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

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 | /** * 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<numsSize;++i){ index[i]=i; } arr=nums; qsort(index, numsSize, sizeof(int), cmpfunc); int* ret=(int*)malloc(sizeof(int)*numsSize); ret[index[0]]=0; for(int i=1;i<numsSize;++i){ if( nums[index[i]] == nums[index[i-1]] ){ ret[index[i]]=ret[index[i-1]]; } else{ ret[index[i]]=i; } } free(index); return ret; } |

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