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

- Using Hash table to solve the problem in O(n) time.
- In our given constraint, nums[i] will be 0 <= nums[i] <= 100,
- So declare hash[101] which holds the number from 0 to 100.

**C# Program**

public class Solution { public int[] SmallerNumbersThanCurrent(int[] nums) { int[] hash = new int[101]; 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)

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