Add solution #3494

Author: JoshCrozier

README.md

@@ -2760,6 +2760,7 @@
 3484|[Design Spreadsheet](./solutions/3484-design-spreadsheet.js)|Medium|
 3487|[Maximum Unique Subarray Sum After Deletion](./solutions/3487-maximum-unique-subarray-sum-after-deletion.js)|Easy|
 3491|[Phone Number Prefix](./solutions/3491-phone-number-prefix.js)|Easy|
+3494|[Find the Minimum Amount of Time to Brew Potions](./solutions/3494-find-the-minimum-amount-of-time-to-brew-potions.js)|Medium|
 3495|[Minimum Operations to Make Array Elements Zero](./solutions/3495-minimum-operations-to-make-array-elements-zero.js)|Hard|
 3496|[Maximize Score After Pair Deletions](./solutions/3496-maximize-score-after-pair-deletions.js)|Medium|
 3506|[Find Time Required to Eliminate Bacterial Strains](./solutions/3506-find-time-required-to-eliminate-bacterial-strains.js)|Hard|

solutions/3494-find-the-minimum-amount-of-time-to-brew-potions.js

@@ -0,0 +1,37 @@
+/**
+ * 3494. Find the Minimum Amount of Time to Brew Potions
+ * https://leetcode.com/problems/find-the-minimum-amount-of-time-to-brew-potions/
+ * Difficulty: Medium
+ *
+ * You are given two integer arrays, skill and mana, of length n and m, respectively.
+ *
+ * In a laboratory, n wizards must brew m potions in order. Each potion has a mana capacity
+ * mana[j] and must pass through all the wizards sequentially to be brewed properly. The
+ * time taken by the ith wizard on the jth potion is timeij = skill[i] * mana[j].
+ *
+ * Since the brewing process is delicate, a potion must be passed to the next wizard
+ * immediately after the current wizard completes their work. This means the timing must
+ * be synchronized so that each wizard begins working on a potion exactly when it arrives.
+ *
+ * Return the minimum amount of time required for the potions to be brewed properly.
+ */
+
+/**
+ * @param {number[]} skill
+ * @param {number[]} mana
+ * @return {number}
+ */
+var minTime = function(skill, mana) {
+  const result = new Array(skill.length + 1).fill(0);
+
+  for (let j = 0; j < mana.length; j++) {
+    for (let i = 0; i < skill.length; i++) {
+      result[i + 1] = Math.max(result[i + 1], result[i]) + mana[j] * skill[i];
+    }
+    for (let i = skill.length - 1; i > 0; i--) {
+      result[i] = result[i + 1] - mana[j] * skill[i];
+    }
+  }
+
+  return result[skill.length];
+};