# LeetCode 1365. How Many Numbers Are Smaller Than the Current Number

## Description

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

## Explanation

Two loops comparison.

## Python Solution

class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
results = [0 for i in range(len(nums))]

i = 0

while i < len(nums):
for j, num in enumerate(nums):
if j != i and num < nums[i]:
results[i] += 1
i += 1

return results
• Time Complexity: O(N^2)
• Space Complexity: O(N)