"""
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]
"""


class Solution:
    def smallerNumbersThanCurrent(self, nums):
        res = []
        for i in nums:
            lower = [j for j in nums if i > j]
            res.append(len(lower))
        return res


if __name__ == '__main__':
    s = Solution()
    print(s.smallerNumbersThanCurrent([8, 1, 2, 2, 3]))
