Add solution #638

Author: JoshCrozier

README.md

@@ -480,6 +480,7 @@
 633|[Sum of Square Numbers](./0633-sum-of-square-numbers.js)|Medium|
 636|[Exclusive Time of Functions](./0636-exclusive-time-of-functions.js)|Medium|
 637|[Average of Levels in Binary Tree](./0637-average-of-levels-in-binary-tree.js)|Easy|
+638|[Shopping Offers](./0638-shopping-offers.js)|Medium|
 643|[Maximum Average Subarray I](./0643-maximum-average-subarray-i.js)|Easy|
 645|[Set Mismatch](./0645-set-mismatch.js)|Medium|
 648|[Replace Words](./0648-replace-words.js)|Medium|

solutions/0638-shopping-offers.js

@@ -0,0 +1,54 @@
+/**
+ * 638. Shopping Offers
+ * https://leetcode.com/problems/shopping-offers/
+ * Difficulty: Medium
+ *
+ * In LeetCode Store, there are n items to sell. Each item has a price. However, there are
+ * some special offers, and a special offer consists of one or more different kinds of items
+ * with a sale price.
+ *
+ * You are given an integer array price where price[i] is the price of the ith item, and an
+ * integer array needs where needs[i] is the number of pieces of the ith item you want to buy.
+ *
+ * You are also given an array special where special[i] is of size n + 1 where special[i][j]
+ * is the number of pieces of the jth item in the ith offer and special[i][n] (i.e., the last
+ * integer in the array) is the price of the ith offer.
+ *
+ * Return the lowest price you have to pay for exactly certain items as given, where you could
+ * make optimal use of the special offers. You are not allowed to buy more items than you want,
+ * even if that would lower the overall price. You could use any of the special offers as many
+ * times as you want.
+ */
+
+/**
+ * @param {number[]} price
+ * @param {number[][]} special
+ * @param {number[]} needs
+ * @return {number}
+ */
+var shoppingOffers = function(price, special, needs) {
+  const map = new Map();
+  return dp(needs);
+
+  function dp(input) {
+    const key = input.join(',');
+    if (map.has(key)) return map.get(key);
+    let minCost = input.reduce((sum, need, i) => sum + need * price[i], 0);
+    for (const offer of special) {
+      const nextNeeds = [...input];
+      let valid = true;
+      for (let i = 0; i < price.length; i++) {
+        if (nextNeeds[i] < offer[i]) {
+          valid = false;
+          break;
+        }
+        nextNeeds[i] -= offer[i];
+      }
+      if (valid) {
+        minCost = Math.min(minCost, offer[price.length] + dp(nextNeeds));
+      }
+    }
+    map.set(key, minCost);
+    return minCost;
+  }
+};