# How Many Numbers Are Smaller Than the Current Number

## 1365. How Many Numbers Are Smaller Than the Current Number (Easy)

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=8 there exist four smaller numbers than it (1, 2, 2 and 3).
For nums=1 does not exist any smaller number than it.
For nums=2 there exist one smaller number than it (1).
For nums=2 there exist one smaller number than it (1).
For nums=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`

Hints:

Brute force for each array element.
In order to improve the time complexity, we can sort the array and get the answer for each array element.

## How Many Numbers Are Smaller Than the Current Number (Java)

```class Solution {
public int[] smallerNumbersThanCurrent(int[] nums) {
int[] arr = 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++;
}
}
arr[i]=count;
}
return arr;
}
}```