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++ 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 nodes
int 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 Code
int 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;
}
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 program for the above approach
from collections import deque
# Function to find the longest distance
# from source to destination with at
# most K intermediate nodes
def 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 Code
if __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
// C# program for the above approach
using 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
// JavaScript implementation of the above C++ code
function 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 usage
var 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));
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
#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
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));
}
}
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 = 0
dst = 2
k = 1
n = 3
print(max_cost(n,edges,src,dst,k))
# This code was contributed by Akshayan Muralikrishnan
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));
}
}
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));
Output
500
Time Complexity: O(E*k) where E is the number of edges
Auxiliary Space: O(n)