feat(algorithms, graphs, cheapest-flight): cheapest flight with k stops

DIRECTORY.md

@@ -137,6 +137,8 @@
       * [Test Alien Dictionary](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/graphs/alien_dictionary/test_alien_dictionary.py)
     * Cat And Mouse
       * [Test Cat And Mouse](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/graphs/cat_and_mouse/test_cat_and_mouse.py)
+    * Cheapest Flights With K Stops
+      * [Test Cheapest Flights With K Stops](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/graphs/cheapest_flights_with_k_stops/test_cheapest_flights_with_k_stops.py)
     * Course Schedule
       * [Test Course Schedule](https://github.com/BrianLusina/PythonSnips/blob/master/algorithms/graphs/course_schedule/test_course_schedule.py)
     * Evaluate Division

algorithms/graphs/cheapest_flights_with_k_stops/README.md

@@ -0,0 +1,114 @@
+# Cheapest Flights Within K Stops
+
+You are given n cities, numbered from 0 to n−1 connected by several flights. You are also given an array flights, where
+each flight is represented as `flights[i]=[fromi ,toi, pricei]` meaning there is a direct flight from city 
+`fromᵢ` to city `toᵢ` with a cost of `priceᵢ`.
+
+You are also given three integers:
+
+- src: The starting city.
+- dst: The destination city.
+- k: The maximum number of stops allowed on the route (i.e., intermediate cities between src and dst).
+
+Your task is to find the minimum possible cost to travel from src to dst using at most k stops (i.e., the route may
+contain up to k + 1 flights). If there is no valid route from src to dst that uses at most k stops, return −1.
+
+## Constraints
+
+- 2 <= n <= 100
+- 0 <= flights.length <= (n * (n - 1) / 2)
+- flights[i].length == 3
+- 0 <= `fromi`, `toi` < n
+- `fromi` != `toi`
+- 1 <= `pricei` <= 10^4
+- There will not be any multiple flights between two cities.
+- 0 <= src, dst, k < n
+- src != dst
+
+## Examples
+
+Example 1
+```text
+Input: n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1
+Output: 700
+Explanation:
+The graph is shown above.
+The optimal path with at most 1 stop from city 0 to 3 is marked in red and has cost 100 + 600 = 700.
+Note that the path through cities [0,1,2,3] is cheaper but is invalid because it uses 2 stops.
+```
+
+Example 2
+```text
+Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
+Output: 200
+Explanation:
+The graph is shown above.
+The optimal path with at most 1 stop from city 0 to 2 is marked in red and has cost 100 + 100 = 200.
+```
+
+Example 3
+```text
+Input: n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
+Output: 500
+Explanation:
+The graph is shown above.
+The optimal path with no stops from city 0 to 2 is marked in red and has cost 500.
+```
+
+## Topics
+
+- Dynamic Programming
+- Depth-First Search
+- Breadth-First Search
+- Graph Theory
+- Heap (Priority Queue)
+- Shortest Path
+
+## Solution
+
+The core intuition behind this solution is to treat the problem as a shortest path in a directed, weighted graph with a
+strict limit on the number of edges (flights), i.e., we are only allowed to use at most k stops, meaning at most k + 1
+flights (edges). Traditional algorithms, such as Standard Dijkstra with a single dist[node], doesn’t enforce the stop
+bound because they always choose the globally cheapest path so far, even if that path exceeds the allowed number of stops.
+To correctly enforce the stop limit, we employ a Bellman–Ford–style dynamic programming approach, which naturally handles
+constraints on the number of edges in a path.
+
+The idea is to repeatedly relax all flights, exactly k + 1 times, where iteration t represents allowing routes that use
+up to t flights (one more than the previous iteration). After each iteration, we have the cheapest costs using at most r
+flights (edges).
+
+To achieve this, we maintain two arrays: `prices`, which stores the best costs found using up to t − 1 flights, and 
+`temp_prices`, which stores the best costs for the current iteration t. During iteration t, every update to a path is
+made only from values in `prices`, not from updates made earlier in the same iteration. This separation ensures that any
+path discovered in iteration t uses at most one more flight than the paths from the previous iteration t − 1, preventing
+us from accidentally chaining multiple flights in the same round.
+
+The algorithm builds valid paths layer by layer, only extending shorter paths into longer ones, guaranteeing that when
+all k + 1 iterations are done, we have explored all possible routes that use at most k stops. And the cheapest such route
+will be stored in prices[dst].
+
+Using the intuition above, we implement the algorithm as follows:
+
+1. Create an array `prices` of length n and initialize all its entries to infinity.
+2. Set `prices[src] = 0` because the cost to reach the starting city from itself is 0 and requires no flights.
+3. Iterate at most `k + 1` times to model the flight limit:
+   - Initialize a new array, `temp_prices`, with a copy of the dist array. Copying ensures we can also keep the best
+     older answers (using fewer flights) instead of forcing exactly t flights.
+   - For each flight:
+     - If `prices[u]` is not equal to infinity, and the candidate cost: `prices[u] + w` is less than the current
+       `temp_prices[v]`:
+       - Set `temp_prices[v]` to `prices[u] + w`.
+   - After processing all flights in this iteration, set `prices` to `temp_prices`.
+4. After completing all `k + 1` iterations, check `prices[dst]`. If `prices[dst]` is still inf, it means there is no
+   valid route from `src` to `dst` that uses at most `k` stops, so we return -1. Otherwise, return `prices[dst]`.
+
+### Time Complexity
+
+The time complexity of this algorithm is O((k+1)×m) because we perform k+1 relaxation rounds, and in each round, we
+iterate over all m flights once. Asymptotically, this simplifies to O(k×m).
+
+### Space Complexity
+
+The space complexity of this algorithm is O(n) because we maintain two arrays, dist and new_dist, each of size n (the
+number of cities). These arrays are reused across iterations, and no other auxiliary data structure grows with the input
+size.

