## Description

https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/

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.

**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`

## Explanation

Two loops comparison.

## Python Solution

```
class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
results = [0 for i in range(len(nums))]
i = 0
while i < len(nums):
for j, num in enumerate(nums):
if j != i and num < nums[i]:
results[i] += 1
i += 1
return results
```

- Time Complexity: O(N^2)
- Space Complexity: O(N)