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Author: hogan-tech

2093-minimum-cost-to-reach-city-with-discounts/2093-minimum-cost-to-reach-city-with-discounts.py

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+# time complexity: O((n*k + e)log(n*k))
+# space complexity: O(n*k + e)
+import heapq
+from typing import List
+
+
+class Solution:
+    def minimumCost(self, n: int, highways: List[List[int]], discounts: int) -> int:
+        graph = [[] for _ in range(n)]
+        for highway in highways:
+            u, v, toll = highway
+            graph[u].append((v, toll))
+            graph[v].append((u, toll))
+        pq = [(0, 0, 0)]
+        dist = [[float("inf")] * (discounts + 1) for _ in range(n)]
+        dist[0][0] = 0
+        visited = [[False] * (discounts + 1) for _ in range(n)]
+        while pq:
+            currentCost, city, discountsUsed = heapq.heappop(pq)
+            if visited[city][discountsUsed]:
+                continue
+            visited[city][discountsUsed] = True
+            for neighbor, toll in graph[city]:
+                if currentCost + toll < dist[neighbor][discountsUsed]:
+                    dist[neighbor][discountsUsed] = currentCost + toll
+                    heapq.heappush(
+                        pq, (dist[neighbor][discountsUsed], neighbor, discountsUsed))
+
+                if discountsUsed < discounts:
+                    newCostWithDiscount = currentCost + toll // 2
+                    if (newCostWithDiscount < dist[neighbor][discountsUsed + 1]):
+                        dist[neighbor][
+                            discountsUsed + 1
+                        ] = newCostWithDiscount
+                        heapq.heappush(
+                            pq, (newCostWithDiscount, neighbor, discountsUsed + 1))
+
+        minCost = min(dist[n - 1])
+        return -1 if minCost == float("inf") else minCost
+
+
+n = 5
+highways = [[0, 1, 4], [2, 1, 3], [1, 4, 11], [3, 2, 3], [3, 4, 2]]
+discounts = 1
+print(Solution().minimumCost(n, highways, discounts))