algorithms/graphs/cheapest_flights_with_k_stops/__init__.py

@@ -0,0 +1,21 @@
+from typing import List
+from math import inf
+
+
+def find_cheapest_price(
+    n: int, flights: List[List[int]], src: int, dst: int, k: int
+) -> int:
+    prices = [inf] * n
+    prices[src] = 0
+
+    for i in range(k + 1):
+        temp_prices = prices.copy()
+
+        for source, destination, price in flights:
+            if prices[source] == inf:
+                continue
+            if prices[source] + price < temp_prices[destination]:
+                temp_prices[destination] = prices[source] + price
+        prices = temp_prices
+
+    return -1 if prices[dst] == inf else prices[dst]

algorithms/graphs/cheapest_flights_with_k_stops/test_cheapest_flights_with_k_stops.py

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+import unittest
+from typing import List
+from parameterized import parameterized
+from algorithms.graphs.cheapest_flights_with_k_stops import find_cheapest_price
+
+CHEAPEST_FLIGHTS_WITH_K_STOPS_TEST_CASES = [
+    (3, [[0, 1, 100], [1, 2, 100], [0, 2, 500]], 0, 2, 1, 200),
+    (3, [[0, 1, 100], [1, 2, 400], [0, 2, 350]], 0, 2, 0, 350),
+    (4, [[0, 1, 100], [1, 2, 100], [2, 3, 100]], 0, 3, 1, -1),
+    (
+        4,
+        [[0, 1, 100], [1, 2, 100], [2, 0, 100], [1, 3, 600], [2, 3, 200]],
+        0,
+        3,
+        1,
+        700,
+    ),
+    (
+        5,
+        [
+            [0, 1, 100],
+            [1, 2, 100],
+            [2, 3, 100],
+            [3, 4, 100],
+            [0, 4, 1000],
+            [0, 2, 500],
+            [1, 3, 250],
+            [2, 4, 200],
+        ],
+        0,
+        4,
+        2,
+        400,
+    ),
+    (3, [[0, 1, 100], [1, 2, 100], [0, 2, 500]], 0, 2, 0, 500),
+]
+
+
+class CheapestFlightsWithKStopsTestCase(unittest.TestCase):
+    @parameterized.expand(CHEAPEST_FLIGHTS_WITH_K_STOPS_TEST_CASES)
+    def test_cheapest_flights_with_k_stops(
+        self,
+        n: int,
+        flights: List[List[int]],
+        src: int,
+        dst: int,
+        k: int,
+        expected: int,
+    ):
+        actual = find_cheapest_price(n, flights, src, dst, k)
+        self.assertEqual(expected, actual)
+
+
+if __name__ == "__main__":
+    unittest.main()