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Author: hogan-tech

3562-maximum-profit-from-trading-stocks-with-discounts/3562-maximum-profit-from-trading-stocks-with-discounts.py

@@ -0,0 +1,85 @@
+# time complexity: O(n * b ^ 2)
+# space complexity: O(n * b)
+from typing import List
+
+
+class Solution:
+    def maxProfit(
+        self,
+        n: int,
+        present: List[int],
+        future: List[int],
+        hierarchy: List[List[int]],
+        budget: int,
+    ) -> int:
+        g = [[] for _ in range(n)]
+        for e in hierarchy:
+            g[e[0] - 1].append(e[1] - 1)
+
+        def dfs(u: int):
+            cost = present[u]
+            dCost = present[u] // 2
+
+            dp0 = [0] * (budget + 1)
+            dp1 = [0] * (budget + 1)
+
+            subProfit0 = [0] * (budget + 1)
+            subProfit1 = [0] * (budget + 1)
+            uSize = cost
+
+            for v in g[u]:
+                child_dp0, child_dp1, vSize = dfs(v)
+                uSize += vSize
+                for i in range(budget, -1, -1):
+                    for sub in range(min(vSize, i) + 1):
+                        if i - sub >= 0:
+                            subProfit0[i] = max(
+                                subProfit0[i],
+                                subProfit0[i - sub] + child_dp0[sub],
+                            )
+                            subProfit1[i] = max(
+                                subProfit1[i],
+                                subProfit1[i - sub] + child_dp1[sub],
+                            )
+
+            for i in range(budget + 1):
+                dp0[i] = subProfit0[i]
+                dp1[i] = subProfit0[i]
+                if i >= dCost:
+                    dp1[i] = max(
+                        subProfit0[i], subProfit1[i - dCost] +
+                        future[u] - dCost
+                    )
+                if i >= cost:
+                    dp0[i] = max(
+                        subProfit0[i], subProfit1[i - cost] + future[u] - cost
+                    )
+
+            return dp0, dp1, uSize
+
+        return dfs(0)[0][budget]
+
+
+n = 2
+present = [1, 2]
+future = [4, 3]
+hierarchy = [[1, 2]]
+budget = 3
+print(Solution().maxProfit(n, present, future, hierarchy, budget))
+n = 2
+present = [3, 4]
+future = [5, 8]
+hierarchy = [[1, 2]]
+budget = 4
+print(Solution().maxProfit(n, present, future, hierarchy, budget))
+n = 3
+present = [4, 6, 8]
+future = [7, 9, 11]
+hierarchy = [[1, 2], [1, 3]]
+budget = 10
+print(Solution().maxProfit(n, present, future, hierarchy, budget))
+n = 3
+present = [5, 2, 3]
+future = [8, 5, 6]
+hierarchy = [[1, 2], [2, 3]]
+budget = 7