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tags: data_structure_python
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# How Many Numbers Are Smaller Than the Current Number <img src="https://img.shields.io/badge/-easy-brightgreen">

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.

<ins>**Example 1:**</ins>

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).

<ins>**Example 2:**</ins>

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

<ins>**Example 3:**</ins>

Input: nums = [7,7,7,7]
Output: [0,0,0,0]

**Constraints:**

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

# Solution

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        # O(n^2) time complexity.
        # O(n) space complexity.
        res = []
        m = len(nums)
        
        for i in range(m):
            count = 0
            for j in range(m):
                if nums[j] < nums[i]:
                    count += 1
            res.append(count)
        return res