# LeetCode 1365. How Many Numbers Are Smaller Than the Current Number

## 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=8 there exist four smaller numbers than it (1, 2, 2 and 3).
For nums=1 does not exist any smaller number than it.
For nums=2 there exist one smaller number than it (1).
For nums=2 there exist one smaller number than it (1).
For nums=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)