# Leetcode - How Many Numbers Are Smaller Than the Current Number Solution

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`

## Solution in Python

```
from collections import Counter
class Solution:
def smallerNumbersThanCurrent(self, nums: List[int]) -> List[int]:
counts = Counter(nums)
smaller_nums = []
total = 0
for x,y in sorted(counts.items()):
total += y
smaller_nums.append((x,total-y))
smaller_nums = dict(smaller_nums)
return [smaller_nums[i] for i in nums]
```