Given a directed weighted graph and two vertices S and D in it, the task is to find the shortest path from S to D with exactly K edges on the path. If no such path exists, print -1.
Examples:
Input: N = 3, K = 2, ed = {{{1, 2}, 5}, {{2, 3}, 3}, {{3, 1}, 4}}, S = 1, D = 3
Output: 8
Explanation: The shortest path with two edges will be 1->2->3Input: N = 3, K = 4, ed = {{{1, 2}, 5}, {{2, 3}, 3}, {{3, 1}, 4}}, S = 1, D = 3
Output: -1
Explanation: No path with edge length 4 exists from node 1 to 3Input: N = 3, K = 5, ed = {{{1, 2}, 5}, {{2, 3}, 3}, {{3, 1}, 4}}, S = 1, D = 3
Output: 20
Explanation: Shortest path will be 1->2->3->1->2->3.
Approach: An O(V^3*K) approach for this problem has already been discussed in the previous article. In this article, an O(E*K) approach is discussed for solving this problem.
The idea is to use dynamic-programming to solve this problem.
Let dp[X][J] be the shortest path from node S to node X using exactly J edges in total. Using this, dp[X][J+1] can be calculated as:
dp[X][J+1] = min(arr[Y][J]+weight[{Y, X}])
for all Y which has an edge from Y to X.
The result for the problem can be computed by following below steps:
- Initialise an array, dis[] with initial value as ‘inf’ except dis[S] as 0.
- For i equals 1 – K, run a loop
- Initialise an array, dis1[] with initial value as ‘inf’.
- For each edge in the graph,
dis1[edge.second] = min(dis1[edge.second], dis[edge.first]+weight(edge))
- If dist[d] in infinity, return -1, else return dist[d].
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>#define inf 100000000using namespace std;// Function to find the smallest path// with exactly K edgesdouble smPath(int s, int d, vector<pair<pair<int, int>, int> > ed, int n, int k){ // Array to store dp int dis[n + 1]; // Initialising the array for (int i = 0; i <= n; i++) dis[i] = inf; dis[s] = 0; // Loop to solve DP for (int i = 0; i < k; i++) { // Initialising next state int dis1[n + 1]; for (int j = 0; j <= n; j++) dis1[j] = inf; // Recurrence relation for (auto it : ed) dis1[it.first.second] = min(dis1[it.first.second], dis[it.first.first] + it.second); for (int i = 0; i <= n; i++) dis[i] = dis1[i]; } // Returning final answer if (dis[d] == inf) return -1; else return dis[d];}// Driver codeint main(){ int n = 4; vector<pair<pair<int, int>, int> > ed; // Input edges ed = { { { 0, 1 }, 10 }, { { 0, 2 }, 3 }, { { 0, 3 }, 2 }, { { 1, 3 }, 7 }, { { 2, 3 }, 7 } }; // Source and Destination int s = 0, d = 3; // Number of edges in path int k = 2; // Calling the function cout << smPath(s, d, ed, n, k);} |
Java
// Java implementation of the above approachimport java.util.ArrayList;import java.util.Arrays;class GFG{static class Pair<K, V>{ K first; V second; public Pair(K first, V second) { this.first = first; this.second = second; }}static int inf = 100000000;// Function to find the smallest path// with exactly K edgesstatic int smPath(int s, int d, ArrayList<Pair<Pair<Integer, Integer>, Integer>> ed, int n, int k) { // Array to store dp int[] dis = new int[n + 1]; // Initialising the array Arrays.fill(dis, inf); dis[s] = 0; // Loop to solve DP for(int i = 0; i < k; i++) { // Initialising next state int[] dis1 = new int[n + 1]; Arrays.fill(dis1, inf); // Recurrence relation for(Pair<Pair<Integer, Integer>, Integer> it : ed) dis1[it.first.second] = Math.min(dis1[it.first.second], dis[it.first.first] + it.second); for(int j = 0; j <= n; j++) dis[j] = dis1[j]; } // Returning final answer if (dis[d] == inf) return -1; else return dis[d];}// Driver codepublic static void main(String[] args) { int n = 4; // Input edges ArrayList<Pair<Pair<Integer, Integer>, Integer>> ed = new ArrayList<>( Arrays.asList( new Pair<Pair<Integer, Integer>, Integer>( new Pair<Integer, Integer>(0, 1), 10), new Pair<Pair<Integer, Integer>, Integer>( new Pair<Integer, Integer>(0, 2), 3), new Pair<Pair<Integer, Integer>, Integer>( new Pair<Integer, Integer>(0, 3), 2), new Pair<Pair<Integer, Integer>, Integer>( new Pair<Integer, Integer>(1, 3), 7), new Pair<Pair<Integer, Integer>, Integer>( new Pair<Integer, Integer>(2, 3), 7) ) ); // Source and Destination int s = 0, d = 3; // Number of edges in path int k = 2; // Calling the function System.out.println(smPath(s, d, ed, n, k));}}// This code is contributed by sanjeev2552 |
