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Author: NinePiece2

Solutions/0803-cheapest-flights-within-k-stops/README.md

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+<p>There are <code>n</code> cities connected by some number of flights. You are given an array <code>flights</code> where <code>flights[i] = [from<sub>i</sub>, to<sub>i</sub>, price<sub>i</sub>]</code> indicates that there is a flight from city <code>from<sub>i</sub></code> to city <code>to<sub>i</sub></code> with cost <code>price<sub>i</sub></code>.</p>
+
+<p>You are also given three integers <code>src</code>, <code>dst</code>, and <code>k</code>, return <em><strong>the cheapest price</strong> from </em><code>src</code><em> to </em><code>dst</code><em> with at most </em><code>k</code><em> stops. </em>If there is no such route, return<em> </em><code>-1</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-3drawio.png" style="width: 332px; height: 392px;" />
+<pre>
+<strong>Input:</strong> n = 4, flights = [[0,1,100],[1,2,100],[2,0,100],[1,3,600],[2,3,200]], src = 0, dst = 3, k = 1
+<strong>Output:</strong> 700
+<strong>Explanation:</strong>
+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.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-1drawio.png" style="width: 332px; height: 242px;" />
+<pre>
+<strong>Input:</strong> n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 1
+<strong>Output:</strong> 200
+<strong>Explanation:</strong>
+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.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/03/18/cheapest-flights-within-k-stops-2drawio.png" style="width: 332px; height: 242px;" />
+<pre>
+<strong>Input:</strong> n = 3, flights = [[0,1,100],[1,2,100],[0,2,500]], src = 0, dst = 2, k = 0
+<strong>Output:</strong> 500
+<strong>Explanation:</strong>
+The graph is shown above.
+The optimal path with no stops from city 0 to 2 is marked in red and has cost 500.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n &lt;= 100</code></li>
+	<li><code>0 &lt;= flights.length &lt;= (n * (n - 1) / 2)</code></li>
+	<li><code>flights[i].length == 3</code></li>
+	<li><code>0 &lt;= from<sub>i</sub>, to<sub>i</sub> &lt; n</code></li>
+	<li><code>from<sub>i</sub> != to<sub>i</sub></code></li>
+	<li><code>1 &lt;= price<sub>i</sub> &lt;= 10<sup>4</sup></code></li>
+	<li>There will not be any multiple flights between two cities.</li>
+	<li><code>0 &lt;= src, dst, k &lt; n</code></li>
+	<li><code>src != dst</code></li>
+</ul>

Solutions/0803-cheapest-flights-within-k-stops/solution.py

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+class Solution:
+    def findCheapestPrice(self, n: int, flights: List[List[int]], src: int, dst: int, k: int) -> int:
+        # graph = defaultdict(list)
+        # for source, dest, price in flights:
+        #     graph[source].append((dest, price))
+
+        # @cache
+        # def dfs(source: int, k: int) -> int:
+        #     if source == dst:
+        #         return 0
+        #     if k <= 0:
+        #         return float('inf')
+
+        #     k -= 1
+        #     ans = float('inf')
+        #     for dest, price in graph[source]:
+        #         ans = min(ans, dfs(dest, k) + price)
+        #     return ans
+
+        # result = dfs(src, k + 1)
+
+        # return -1 if result == float('inf') else result
+        
+        # Bellman-Ford approach
+        dist = [float('inf')] * n
+        dist[src] = 0
+
+        for _ in range(k + 1):
+            changed = False
+            new_dists = dist[:]
+
+            for source, dest, price in flights:
+                if dist[source] != float('inf') and new_dists[dest] > dist[source] + price:
+                    new_dists[dest] = dist[source] + price
+                    changed = True
+
+            dist = new_dists
+            if not changed:
+                break
+        
+        return -1 if dist[dst] == float('inf') else dist[dst]