Add solution #3075

Author: JoshCrozier

README.md

@@ -2158,6 +2158,7 @@
 3070|[Count Submatrices with Top-Left Element and Sum Less Than k](./solutions/3070-count-submatrices-with-top-left-element-and-sum-less-than-k.js)|Medium|
 3071|[Minimum Operations to Write the Letter Y on a Grid](./solutions/3071-minimum-operations-to-write-the-letter-y-on-a-grid.js)|Medium|
 3074|[Apple Redistribution into Boxes](./solutions/3074-apple-redistribution-into-boxes.js)|Easy|
+3075|[Maximize Happiness of Selected Children](./solutions/3075-maximize-happiness-of-selected-children.js)|Medium|
 3105|[Longest Strictly Increasing or Strictly Decreasing Subarray](./solutions/3105-longest-strictly-increasing-or-strictly-decreasing-subarray.js)|Easy|
 3108|[Minimum Cost Walk in Weighted Graph](./solutions/3108-minimum-cost-walk-in-weighted-graph.js)|Hard|
 3110|[Score of a String](./solutions/3110-score-of-a-string.js)|Easy|

solutions/3075-maximize-happiness-of-selected-children.js

@@ -0,0 +1,35 @@
+/**
+ * 3075. Maximize Happiness of Selected Children
+ * https://leetcode.com/problems/maximize-happiness-of-selected-children/
+ * Difficulty: Medium
+ *
+ * You are given an array happiness of length n, and a positive integer k.
+ *
+ * There are n children standing in a queue, where the ith child has happiness value
+ * happiness[i]. You want to select k children from these n children in k turns.
+ *
+ * In each turn, when you select a child, the happiness value of all the children that
+ * have not been selected till now decreases by 1. Note that the happiness value cannot
+ * become negative and gets decremented only if it is positive.
+ *
+ * Return the maximum sum of the happiness values of the selected children you can
+ * achieve by selecting k children.
+ */
+
+/**
+ * @param {number[]} happiness
+ * @param {number} k
+ * @return {number}
+ */
+var maximumHappinessSum = function(happiness, k) {
+  const sortedHappiness = happiness.sort((a, b) => b - a);
+  let result = 0;
+
+  for (let i = 0; i < k; i++) {
+    const currentHappiness = sortedHappiness[i] - i;
+    if (currentHappiness <= 0) break;
+    result += currentHappiness;
+  }
+
+  return result;
+};