# 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=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`

## Solution

1. Using Hash table to solve the problem in O(n) time.
2. In our given constraint, nums[i] will be 0 <= nums[i] <= 100,
3. So declare hash which holds the number from 0 to 100.

C# Program

```public class Solution {
public int[] SmallerNumbersThanCurrent(int[] nums) {
int[] hash = new int;
for(int i=0;i<nums.Length;i++)
{
hash[nums[i]]++;
}
for(int i=1;i<=100;i++)
{
hash[i] = hash[i] + hash[i-1];
}
for(int i=0;i<nums.Length;i++)
{
if(nums[i] != 0)
nums[i] = hash[nums[i]-1];

}
return nums;
}
}
```

Time Complexity: O(n)

Space Complexity: O(1)