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 = 0Output:500Explanation:

Path 0 → 2:The path with maximum weight and at most 0 intermediate nodes is of weight 500.

Input:N = 3, Edges[][] = {{0, 1, 100}, {1, 2, 100}, {0, 2, 500}}, src = 0, dst = 2, K = 0Output:500Explanation:

Path 0 → 2:The path with maximum weight and at most 1 intermediate node 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.

- Pop the front element of the queue and store it in a variable, say
- Increase the value of
**lvl**by**1**.

- Store the size of the queue in a variable, say
- 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 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;` `}` |

## Python3

`# 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` |

**Output:**

500

**Time Complexity:** O(N + E)**Auxiliary Space:** O(N)