Create minimum-cost-to-reach-city-with-discounts.py

Author: kamyu104

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

@@ -0,0 +1,38 @@
+# Time:  O((|E| + |V|) * log|V|) = O(|E| * log|V|) by using binary heap,
+#        if we can further to use Fibonacci heap, it would be O(|E| + |V| * log|V|)
+# Space: O(|E| + |V|) = O(|E|)
+
+import collections
+import heapq
+
+
+class Solution(object):
+    def minimumCost(self, n, highways, discounts):
+        """
+        :type n: int
+        :type highways: List[List[int]]
+        :type discounts: int
+        :rtype: int
+        """
+        adj = [[] for _ in xrange(n)]
+        for u, v, w in highways:
+            adj[u].append((v, w))
+            adj[v].append((u, w))
+        src, dst = 0, n-1
+        best = collections.defaultdict(lambda: collections.defaultdict(lambda: float("inf")))
+        best[src][discounts] = 0
+        min_heap = [(0, src, discounts)]
+        while min_heap:
+            result, u, k = heapq.heappop(min_heap)
+            if best[u][k] < result:
+                continue
+            if u == dst:
+                return result
+            for v, w in adj[u]:
+                if result+w < best[v][k]:
+                    best[v][k] = result+w                    
+                    heapq.heappush(min_heap, (result+w, v, k))
+                if k > 0 and result+w//2 < best[v][k-1]:
+                    best[v][k-1] = result+w//2                   
+                    heapq.heappush(min_heap, (result+w//2, v, k-1))
+        return -1