diff --git a/Computational_Algorithms/Graphs/dijkstrasApplication.cpp b/Computational_Algorithms/Graphs/dijkstrasApplication.cpp
deleted file mode 100644
index b9fbe7b..0000000
--- a/Computational_Algorithms/Graphs/dijkstrasApplication.cpp
+++ /dev/null
@@ -1,92 +0,0 @@
-/*
-Dijkstra is a very interesting algorithm that can give us the shortest path reach a certain position.
-Here, we will be considering an example which will be indicating the implemention of Dijkstra's algorithm.
-The example is as stated:
-**Cheapest Flights Within K Stops
-We have n cities and m edges which are connected by some number of flights. 
-1. We have flights array which contains element in the form [fromi, toi, price[i]] i.e we have the subarray in which we have the
-information that we have to go from a given city to another city and the price that is costed by moving from a given city to another city.
-2. We also have three integers src, dst, and k, return the cheapest price from src to dst with at most k stops. 
-3. If we don't find such route, then we can return -1 which signify that we don't have any route that can satisfy all our conditions.
-*/
-#include <bits/stdc++.h>
-using namespace std;
-
-class Solution {
-  public:
-    int CheapestFLight(int n, vector<vector<int>>& flights, int src, int dst, int K)  {
-        /* we are given the flights so to solve it using Dijkstra's algorithm we have to make an adjacency list to do the further
-         procedure.*/
-        vector<pair<int, int>> adj[n];
-        for (auto it : flights)
-        {
-            adj[it[0]].push_back({it[1], it[2]});
-        }
-
-        /* Now we will create a queue which stores the node and their distances from the 
-        source in the form of {stops, {node, dist}} with ‘stops’ indicating 
-        the number of nodes between src and current node.*/
-        queue<pair<int, pair<int, int>>> q;
-        
-        q.push({0, {src, 0}});
-
-        // Here we are storing the updated distance from our calculations in the code given below.
-        vector<int> dist(n, 1e9);
-        dist[src] = 0;
-
-        // We first iterate through the graph using a queue like in Dijkstra with popping out the element with min stops first.
-        while (!q.empty())
-        {
-            auto it = q.front();
-            q.pop();
-            int stops = it.first;
-            int node = it.second.first;
-            int cost = it.second.second;
-
-            // We stop the process when we the min number of steps we are given.
-            if (stops > K)
-                continue;
-            for (auto iter : adj[node])
-            {
-                int adjNode = iter.first;
-                int edW = iter.second;
-
-                // We only update the queue if the new calculated dist is
-                //less than the prev and the stops are also within limits.
-                if (cost + edW < dist[adjNode] && stops <= K)
-                {
-                    dist[adjNode] = cost + edW;
-                    q.push({stops + 1, {adjNode, cost + edW}});
-                }
-            }
-        }
-       // if after our calculations we still have dist[dist] = 1e9 then we return -1 it signify we can't reach that destination with our
-       // given conditions
-        if (dist[dst] == 1e9)
-            return -1;
-        // otherwise we return the dist[dst] that is the min cost we have to pay to reach from one city to another with min k stops
-        return dist[dst];
-    }
-};
-
-
-int main() {
-        int n; cin>>n;
-        int edge; cin>>edge;
-        vector<vector<int>> flights;
-        
-        for(int i=0; i<edge; ++i){
-            vector<int> temp;
-            for(int j=0; j<3; ++j){
-                int x; cin>>x;
-                temp.push_back(x);
-            }
-            flights.push_back(temp);
-        }
-        
-        int src,dst,k;
-        cin>>src>>dst>>k;
-        Solution obj;
-        cout<<obj.CheapestFLight(n,flights,src,dst,k)<<"\n";
-    return 0;
-}