Add solution #2141

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,773 LeetCode solutions in JavaScript
+# 1,774 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -1641,6 +1641,7 @@
 2138|[Divide a String Into Groups of Size k](./solutions/2138-divide-a-string-into-groups-of-size-k.js)|Easy|
 2139|[Minimum Moves to Reach Target Score](./solutions/2139-minimum-moves-to-reach-target-score.js)|Medium|
 2140|[Solving Questions With Brainpower](./solutions/2140-solving-questions-with-brainpower.js)|Medium|
+2141|[Maximum Running Time of N Computers](./solutions/2141-maximum-running-time-of-n-computers.js)|Hard|
 2145|[Count the Hidden Sequences](./solutions/2145-count-the-hidden-sequences.js)|Medium|
 2154|[Keep Multiplying Found Values by Two](./solutions/2154-keep-multiplying-found-values-by-two.js)|Easy|
 2161|[Partition Array According to Given Pivot](./solutions/2161-partition-array-according-to-given-pivot.js)|Medium|

solutions/2141-maximum-running-time-of-n-computers.js

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+/**
+ * 2141. Maximum Running Time of N Computers
+ * https://leetcode.com/problems/maximum-running-time-of-n-computers/
+ * Difficulty: Hard
+ *
+ * You have n computers. You are given the integer n and a 0-indexed integer array batteries
+ * where the ith battery can run a computer for batteries[i] minutes. You are interested in
+ * running all n computers simultaneously using the given batteries.
+ *
+ * Initially, you can insert at most one battery into each computer. After that and at any
+ * integer time moment, you can remove a battery from a computer and insert another battery
+ * any number of times. The inserted battery can be a totally new battery or a battery from
+ * another computer. You may assume that the removing and inserting processes take no time.
+ *
+ * Note that the batteries cannot be recharged.
+ *
+ * Return the maximum number of minutes you can run all the n computers simultaneously.
+ */
+
+/**
+ * @param {number} n
+ * @param {number[]} batteries
+ * @return {number}
+ */
+var maxRunTime = function(n, batteries) {
+  let left = 1;
+  let right = Math.floor(batteries.reduce((sum, battery) => sum + battery, 0) / n);
+
+  while (left < right) {
+    const mid = Math.floor((left + right + 1) / 2);
+    const total = batteries.reduce((sum, battery) => sum + Math.min(battery, mid), 0);
+
+    if (total >= n * mid) {
+      left = mid;
+    } else {
+      right = mid - 1;
+    }
+  }
+
+  return left;
+};