diff --git a/cpp/neetcode_150/12_advanced_graphs/cheapest_flights_within_k_stops.cpp b/cpp/neetcode_150/12_advanced_graphs/cheapest_flights_within_k_stops.cpp
index ab62525ff..a53938b54 100644
--- a/cpp/neetcode_150/12_advanced_graphs/cheapest_flights_within_k_stops.cpp
+++ b/cpp/neetcode_150/12_advanced_graphs/cheapest_flights_within_k_stops.cpp
@@ -2,77 +2,31 @@
     Given cities connected by flights [from,to,price], also given src, dst, & k:
     Return cheapest price from src to dst with at most k stops
 
-    Dijkstra's but modified, normal won't work b/c will discard heap nodes w/o finishing
-    Modify: need to re-consider a node if dist from source is shorter than what we recorded
-    But, if we encounter node already processed but # of stops from source is lesser,
-    Need to add it back to the heap to be considered again
-
-    Time: O(V^2 log V) -> V = number of cities
-    Space: O(V^2)
+    Minimized implementation of the Bellman-Ford algorithm.
+    Loop for #maximum-stops times and, for each flight, update the array of distances if 
+    there's a new path from the destination to the source having lower cost than the current.
+    Time: O(k*n)
+    Space: O(n)
 */
 
 class Solution {
 public:
     int findCheapestPrice(int n, vector<vector<int>>& flights, int src, int dst, int k) {
-        // build adjacency matrix
-        vector<vector<int>> adj(n, vector<int>(n));
-        for (int i = 0; i < flights.size(); i++) {
-            vector<int> flight = flights[i];
-            adj[flight[0]][flight[1]] = flight[2];
-        }
-        
-        // shortest distances
-        vector<int> distances(n, INT_MAX);
-        distances[src] = 0;
-        // shortest steps
-        vector<int> currStops(n, INT_MAX);
-        currStops[src] = 0;
-        
-        // priority queue -> (cost, node, stops)
-        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> pq;
-        pq.push({0, src, 0});
-        
-        while (!pq.empty()) {
-            int cost = pq.top()[0];
-            int node = pq.top()[1];
-            int stops = pq.top()[2];
-            pq.pop();
-            
-            // if destination is reached, return cost to get here
-            if (node == dst) {
-                return cost;
-            }
-            
-            // if no more steps left, continue
-            if (stops == k + 1) {
-                continue;
-            }
-            
-            // check & relax all neighboring edges
-            for (int neighbor = 0; neighbor < n; neighbor++) {
-                if (adj[node][neighbor] > 0) {
-                    int currCost = cost;
-                    int neighborDist = distances[neighbor];
-                    int neighborWeight = adj[node][neighbor];
-                    
-                    // check if better cost
-                    int currDist = currCost + neighborWeight;
-                    if (currDist < neighborDist || stops + 1 < currStops[neighbor]) {
-                        pq.push({currDist, neighbor, stops + 1});
-                        distances[neighbor] = currDist;
-                        currStops[neighbor] = stops;
-                    } else if (stops < currStops[neighbor]) {
-                        // check if better steps
-                        pq.push({currDist, neighbor, stops + 1});
-                    }
-                    currStops[neighbor] = stops;
-                }
+        vector<int> dist(n, INT_MAX);
+        dist[src] = 0;
+
+        for (int stops = 0; stops <= k; ++stops){
+            vector<int> tmp = dist;
+            for (auto& flight : flights){
+                int s = flight[0], d = flight[1], p = flight[2];
+                if (dist[s] == INT_MAX)
+                    continue;
+                if (tmp[d] > dist[s] + p)
+                    tmp[d] = dist[s] + p;                
             }
+            dist = tmp;
         }
-        
-        if (distances[dst] == INT_MAX) {
-            return -1;
-        }
-        return distances[dst];
+
+        return dist[dst] == INT_MAX ? -1 : dist[dst];
     }
 };