315. Count of Smaller Numbers After Self

You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i]is the number of smaller elements to the right of nums[i].

Example:

Given nums = [5, 2, 6, 1]

To the right of 5 there are 2 smaller elements (2 and 1).
To the right of 2 there is only 1 smaller element (1).
To the right of 6 there is 1 smaller element (1).
To the right of 1 there is 0 smaller element.

Return the array [2, 1, 1, 0].

class Solution {
    
public:
    vector<int> countSmaller(vector<int>& nums) {
         std::vector<std::pair<int, int> > vec;
         std::vector<int> count;
         for(int i=0; i<nums.size(); i++)
         {
             vec.push_back(std::make_pair(nums[i], i));
             count.push_back(0);
         }
        merge_sort(vec, count);
        return count;
    }
private:
    void merge_sort_two_vec(std::vector<std::pair<int, int> > &sub_vec1, std::vector<std::pair<int, int> > &sub_vec2,                               std::vector<std::pair<int, int> > &vec, std::vector<int> &count)
    {
        int i = 0;
        int j = 0;
        while(i<sub_vec1.size()&&j<sub_vec2.size()){
            if(sub_vec1[i].first <= sub_vec2[j].first)
            {
                count[sub_vec1[i].second] += j;
                vec.push_back(sub_vec1[i]);
                i++;
            }else
            {
                vec.push_back(sub_vec2[j]);
                j++;
            }
        }
        
        for(; i<sub_vec1.size(); i++)
        {
            count[sub_vec1[i].second] += j;
            vec.push_back(sub_vec1[i]);
        }
        
        for(; j<sub_vec2.size(); j++)
        {
            vec.push_back(sub_vec2[j]);
        }
    }
    
    void merge_sort(std::vector<std::pair<int, int> > &vec, std::vector<int> &count)
    {
        if(vec.size()<2)
            return;
        int mid = vec.size()/2;
        std::vector<std::pair<int, int> > sub_vec1;
        std::vector<std::pair<int, int> > sub_vec2;
        
        for(int i=0; i<mid; i++)
        {
            sub_vec1.push_back(vec[i]);
        }
        
        for(int i=mid; i<vec.size(); i++)
        {
            sub_vec2.push_back(vec[i]);
        }
        merge_sort(sub_vec1, count);
        merge_sort(sub_vec2, count);
        
        vec.clear();
        merge_sort_two_vec(sub_vec1, sub_vec2, vec, count);
    }
};