###### tags: `leetcode` `easy` `Array` `Hash Table` `Sorting` `Counting`
# [1365. How Many Numbers Are Smaller Than the Current Number](https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/)

## Description

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.

## Examples
### 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:

- $2 \leq nums.length \leq 500$
- $0 \leq nums[i] \leq 100$

## Code

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* smallerNumbersThanCurrent(int* nums, int numsSize, int* returnSize) {
    int* returnNums = calloc(numsSize, sizeof(int));
    int i, count[101] = { 0 }, all[101] = { 0 };
    for (i = 0; i < numsSize; i++)
        count[nums[i]]++;
    all[1] = count[0];
    for (i = 1; i <= 100; i++)
        all[i] = count[i - 1] + all[i - 1];
    for (i = 0; i < numsSize; i++)
        returnNums[i] = all[nums[i]];
    *returnSize = numsSize;
    return returnNums;
}

## Complexity

|Space |Time |
|- |- |
|$O(N)$|$O(N)$|

## Result

- Runtime : 18 ms, 89.38%
- Memory usage : 6.7 MB, 94.14%