Python3
# Python3 implementation of the above approachinf = 100000000# Function to find the smallest path# with exactly K edgesdef smPath(s, d, ed, n, k): # Array to store dp dis = [inf] * (n + 1) dis[s] = 0 # Loop to solve DP for i in range(k): # Initialising next state dis1 = [inf] * (n + 1) # Recurrence relation for it in ed: dis1[it[1]] = min(dis1[it[1]], dis[it[0]]+ it[2]) for i in range(n + 1): dis[i] = dis1[i] # Returning final answer if (dis[d] == inf): return -1 else: return dis[d]# Driver codeif __name__ == '__main__': n = 4 # Input edges ed = [ [0, 1 ,10], [ 0, 2 ,3], [ 0, 3 ,2], [ 1, 3 ,7], [ 2, 3 ,7] ] # Source and Destination s = 0 d = 3 # Number of edges in path k = 2 # Calling the function print(smPath(s, d, ed, n, k))# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the above approachusing System;using System.Linq;using System.Collections.Generic;public class GFG{static int inf = 100000000;// Function to find the smallest path// with exactly K edgespublic static int smPath(int s, int d, List<KeyValuePair<KeyValuePair<int, int>, int>> ed, int n, int k) { // Array to store dp // Initialising the array int []dis = Enumerable.Repeat(inf, n+1).ToArray(); dis[s] = 0; // Loop to solve DP for(int i = 0; i < k; i++) { // Initialising next state int []dis1 = Enumerable.Repeat(inf, n+1).ToArray(); // Recurrence relation foreach(KeyValuePair<KeyValuePair<int, int>, int> it in ed) dis1[it.Key.Value] = Math.Min(dis1[it.Key.Value], dis[it.Key.Key] + it.Value); for(int j = 0; j <= n; j++) dis[j] = dis1[j]; } // Returning final answer if (dis[d] == inf) return -1; else return dis[d];}// Driver codepublic static void Main(string[] args) { int n = 4; // Input edges List<KeyValuePair<KeyValuePair<int, int>, int>> ed = new List<KeyValuePair<KeyValuePair<int, int>, int>>(){ new KeyValuePair<KeyValuePair<int, int>, int>( new KeyValuePair<int, int>(0, 1), 10), new KeyValuePair<KeyValuePair<int, int>, int>( new KeyValuePair<int, int>(0, 2), 3), new KeyValuePair<KeyValuePair<int, int>, int>( new KeyValuePair<int, int>(0, 3), 2), new KeyValuePair<KeyValuePair<int, int>, int>( new KeyValuePair<int, int>(1, 3), 7), new KeyValuePair<KeyValuePair<int, int>, int>( new KeyValuePair<int, int>(2, 3), 7) }; // Source and Destination int s = 0, d = 3; // Number of edges in path int k = 2; // Calling the function Console.Write(smPath(s, d, ed, n, k));}}// This code is contributed by rrrtnx. |
Javascript
<script>// Javascript implementation of the above approachlet inf = 100000000;// Function to find the smallest path// with exactly K edgesfunction smPath(s,d,ed,n,k){ // Array to store dp let dis = new Array(n + 1); // Initialising the array for(let i=0;i<(n+1);i++) { dis[i]=inf; } dis[s] = 0; // Loop to solve DP for(let i = 0; i < k; i++) { // Initialising next state let dis1 = new Array(n + 1); for(let i=0;i<(n+1);i++) { dis1[i]=inf; } // Recurrence relation for(let it=0;it< ed.length;it++) dis1[ed[it][1]] = Math.min(dis1[ed[it][1]], dis[ed[it][0]] + ed[it][2]); document.write() for(let j = 0; j <= n; j++) dis[j] = dis1[j]; } // Returning final answer if (dis[d] == inf) return -1; else return dis[d];}// Driver codelet n = 4; // Input edgeslet ed = [ [0, 1 ,10], [ 0, 2 ,3], [ 0, 3 ,2], [ 1, 3 ,7], [ 2, 3 ,7] ];// Source and Destinationlet s = 0, d = 3;// Number of edges in pathlet k = 2;// Calling the functiondocument.write(smPath(s, d, ed, n, k));// This code is contributed by patel2127</script> |
Output:
10
Time complexity: O(E*K)
Space complexity: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
