1 }); while (!pq.isEmpty()) { // Pop the top element // of the queue var top = pq.poll(); // If destination is reached if (top[1] == dst) { // Return the cost return top[0]; } // If all stops are exhausted if (top[2] == 0) { continue; } // Find the neighbour with minimum cost if (graph.get(top[1]) != null) { for (var neighbor : graph.get(top[1])) { // Pruning if (costs[neighbor[0]][top[2] - 1] < neighbor[1] + top[0]) { continue; } // Update cost costs[neighbor[0]][top[2] - 1] = neighbor[1] + top[0]; // Update priority queue pq.offer(new int[] { neighbor[1] + top[0], neighbor[0], top[2] - 1 }); } } } // If no path exists return -1; } // Driver Code public static void main(String[] args) { int n = 3, src = 0, dst = 2, k = 1; var edges = new ArrayList<List<Integer> >(); edges.add(List.of(0, 1, 100)); edges.add(List.of(1, 2, 100)); edges.add(List.of(0, 2, 500)); // Function Call to find the path // from src to dist via k nodes // having least sum of weights System.out.println( LeastWeightedSumPath(n, edges, src, dst, k)); }}// This code is contributed by phasing17 |
Python3
# Python3 program for the above approach# Function to find the minimum cost path# from the source vertex to destination# vertex via K intermediate verticesdef leastWeightedSumPath(n, edges, src, dst, K): graph = [[] for i in range(3)] # Generate the adjacency list for edge in edges: graph[edge[0]].append([edge[1], edge[2]]) # Initialize the minimum priority queue pq = [] # Stores the minimum cost to # travel between vertices via # K intermediate nodes costs = [[10**9 for i in range(K + 2)] for i in range(n)] costs[src][K + 1] = 0 # Push the starting vertex, # cost to reach and the number # of remaining vertices pq.append([ 0, src, K + 1 ]) pq = sorted(pq)[::-1] while (len(pq) > 0): # Pop the top element # of the stack top = pq[-1] del pq[-1] # If destination is reached if (top[1] == dst): # Return the cost return top[0] # If all stops are exhausted if (top[2] == 0): continue # Find the neighbour with minimum cost for neighbor in graph[top[1]]: # Pruning if (costs[neighbor[0]][top[2] - 1] < neighbor[1] + top[0]): continue # Update cost costs[neighbor[0]][top[2] - 1] = neighbor[1]+ top[0] # Update priority queue pq.append([neighbor[1]+ top[0],neighbor[0], top[2] - 1]) pq = sorted(pq)[::-1] # If no path exists return -1# Driver Codeif __name__ == '__main__': n, src, dst, k = 3, 0, 2, 1 edges = [ [ 0, 1, 100 ], [ 1, 2, 100 ], [ 0, 2, 500 ] ] # Function Call to find the path # from src to dist via k nodes # having least sum of weights print (leastWeightedSumPath(n, edges, src, dst, k))# This code is contributed by mohit kumar 29. |
C#
using System;using System.Collections.Generic;class GFG { // Function to find the minimum cost path // from the source vertex to destination // vertex via K intermediate vertices static int LeastWeightedSumPath(int n, List<List<int>> edges, int src, int dst, int K) { // Initialize the adjacency list var graph = new Dictionary<int, List<Tuple<int, int>>>(); // Generate the adjacency list for (int i = 0; i < edges.Count; i++) { var edge = edges[i]; if (!graph.ContainsKey(edge[0])) { graph[edge[0]] = new List<Tuple<int, int>>(); } graph[edge[0]].Add(new Tuple<int, int>(edge[1], edge[2])); } // Initialize the minimum priority queue var pq = new SortedSet<Tuple<int, int, int>>( Comparer<Tuple<int, int, int>>.Create((i1, i2) => i1.Item1.CompareTo(i2.Item1))); // Stores the minimum cost to // travel between vertices via // K intermediate nodes var costs = new int[n, K + 2]; for (int i = 0; i < n; i++) { for (int j = 0; j < K + 2; j++) { costs[i, j] = int.MaxValue; } } costs[src, K + 1] = 0; // Push the starting vertex, // cost to reach and the number // of remaining vertices pq.Add(new Tuple<int, int, int>(0, src, K + 1)); while (pq.Count > 0) { // Pop the top element // of the stack var top = pq.Min; pq.Remove(top); // If destination is reached if (top.Item2 == dst) { // Return the cost return top.Item1; } // If all stops are exhausted if (top.Item3 == 0) { continue; } // Find the neighbour with minimum cost if (graph.ContainsKey(top.Item2)) { foreach (var neighbor in graph[top.Item2]) { // Pruning if (costs[neighbor.Item1, top.Item3 - 1] < neighbor.Item2 + top.Item1) { continue; } // Update cost costs[neighbor.Item1, top.Item3 - 1] = neighbor.Item2 + top.Item1; // Update priority queue pq.Add(new Tuple<int, int, int>(neighbor.Item2 + top.Item1, neighbor.Item1, top.Item3 - 1)); } } } // If no path exists return -1; } // Driver Code static void Main() { int n = 3, src = 0, dst = 2, k = 1; var edges = new List<List<int>>() { new List<int> { 0, 1, 100 }, new List<int> { 1, 2, 100 }, new List<int> { 0, 2, 500 } }; // Function Call to find the path // from src to dist via k nodes // having least sum of weights Console.WriteLine(LeastWeightedSumPath(n, edges, src, dst, k)); }} |
Javascript
<script>// Javascript program for the above approach// Function to find the minimum cost path// from the source vertex to destination// vertex via K intermediate verticesfunction leastWeightedSumPath(n, edges, src, dst, K){ // Initialize the adjacency list var graph = new Map(); // Generate the adjacency list for (var edge of edges) { if(!graph.has(edge[0])) graph.set(edge[0], new Array()) var tmp = graph.get(edge[0]); tmp.push([edge[1], edge[2]]); graph.set(edge[0],tmp); } // Initialize the minimum priority queue var pq = []; // Stores the minimum cost to // travel between vertices via // K intermediate nodes var costs = Array.from(Array(n+1), ()=>Array(K+2).fill(1000000000)); costs[src][K + 1] = 0; // Push the starting vertex, // cost to reach and the number // of remaining vertices pq.push([ 0, src, K + 1 ]); while (pq.length!=0) { // Pop the top element // of the stack var top = pq[pq.length-1]; pq.pop(); // If destination is reached if (top[1] == dst) // Return the cost return top[0]; // If all stops are exhausted if (top[2] == 0) continue; if(graph.has(top[1])) { // Find the neighbour with minimum cost for (var neighbor of graph.get(top[1])) { // Pruning if (costs[neighbor[0]][top[2] - 1] < neighbor[1] + top[0]) { continue; } // Update cost costs[neighbor[0]][top[2] - 1] = neighbor[1] + top[0]; // Update priority queue pq.push([neighbor[1] + top[0], neighbor[0], top[2] - 1 ]); } } pq.sort((a,b)=>{ if(a[0]==b[0]) return b[1]-a[1] return b[0]-a[0]; }); } // If no path exists return -1;}// Driver Codevar n = 3, src = 0, dst = 2, k = 1;var edges = [ [ 0, 1, 100 ], [ 1, 2, 100 ], [ 0, 2, 500 ] ];// Function Call to find the path// from src to dist via k nodes// having least sum of weightsdocument.write(leastWeightedSumPath(n, edges, src, dst, k));// This code is contributed by rrrtnx.</script> |
Output:
200
Time Complexity : O(N * log N)
Auxiliary Space: O(N)
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