Add solution #2462

Author: JoshCrozier

README.md

@@ -539,6 +539,7 @@
 2425|[Bitwise XOR of All Pairings](./2425-bitwise-xor-of-all-pairings.js)|Medium|
 2427|[Number of Common Factors](./2427-number-of-common-factors.js)|Easy|
 2429|[Minimize XOR](./2429-minimize-xor.js)|Medium|
+2462|[Total Cost to Hire K Workers](./2462-total-cost-to-hire-k-workers.js)|Medium|
 2469|[Convert the Temperature](./2469-convert-the-temperature.js)|Easy|
 2482|[Difference Between Ones and Zeros in Row and Column](./2482-difference-between-ones-and-zeros-in-row-and-column.js)|Medium|
 2490|[Circular Sentence](./2490-circular-sentence.js)|Easy|

solutions/2462-total-cost-to-hire-k-workers.js

@@ -0,0 +1,62 @@
+/**
+ * 2462. Total Cost to Hire K Workers
+ * https://leetcode.com/problems/total-cost-to-hire-k-workers/
+ * Difficulty: Medium
+ *
+ * You are given a 0-indexed integer array costs where costs[i] is the cost of hiring
+ * the ith worker.
+ *
+ * You are also given two integers k and candidates. We want to hire exactly k workers
+ * according to the following rules:
+ * - You will run k sessions and hire exactly one worker in each session.
+ * - In each hiring session, choose the worker with the lowest cost from either the first
+ *   candidates workers or the last candidates workers. Break the tie by the smallest index.
+ *   - For example, if costs = [3,2,7,7,1,2] and candidates = 2, then in the first hiring
+ *     session, we will choose the 4th worker because they have the lowest cost [3,2,7,7,1,2].
+ *   - In the second hiring session, we will choose 1st worker because they have the same
+ *     lowest cost as 4th worker but they have the smallest index [3,2,7,7,2]. Please note
+ *     that the indexing may be changed in the process.
+ * - If there are fewer than candidates workers remaining, choose the worker with the lowest
+ *   cost among them. Break the tie by the smallest index.
+ * - A worker can only be chosen once.
+ *
+ * Return the total cost to hire exactly k workers.
+ */
+
+/**
+ * @param {number[]} costs
+ * @param {number} k
+ * @param {number} candidates
+ * @return {number}
+ */
+var totalCost = function(costs, k, candidates) {
+  const queue = new PriorityQueue({ compare: (a, b) => {
+    return a[0] === b[0] ? a[1] - b[1] : a[0] - b[0];
+  }});
+
+  costs.forEach((cost, index) => {
+    if (index < candidates || index >= costs.length - candidates) {
+      queue.enqueue([cost, index]);
+    }
+  });
+
+  let result = 0;
+  for (let i = 0, count = candidates, diff = costs.length - candidates - 1; i < k; i++) {
+    const worker = queue.dequeue();
+    result += worker[0];
+
+    if (count <= diff) {
+      let status = null;
+      if (worker[1] < count) {
+        status = [costs[count], count];
+        count++;
+      } else {
+        status = [costs[diff], diff];
+        diff--;
+      }
+      queue.enqueue(status);
+    }
+  }
+
+  return result;
+};