1365.md

Employ exhaustive brute-force search to determine the number of pairs of indices (i, j) which exists such that i ≠ j and nums[j] < nums[i].

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        nums_length = len(nums)
        output = [0] * nums_length
        
        for i in range(nums_length):
            for j in range(nums_length):
                if j != i and nums[j] < nums[i]:
                    output[i] += 1
                    
        return output

Sort the input array supplied. For each i in the range 0 through |nums|, determine the number of elements contained within the input array supplied whose value is less than nums[i] - this is equal to the index of nums[i] within the sorted array.

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        
        if len(nums) >= 2:
            sorted_nums = sorted(nums)
            for i in range(len(nums)):
                nums[i] = sorted_nums.index(nums[i])
        
        return nums  

Sort the input array supplied. Construct a mapping between each number contained within the input array and the index at which that number first appears within the sorted input array. For each i in the range 0 through |nums|, determine the number of elements contained within the input array supplied whose value is less than nums[i] - this is equal to the value given the the value_by_first_index mapping for the key nums[i].

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        
        if len(nums) >= 2:
            
            value_by_first_index = {}
            for i, v in enumerate(sorted(nums)):
                if v not in value_by_first_index:
                    value_by_first_index[v] = i
                    
            for i, v in enumerate(nums):
                nums[i] = value_by_first_index[v]
        
        return nums 

This can be further rewritten to employ Python list comprehension:

class Solution:
    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        
        if len(nums) < 2:
            return nums
        
        else:
            
            value_by_first_index = {}
            for i, v in enumerate(sorted(nums)):
                if v not in value_by_first_index:
                    value_by_first_index[v] = i
                    
            return [value_by_first_index[v] for v in nums]