'''
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

'''

def smallerNumbersThanCurrent(nums):
    result = []
    l = nums
    for num in nums:
        sublist = []

        for n in l:
            if n < num:
                sublist.append(n)

        result.append(len(sublist))

    return result

assert smallerNumbersThanCurrent([6,5,4,8]) == [2,1,0,3]
assert smallerNumbersThanCurrent([7,7,7,7]) == [0,0,0,0]