Given a directed weighted graph consisting of N vertices and an array Edges[][], with each row representing two vertices connected by an edge and the weight of that edge, the task is to find the path with the maximum sum of weights from a given source vertex src to a given destination vertex dst, made up of at most K intermediate vertices. If no such path exists, then print -1.
Examples:
Input: N = 3, Edges[][] = {{0, 1, 100}, {1, 2, 100}, {0, 2, 500}}, src = 0, dst = 2, K = 0
Output: 500
Explanation:
Path 0 → 2: The path with maximum weight and at most 0 intermediate nodes is of weight 500.
Approach: The given problem can be solved by using BFS(Breadth-First Search) Traversal. Follow the steps below to solve the problem:
- Initialize the variable, say ans, to store the maximum distance between the source and the destination node having at most K intermediates nodes.
- Initialize an adjacency list of the graph using the edges.
- Initialize an empty queue and push the source vertex into it. Initialize a variable, say lvl, to store the number of nodes present in between src and dst.
- While the queue is not empty and lvl is less than K + 2 perform the following steps:
- Store the size of the queue in a variable, say S.
- Iterate over the range [1, S] and perform the following steps:
- Pop the front element of the queue and store it in a variable, say T.
- If T is the dst vertex, then update the value of ans as the maximum of ans and the current distance T.second.
- Traverse through all the neighbors of the current popped node and check if the distance of its neighbor is greater than the current distance or not. If found to be true, then push it in the queue and update its distance.
- Increase the value of lvl by 1.
- After completing the above steps, print the value of ans as the resultant maximum distance.
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 longest distance// from source to destination with at// most K intermediate nodesint findShortestPath( int n, vector<vector<int> >& edges, int src, int dst, int K){ // Initialize the adjacency list vector<vector<pair<int, int> > > adjlist( n, vector<pair<int, int> >()); // Initialize a queue to perform BFS queue<pair<int, int> > q; unordered_map<int, int> mp; // Store the maximum distance of // every node from source vertex int ans = INT_MIN; // Initialize adjacency list for (int i = 0; i < edges.size(); i++) { auto edge = edges[i]; adjlist[edge[0]].push_back( make_pair(edge[1], edge[2])); } // Push the first element into queue q.push({ src, 0 }); int level = 0; // Iterate until the queue becomes empty // and the number of nodes between src // and dst vertex is at most to K while (!q.empty() && level < K + 2) { // Current size of the queue int sz = q.size(); for (int i = 0; i < sz; i++) { // Extract the front // element of the queue auto pr = q.front(); // Pop the front element // of the queue q.pop(); // If the dst vertex is reached if (pr.first == dst) ans = max(ans, pr.second); // Traverse the adjacent nodes for (auto pr2 : adjlist[pr.first]) { // If the distance is greater // than the current distance if (mp.find(pr2.first) == mp.end() || mp[pr2.first] > pr.second + pr2.second) { // Push it into the queue q.push({ pr2.first, pr.second + pr2.second }); mp[pr2.first] = pr.second + pr2.second; } } } // Increment the level by 1 level++; } // Finally, return the maximum distance return ans != INT_MIN ? ans : -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 } }; cout << findShortestPath(n, edges, src, dst, k); return 0;} |
Java
