Add solution #857

Author: JoshCrozier

README.md

@@ -664,6 +664,7 @@
 854|[K-Similar Strings](./0854-k-similar-strings.js)|Hard|
 855|[Exam Room](./0855-exam-room.js)|Medium|
 856|[Score of Parentheses](./0856-score-of-parentheses.js)|Medium|
+857|[Minimum Cost to Hire K Workers](./0857-minimum-cost-to-hire-k-workers.js)|Hard|
 867|[Transpose Matrix](./0867-transpose-matrix.js)|Easy|
 868|[Binary Gap](./0868-binary-gap.js)|Easy|
 872|[Leaf-Similar Trees](./0872-leaf-similar-trees.js)|Easy|

solutions/0857-minimum-cost-to-hire-k-workers.js

@@ -0,0 +1,48 @@
+/**
+ * 857. Minimum Cost to Hire K Workers
+ * https://leetcode.com/problems/minimum-cost-to-hire-k-workers/
+ * Difficulty: Hard
+ *
+ * There are n workers. You are given two integer arrays quality and wage where quality[i] is the
+ * quality of the ith worker and wage[i] is the minimum wage expectation for the ith worker.
+ *
+ * We want to hire exactly k workers to form a paid group. To hire a group of k workers, we must
+ * pay them according to the following rules:
+ * 1. Every worker in the paid group must be paid at least their minimum wage expectation.
+ * 2. In the group, each worker's pay must be directly proportional to their quality. This means
+ *    if a worker’s quality is double that of another worker in the group, then they must be paid
+ *    twice as much as the other worker.
+ *
+ * Given the integer k, return the least amount of money needed to form a paid group satisfying
+ * the above conditions. Answers within 10-5 of the actual answer will be accepted.
+ */
+
+/**
+ * @param {number[]} quality
+ * @param {number[]} wage
+ * @param {number} k
+ * @return {number}
+ */
+var mincostToHireWorkers = function(quality, wage, k) {
+  const workers = quality.map((q, i) => ({ ratio: wage[i] / q, quality: q }))
+    .sort((a, b) => a.ratio - b.ratio);
+
+  let result = Infinity;
+  let qualitySum = 0;
+  const maxHeap = new MaxPriorityQueue();
+
+  for (const worker of workers) {
+    maxHeap.enqueue(worker.quality);
+    qualitySum += worker.quality;
+
+    if (maxHeap.size() > k) {
+      qualitySum -= maxHeap.dequeue();
+    }
+
+    if (maxHeap.size() === k) {
+      result = Math.min(result, qualitySum * worker.ratio);
+    }
+  }
+
+  return result;
+};