# LeetCode 1365. 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[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

 1234567891011121314151617181920212223242526272829 /**  * Note: The returned array must be malloced, assume caller calls free().  */ int* arr; int cmpfunc (const void * a, const void * b) {    return  arr[*(int*)a] - arr[*(int*)b ]; } int* smallerNumbersThanCurrent(int* nums, int numsSize, int* returnSize){     *returnSize=numsSize;     int* index=(int*)malloc(sizeof(int)*numsSize);     for(int i=0;i

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