Add solution #1723

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,501 LeetCode solutions in JavaScript
+# 1,502 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1328,6 +1328,7 @@
 1720|[Decode XORed Array](./solutions/1720-decode-xored-array.js)|Easy|
 1721|[Swapping Nodes in a Linked List](./solutions/1721-swapping-nodes-in-a-linked-list.js)|Medium|
 1722|[Minimize Hamming Distance After Swap Operations](./solutions/1722-minimize-hamming-distance-after-swap-operations.js)|Medium|
+1723|[Find Minimum Time to Finish All Jobs](./solutions/1723-find-minimum-time-to-finish-all-jobs.js)|Hard|
 1726|[Tuple with Same Product](./solutions/1726-tuple-with-same-product.js)|Medium|
 1732|[Find the Highest Altitude](./solutions/1732-find-the-highest-altitude.js)|Easy|
 1748|[Sum of Unique Elements](./solutions/1748-sum-of-unique-elements.js)|Easy|

solutions/1723-find-minimum-time-to-finish-all-jobs.js

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+/**
+ * 1723. Find Minimum Time to Finish All Jobs
+ * https://leetcode.com/problems/find-minimum-time-to-finish-all-jobs/
+ * Difficulty: Hard
+ *
+ * You are given an integer array jobs, where jobs[i] is the amount of time it takes to complete
+ * the ith job.
+ *
+ * There are k workers that you can assign jobs to. Each job should be assigned to exactly one
+ * worker. The working time of a worker is the sum of the time it takes to complete all jobs
+ * assigned to them. Your goal is to devise an optimal assignment such that the maximum working
+ * time of any worker is minimized.
+ *
+ * Return the minimum possible maximum working time of any assignment.
+ */
+
+/**
+ * @param {number[]} jobs
+ * @param {number} k
+ * @return {number}
+ */
+var minimumTimeRequired = function(jobs, k) {
+  const n = jobs.length;
+  const workerTimes = new Array(k).fill(0);
+  let result = Infinity;
+
+  jobs.sort((a, b) => b - a);
+  backtrack(0, 0);
+
+  return result;
+
+  function backtrack(jobIndex, maxTime) {
+    if (jobIndex === n) {
+      result = Math.min(result, maxTime);
+      return;
+    }
+
+    if (maxTime >= result) return;
+
+    for (let i = 0; i < k; i++) {
+      if (workerTimes[i] + jobs[jobIndex] >= result) continue;
+
+      workerTimes[i] += jobs[jobIndex];
+      backtrack(jobIndex + 1, Math.max(maxTime, workerTimes[i]));
+      workerTimes[i] -= jobs[jobIndex];
+
+      if (workerTimes[i] === 0) break;
+    }
+  }
+};