Most frequently asked interview question #7 with solution

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.

Input: nums = [8,1,2,2,3]
Output: [4,0,1,1,3]

Solution in C++:

    vector<int> smallerNumbersThanCurrent(vector<int>& nums) {
       
        int count[101] = {0};
        int vect_size  = nums.size();
       
        for(int iter = 0; iter < vect_size; ++iter){
           
            ++count[nums.at(iter)];
        }
        int temp_count = 0;
        bool first_one = true;
        for(int iter = 0; iter < 101; ++iter){

            //Case: Count is not equal to -1
            if(count[iter] != 0){
                if(first_one){
                    first_one = false;
                    temp_count = count[iter];
                    count[iter] = 0;
                    continue;
                }
                    int temp = 0 ;
                temp        = count[iter];
                count[iter] = temp_count;
                temp_count += temp;
            }
        }
        vector <int> ret;
        for(auto iter = 0; iter < vect_size; ++iter){
            ret.push_back(count[nums.at(iter)]);
        }
        return ret;
    }