1365.py

# 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
"""

from typing import List


class Solution:
    # O(n ^2)
    # def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
    #   table = {}
    #   result = [0] * len(nums)

    #   for i in range(len(nums)):
    #     for j in range(len(nums)):
    #       if nums[i] < nums[j] and i != j:
    #         result[j] += 1

    #   return result

    def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
        table = {}

        for i, number in enumerate(sorted(nums)):
            if table.get(number) is not None:
                continue
            table[number] = i

        return [table[nums[i]] for i in range(len(nums))]


print(Solution().smallerNumbersThanCurrent(nums=[6, 5, 4, 8]))