Create a sorted vector of nums, named clac. Starting from nums[0], binary search the position of nums[0] in calc(lower_bound). The position is the number of smaller numbers in nums on the right. Remove the value in calc and proceed to next value nums[1]. Vise versa

```
class Solution {
public:
vector<int> countSmaller(vector<int>& nums) {
vector<int> calc(nums);
vector<int> retval(nums.size());
sort(calc.begin(), calc.end());
vector<int>::iterator it;
for (int idx = 0; idx < nums.size(); idx++) {
it = lower_bound(calc.begin(), calc.end(), nums[idx]);
retval[idx] = it - calc.begin();
calc.erase(it);
}
return retval;
}
};
```