import java.util.*;class Main { // Function to find the longest distance // from source to destination with at // most K intermediate nodes public static int findShortestPath(int n, int[][] edges, int src, int dst, int K) { // Initialize the adjacency list List<List<int[]> > adjlist = new ArrayList<>(); for (int i = 0; i < n; i++) { adjlist.add(new ArrayList<int[]>()); } // Initialize a queue to perform BFS Queue<int[]> q = new LinkedList<>(); Map<Integer, Integer> mp = new HashMap<>(); // Store the maximum distance of // every node from source vertex int ans = Integer.MIN_VALUE; // Initialize adjacency list for (int[] edge : edges) { adjlist.get(edge[0]).add( new int[] { edge[1], edge[2] }); } // Push the first element into queue q.add(new int[] { src, 0 }); int level = 0; // Iterate until the queue becomes empty // and the number of nodes between src // and dst vertex is at most to K while (!q.isEmpty() && level < K + 2) { // Current size of the queue int sz = q.size(); for (int i = 0; i < sz; i++) { // Extract the front // element of the queue int[] pr = q.poll(); // If the dst vertex is reached if (pr[0] == dst) ans = Math.max(ans, pr[1]); // Traverse the adjacent nodes for (int[] pr2 : adjlist.get(pr[0])) { // If the distance is greater // than the current distance if (!mp.containsKey(pr2[0]) || mp.get(pr2[0]) > pr[1] + pr2[1]) { // Push it into the queue q.add(new int[] { pr2[0], pr[1] + pr2[1] }); mp.put(pr2[0], pr[1] + pr2[1]); } } } // Increment the level by 1 level++; } // Finally, return the maximum distance return ans != Integer.MIN_VALUE ? ans : -1; } // Driver Code public static void main(String[] args) { int n = 3, src = 0, dst = 2, k = 1; int[][] edges = { { 0, 1, 100 }, { 1, 2, 100 }, { 0, 2, 500 } }; System.out.println( findShortestPath(n, edges, src, dst, k)); }} |
Python3
# Python3 program for the above approachfrom collections import deque# Function to find the longest distance# from source to destination with at# most K intermediate nodesdef findShortestPath(n, edges, src, dst, K): # Initialize the adjacency list adjlist = [[] for i in range(n)] # Initialize a queue to perform BFS q = deque() mp = {} # Store the maximum distance of # every node from source vertex ans = -10**9 # Initialize adjacency list for i in range(len(edges)): edge = edges[i] adjlist[edge[0]].append([edge[1], edge[2]]) # Push the first element into queue q.append([src, 0]) level = 0 # Iterate until the queue becomes empty # and the number of nodes between src # and dst vertex is at most to K while (len(q) > 0 and level < K + 2): # Current size of the queue sz = len(q) for i in range(sz): # Extract the front # element of the queue pr = q.popleft() # Pop the front element # of the queue # q.pop() # If the dst vertex is reached if (pr[0] == dst): ans = max(ans, pr[1]) # Traverse the adjacent nodes for pr2 in adjlist[pr[0]]: # If the distance is greater # than the current distance if ((pr2[0] not in mp) or mp[pr2[0]] > pr[1] + pr2[1]): # Push it into the queue q.append([pr2[0], pr[1] + pr2[1]]) mp[pr2[0]] = pr[1] + pr2[1] # Increment the level by 1 level += 1 # Finally, return the maximum distance return ans if ans != -10**9 else -1# Driver Codeif __name__ == '__main__': n, src, dst, k = 3, 0, 2, 1 edges= [ [ 0, 1, 100 ], [ 1, 2, 100 ], [ 0, 2, 500 ] ] print(findShortestPath(n, edges,src, dst, k))# This code is contributed by mohit kumar 29 |
Javascript
// JavaScript implementation of the above C++ codefunction findShortestPath(n, edges, src, dst, k) { // Initialize the adjacency list var adjlist = new Array(n).fill(null).map(() => []); // Initialize a queue to perform BFS var q = []; var mp = new Map(); // Store the maximum distance of // every node from source vertex var ans = Number.MIN_SAFE_INTEGER; // Initialize adjacency list for (var i = 0; i < edges.length; i++) { var edge = edges[i]; adjlist[edge[0]].push([edge[1], edge[2]]); } // Push the first element into queue q.push([src, 0]); var level = 0; // Iterate until the queue becomes empty // and the number of nodes between src // and dst vertex is at most to K while (q.length > 0 && level < k + 2) { // Current size of the queue var sz = q.length; for (var i = 0; i < sz; i++) { // Extract the front // element of the queue var pr = q.shift(); // If the dst vertex is reached if (pr[0] === dst) { ans = Math.max(ans, pr[1]); } // Traverse the adjacent nodes for (var j = 0; j < adjlist[pr[0]].length; j++) { var pr2 = adjlist[pr[0]][j]; // If the distance is greater // than the current distance if (mp.get(pr2[0]) === undefined || mp.get(pr2[0]) > pr[1] + pr2[1]) { // Push it into the queue q.push([pr2[0], pr[1] + pr2[1]]); mp.set(pr2[0], pr[1] + pr2[1]); } } } // Increment the level by 1 level++; } // Finally, return the maximum distance return ans !