nums-smaller-than-current.py

# 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.

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        temp = sorted((nums))
        d={}

        for i, num in enumerate(temp):
            if num not in d:
                d[num] = i

        ret = []

        for i in nums:
            ret.append(d[i])
        
        return ret