[Leet Code] How Many Numbers Are Smaller Than The Current Number

Leetcode: https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/submissions/

Explanation of the problem:

You are given an array of numbers, and for each number, you have to find out how many numbers are smaller than that given number in the array.

Solution:

class Solution(object):
    def smallerNumbersThanCurrent(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        answer = []
        count = 0
        
        for i in range(len(nums)):
            for j in range(len(nums)):
                if nums[j] < nums[i]:
                        count = count + 1
            answer.append(count)
            count = 0
        
        return answer

Due to having to loop over the array twice, we will need two for loops.

for i in range(len(nums)):
    for j in range(len(nums)):

Then it is fairly simple logic of: if the inner loop of values is less than the number specified, then we increment the counter by 1.

Then we just return the answer!