LeetCode : 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]

class Solution {
    public int[] smallerNumbersThanCurrent(int[] nums) {
        int[] result = new int[nums.length];
        // 첫 배열 
        for(int i = 0; i < nums.length; i++) {
            int count = 0;
            // 비교 루프
            for(int j = 0; j < nums.length; j++) {
                if(nums[i] > nums[j]) {
                    count++;
                }                
            }
            result[i] = count;
        }
        return result;
    }
}