1365. How Many Numbers Are Smaller Than the Current Number

Question 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] Solution class Solution { public: vector smallerNumbersThanCurrent(vector& nums) { vectorv; for(int i=0;inums[j]) x++; } v.push_back(x); } return v; } };