Add solution #2542

Author: JoshCrozier

README.md

@@ -541,6 +541,7 @@
 2493|[Divide Nodes Into the Maximum Number of Groups](./2493-divide-nodes-into-the-maximum-number-of-groups.js)|Hard|
 2529|[Maximum Count of Positive Integer and Negative Integer](./2529-maximum-count-of-positive-integer-and-negative-integer.js)|Easy|
 2535|[Difference Between Element Sum and Digit Sum of an Array](./2535-difference-between-element-sum-and-digit-sum-of-an-array.js)|Easy|
+2542|[Maximum Subsequence Score](./2542-maximum-subsequence-score.js)|Medium|
 2618|[Check if Object Instance of Class](./2618-check-if-object-instance-of-class.js)|Medium|
 2619|[Array Prototype Last](./2619-array-prototype-last.js)|Easy|
 2620|[Counter](./2620-counter.js)|Easy|

solutions/2542-maximum-subsequence-score.js

@@ -0,0 +1,44 @@
+/**
+ * 2542. Maximum Subsequence Score
+ * https://leetcode.com/problems/maximum-subsequence-score/
+ * Difficulty: Medium
+ *
+ * You are given two 0-indexed integer arrays nums1 and nums2 of equal length n and a positive
+ * integer k. You must choose a subsequence of indices from nums1 of length k.
+ *
+ * For chosen indices i0, i1, ..., ik - 1, your score is defined as:
+ * - The sum of the selected elements from nums1 multiplied with the minimum of the selected
+ *   elements from nums2.
+ * - It can defined simply as: (nums1[i0] + nums1[i1] +...+ nums1[ik - 1]) * min(nums2[i0],
+ *   nums2[i1], ... ,nums2[ik - 1]).
+ *
+ * Return the maximum possible score.
+ *
+ * A subsequence of indices of an array is a set that can be derived from the set {0, 1, ..., n-1}
+ * by deleting some or no elements.
+ */
+
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @param {number} k
+ * @return {number}
+ */
+var maxScore = function(nums1, nums2, k) {
+  const zipped = nums1.map((num1, i) => [num1, nums2[i]]).sort((a, b) => b[1] - a[1]);
+  const heap = new MinPriorityQueue();
+  let result = 0;
+  let sum = 0;
+
+  for (const [num, min] of zipped) {
+    heap.enqueue(num);
+    sum += num;
+
+    if (heap.size() == k) {
+      result = Math.max(result, sum * min);
+      sum -= heap.dequeue().element;
+    }
+  }
+
+  return result;
+};