Add solution #1283

Author: JoshCrozier

README.md

@@ -1,4 +1,4 @@
-# 1,396 LeetCode solutions in JavaScript
+# 1,397 LeetCode solutions in JavaScript
 
 [https://leetcodejavascript.com](https://leetcodejavascript.com)
 
@@ -975,6 +975,7 @@
 1277|[Count Square Submatrices with All Ones](./solutions/1277-count-square-submatrices-with-all-ones.js)|Medium|
 1278|[Palindrome Partitioning III](./solutions/1278-palindrome-partitioning-iii.js)|Hard|
 1282|[Group the People Given the Group Size They Belong To](./solutions/1282-group-the-people-given-the-group-size-they-belong-to.js)|Medium|
+1283|[Find the Smallest Divisor Given a Threshold](./solutions/1283-find-the-smallest-divisor-given-a-threshold.js)|Medium|
 1284|[Minimum Number of Flips to Convert Binary Matrix to Zero Matrix](./solutions/1284-minimum-number-of-flips-to-convert-binary-matrix-to-zero-matrix.js)|Hard|
 1286|[Iterator for Combination](./solutions/1286-iterator-for-combination.js)|Medium|
 1287|[Element Appearing More Than 25% In Sorted Array](./solutions/1287-element-appearing-more-than-25-in-sorted-array.js)|Easy|

solutions/1283-find-the-smallest-divisor-given-a-threshold.js

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+/**
+ * 1283. Find the Smallest Divisor Given a Threshold
+ * https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/
+ * Difficulty: Medium
+ *
+ * Given an array of integers nums and an integer threshold, we will choose a positive integer
+ * divisor, divide all the array by it, and sum the division's result. Find the smallest divisor
+ * such that the result mentioned above is less than or equal to threshold.
+ *
+ * Each result of the division is rounded to the nearest integer greater than or equal to that
+ * element. (For example: 7/3 = 3 and 10/2 = 5).
+ *
+ * The test cases are generated so that there will be an answer.
+ */
+
+/**
+ * @param {number[]} nums
+ * @param {number} threshold
+ * @return {number}
+ */
+var smallestDivisor = function(nums, threshold) {
+  let left = 1;
+  let right = Math.max(...nums);
+
+  while (left < right) {
+    const mid = Math.floor((left + right) / 2);
+    const sum = nums.reduce((acc, num) => acc + Math.ceil(num / mid), 0);
+
+    if (sum <= threshold) {
+      right = mid;
+    } else {
+      left = mid + 1;
+    }
+  }
+
+  return left;
+};