Add solution #1365

Author: JoshCrozier

README.md

@@ -142,6 +142,7 @@
 1351|[Count Negative Numbers in a Sorted Matrix](./1351-count-negative-numbers-in-a-sorted-matrix.js)|Easy|
 1356|[Sort Integers by The Number of 1 Bits](./1356-sort-integers-by-the-number-of-1-bits.js)|Easy|
 1360|[Number of Days Between Two Dates](./1360-number-of-days-between-two-dates.js)|Easy|
+1365|[How Many Numbers Are Smaller Than the Current Number](./1365-how-many-numbers-are-smaller-than-the-current-number.js)|Easy|
 1380|[Lucky Numbers in a Matrix](./1380-lucky-numbers-in-a-matrix.js)|Easy|
 1472|[Design Browser History](./1472-design-browser-history.js)|Medium|
 1598|[Crawler Log Folder](./1598-crawler-log-folder.js)|Easy|

solutions/1365-how-many-numbers-are-smaller-than-the-current-number.js

@@ -0,0 +1,20 @@
+/**
+ * 1365. How Many Numbers Are Smaller Than the Current Number
+ * https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/
+ * Difficulty: Easy
+ *
+ * Given the array nums, for each nums[i] find out how many numbers in the array are smaller
+ * than it. That is, for each nums[i] you have to count the number of valid j's such that
+ * j != i and nums[j] < nums[i].
+ *
+ * Return the answer in an array.
+ */
+
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var smallerNumbersThanCurrent = function(nums) {
+  const sorted = nums.slice().sort((a, b) => a - b);
+  return nums.map(num => sorted.indexOf(num));
+};