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[]temp=new int[nums.length];
        for(int i=0;i<nums.length;i++){
            int count=0;
            for(int j=0;j<nums.length;j++){
                if(nums[i]>nums[j]){
                    count++;
                }
            }
            temp[i]=count;
        }
        return temp;
    }
}