leetcode/docs/(welcome)/array/1365.mdx

---
title: LC-1365
description: How Many Numbers Are Smaller Than the Current Number
---

## Examples

| Input            | Output               |
| ---------------- | -------------------- |
| `[8,1,2,2,3]`    | `[4,0,1,1,3]`        |
| `[6,5,4,8]`      | `[2,1,0,3]`          |
| `[7,7,7,7]`      | `[0,0,0,0]`          |

## Hints

<Accordion type="single">
  <AccordionItem title="Constraints">
    Constraints:
     • `2 <= nums.length <= 500`
     • `0 <= nums[i] <= 100`
  </AccordionItem>
  <AccordionItem title="Problem Statement">
    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]`.
  </AccordionItem>
</Accordion>

## Solution

```js
/**
 * @param {number[]} nums
 * @return {number[]}
 */

var smallerNumbersThanCurrent = function (nums) {
  // Default for Hammer
  const freq = {};
  for (let n of nums) freq[n] = (freq[n] || 0) + 1;
  const sorted = Object.keys(freq).sort((a, b) => a - b);

  // Problem Specific
  let count = 0;
  let smaller = {};

  // Iterate over sorted.
  for (let num of sorted) {
    smaller[num] = count;
    count += freq[num];
  }

  // Map output
  return nums.map((n) => smaller[n]);
};
```