# [Solved] 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.

## Question

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`

## Python Solution

```class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
n = sorted(nums)
s=[]
for i in nums:
c=0
for j in n:
if j<i:
c+=1
else:
break
s.append(c)
return (s)```