== Number.MIN_SAFE_INTEGER ? ans : -1;}// Example usagevar n = 3, src = 0, dst = 2, k = 1;var edges = [[0, 1, 100], [1, 2, 100], [0, 2, 500]];console.log(findShortestPath(n, edges, src, dst, k)); |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG { // Function to find the longest distance // from source to destination with at // most K intermediate nodes static int FindShortestPath(int n, int[][] edges, int src, int dst, int K) { // Initialize the adjacency list List<int[]>[] adjlist = new List<int[]>[n]; for (int i = 0; i < n; i++) { adjlist[i] = new List<int[]>(); } // Initialize a queue to perform BFS Queue<int[]> q = new Queue<int[]>(); Dictionary<int, int> mp = new Dictionary<int, int>(); // Store the maximum distance of // every node from source vertex int ans = -1000000000; // Initialize adjacency list for (int i = 0; i < edges.Length; i++) { int[] edge = edges[i]; adjlist[edge[0]].Add(new int[] {edge[1], edge[2]}); } // Push the first element into queue q.Enqueue(new int[] {src, 0}); int level = 0; // Iterate until the queue becomes empty // and the number of nodes between src // and dst vertex is at most to K while (q.Count > 0 && level < K + 2) { // Current size of the queue int sz = q.Count; for (int i = 0; i < sz; i++) { // Extract the front // element of the queue int[] pr = q.Dequeue(); // If the dst vertex is reached if (pr[0] == dst) { ans = Math.Max(ans, pr[1]); } // Traverse the adjacent nodes foreach (int[] pr2 in adjlist[pr[0]]) { // If the distance is greater // than the current distance if (!mp.ContainsKey(pr2[0]) || mp[pr2[0]] > pr[1] + pr2[1]) { // Push it into the queue q.Enqueue(new int[] {pr2[0], pr[1] + pr2[1]}); mp[pr2[0]] = pr[1] + pr2[1]; } } } // Increment the level by 1 level++; } // Finally, return the maximum distance return ans != -1000000000 ? ans : -1; } // Driver Code public static void Main() { int n = 3, src = 0, dst = 2, k = 1; int[][] edges = new int[][] { new int[] {0, 1, 100}, new int[] {1, 2, 100}, new int[] {0, 2, 500} }; Console.WriteLine(FindShortestPath(n, edges, src, dst, k)); }}// This code is contributed by codebraxnzt |
Output
500
Time Complexity: O(N + E)
Auxiliary Space: O(N)
Alternate approach: Modification of Bellman Ford algorithm after modifying the weights
If all the weights of the given graph are made negative of the original weights, the path taken to minimize the sum of weights with at most k nodes in middle will give us the path we need. Hence this question is similar to this problem. Below is the code implementation of the problem
C++
#include <bits/stdc++.h>using namespace std;int max_cost(int n, vector<vector<int> >& edges, int src, int dst, int k){ // We use 2 arrays for this algorithm // temp is the shortest distances array in current pass vector<int> temp(n, INT_MAX); temp[src] = 0; for (int i = 0; i <= k; i++) { // c is the shortest distances array in previous // pass For every iteration current pass becomes the // previous vector<int> c(temp); for (auto edge : edges) { int a = edge[0], b = edge[1], d = edge[2]; // Updating the current array using previous // array Subtracting d is same as adding -d temp[b] = min(temp[b], c[a] == INT_MAX ? INT_MAX : c[a] - d); } } // Checking is dst is reachable from src or not if (temp[dst] != INT_MAX) { // Returning the negative value of the shortest // distance to return the longest distance return -temp[dst]; } return -1;}int main(){ vector<vector<int> > edges = { { 0, 1, 100 }, { 1, 2, 100 }, { 0, 2, 500 }, }; int src = 0; int dst = 2; int k = 1; int n = 3; cout << max_cost(n, edges, src, dst, k) << endl; return 0;}// This code was contributed Prajwal Kandekar |
Java
