leetcode/apps/docs/content/(array)/1365-how-many-numbers-are-smaller-than-the-current-number.mdx

---
title: '1365. How Many Numbers Are Smaller Than the Current Number'
description: 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]
sidebar:
  label: 'How Many Numbers Are Smaller Than the Current Number'
  badge: 'Easy'
---

<Badge variant="accent">Array</Badge>

### Example 1:
- Input: `nums = [8,1,2,2,3]`
- Output: `[4,0,1,1,3]`
- Explanation: For `nums[0]=8` there exist four smaller numbers than it (`1`, `2`, `2` and `3`). For `nums[1]=1` does not exist any smaller number than it. For `nums[2]=2` there exist one smaller number than it (`1`). For `nums[3]=2` there exist one smaller number than it (`1`). For `nums[4]=3` there exist three smaller numbers than it (`1`, `2` and `2`).

### Example 2:
- Input: `nums = [6,5,4,8]`
- Output: `[2,1,0,3]`

### Example 3:
- Input: `nums = [7,7,7,7]`
- Output: `[0,0,0,0]`

### Constraints:

- `2 <= nums.length <= 500`
- `0 <= nums[i] <= 100`

## Approach

```mermaid
flowchart TD
  S(["smallerNumbersThanCurrent(nums)"]) --> F["freq: count every value in nums"]
  F --> O["sorted = keys of freq, ascending"]
  O --> I["count = 0, smaller = {}"]
  I --> L{"more num in sorted?"}
  L -- yes --> A["smaller[num] = count — everything already passed is smaller"]
  A --> B["count += freq[num]"]
  B --> L
  L -- no --> M(["return nums.map(n => smaller[n])"])
```

## Solution

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

var smallerNumbersThanCurrent = function (nums) {
  // Step 1: Begin by initializing a [Frequency Map]()
  const freq = {};
  for (let n of nums) freq[n] = (freq[n] || 0) + 1;

  // Step 2: Sort the numbers by ascending order
  const sorted = Object.keys(freq).sort((a, b) => a - b);

  // Step 3: Init a count of numbers smaller than the active number
  let count = 0;

  // Step 4: Init a map to track number of values smaller for each number
  const smaller = {};

  // Step 5: Iterate over the sorted list
  for (let num of sorted) {
    // Set count for active number
    smaller[num] = count;

    // Update the count by frequency
    count += freq[num];
  }

  // Step 6: Use original list and find number of smaller values than it
  return nums.map((n) => smaller[n]);
};
```