"""
1365. How Many Numbers Are Smaller Than the Current Number
Difficulty: Easy
https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/

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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
"""

from collections import Counter


class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        freq = Counter(nums)
        sorted_nums = sorted(freq.keys())
        smaller = {}
        count = 0
        for num in sorted_nums:
            smaller[num] = count
            count += freq[num]

        return [smaller[num] for num in nums]