import java.util.*;public class Main { static int max_cost(int n, List<List<Integer>> edges, int src, int dst, int k) { // We use 2 arrays for this algorithm // temp is the shortest distances array in current pass int[] temp = new int[n]; Arrays.fill(temp, Integer.MAX_VALUE); temp[src] = 0; for (int i = 0; i <= k; i++) { // c is the shortest distances array in previous // pass For every iteration current pass becomes the // previous int[] c = temp.clone(); for (List<Integer> edge : edges) { int a = edge.get(0), b = edge.get(1), d = edge.get(2); // Updating the current array using previous // array Subtracting d is same as adding -d temp[b] = Math.min(temp[b], c[a] == Integer.MAX_VALUE ? Integer.MAX_VALUE : c[a] - d); } } // Checking if dst is reachable from src or not if (temp[dst] != Integer.MAX_VALUE) { // Returning the negative value of the shortest // distance to return the longest distance return -temp[dst]; } return -1; } public static void main(String[] args) { List<List<Integer>> edges = Arrays.asList( Arrays.asList(0, 1, 100), Arrays.asList(1, 2, 100), Arrays.asList(0, 2, 500) ); int src = 0; int dst = 2; int k = 1; int n = 3; System.out.println(max_cost(n, edges, src, dst, k)); }} |
Python3
def max_cost(n, edges, src, dst, k): # We use 2 arrays for this algorithm # temp is the shortest distances array in current pass temp=[0 if i==src else float("inf") for i in range(n)] for _ in range(k+1): # c is the shortest distances array in previous pass # For every iteration current pass becomes the previous c=temp.copy() for a,b,d in edges: # Updating the current array using previous array # Subtracting d is same as adding -d temp[b]=min(temp[b],c[a]-d) # Checking is dst is reachable from src or not if temp[dst]!=float("inf"): # Returning the negative value of the shortest distance to return the longest distance return -temp[dst] return -1 edges = [ [0, 1, 100], [1, 2, 100], [0, 2, 500],]src = 0dst = 2k = 1n = 3 print(max_cost(n,edges,src,dst,k))# This code was contributed by Akshayan Muralikrishnan |
Javascript
function max_cost(n, edges, src, dst, k) { // We use 2 arrays for this algorithm // temp is the shortest distances array in current pass let temp = new Array(n).fill(Number.MAX_SAFE_INTEGER); temp[src] = 0; for (let i = 0; i <= k; i++) { // c is the shortest distances array in previous // pass For every iteration current pass becomes the // previous let c = [...temp]; for (let j = 0; j < edges.length; j++) { let [a, b, d] = edges[j]; // Updating the current array using previous // array Subtracting d is same as adding -d temp[b] = Math.min(temp[b], (c[a] === Number.MAX_SAFE_INTEGER) ? Number.MAX_SAFE_INTEGER : c[a] - d); } } // Checking is dst is reachable from src or not if (temp[dst] !== Number.MAX_SAFE_INTEGER) { // Returning the negative value of the shortest // distance to return the longest distance return -temp[dst]; } return -1;}let edges = [ [0, 1, 100], [1, 2, 100], [0, 2, 500]];let src = 0;let dst = 2;let k = 1;let n = 3;console.log(max_cost(n, edges, src, dst, k)); |
C#
using System;using System.Collections.Generic;using System.Linq;public class MainClass{ static int MaxCost(int n, List<List<int>> edges, int src, int dst, int k) { // We use 2 arrays for this algorithm // temp is the shortest distances array in current pass int[] temp = new int[n]; Array.Fill(temp, int.MaxValue); temp[src] = 0; for (int i = 0; i <= k; i++) { // c is the shortest distances array in previous // pass For every iteration current pass becomes the // previous int[] c = (int[])temp.Clone(); foreach (var edge in edges) { int a = edge[0], b = edge[1], d = edge[2]; // Updating the current array using previous // array Subtracting d is same as adding -d temp[b] = Math.Min(temp[b], c[a] == int.MaxValue ? int.MaxValue : c[a] - d); } } // Checking if dst is reachable from src or not if (temp[dst] != int.MaxValue) { // Returning the negative value of the shortest // distance to return the longest distance return -temp[dst]; } return -1; } public static void Main() { List<List<int>> edges = new List<List<int>>() { new List<int>(){0, 1, 100}, new List<int>(){1, 2, 100}, new List<int>(){0, 2, 500} }; int src = 0; int dst = 2; int k = 1; int n = 3; Console.WriteLine(MaxCost(n, edges, src, dst, k)); }} |
Output
500
Time Complexity: O(E*k) where E is the number of edges
Auxiliary Space: O(n)
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