<p>Given an integer array <code>nums</code>, return<em> an integer array </em><code>counts</code><em> where </em><code>counts[i]</code><em> is the number of smaller elements to the right of </em><code>nums[i]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,2,6,1]
<strong>Output:</strong> [2,1,1,0]
<strong>Explanation:</strong>
To the right of 5 there are <b>2</b> smaller elements (2 and 1).
To the right of 2 there is only <b>1</b> smaller element (1).
To the right of 6 there is <b>1</b> smaller element (1).
To the right of 1 there is <b>0</b> smaller element.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1]
<strong>Output:</strong> [0]
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [-1,-1]
<strong>Output:</strong> [0,0]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    void merge(vector<pair<int, int>>& nums, int l, int r, vector<int>& ans)
    {
        vector<pair<int, int>> left, right;
        int mid = (l + r)/2;

        for(int i = l; i <= mid; i++)
            left.push_back(nums[i]);

        for(int i = mid + 1; i <= r; i++)
            right.push_back(nums[i]);

        int leftLen = left.size(), rightLen = right.size();
        int i = 0, j = 0, k = l;

        while(i < leftLen && j < rightLen)
        {
            if(left[i].first <= right[j].first)
            {
                nums[k] = left[i];
                ans[left[i].second] += j;
                i++;
                k++;
            }

            else
            {
                nums[k] = right[j];
                j++;
                k++;
            }
        }

        while(i < leftLen)
        {
            nums[k] = left[i];
            ans[left[i].second] += j;
            i++;
            k++;
        }

        while(j < rightLen)
        {
            nums[k] = right[j];
            j++;
            k++;
        }

        return;
    }

    void mergeSort(vector<pair<int, int>>& nums, int l, int r, vector<int>& ans)
    {
        if(l >= r)
            return;

        int mid = (l + r)/2;

        mergeSort(nums, l, mid, ans);
        mergeSort(nums, mid + 1, r, ans);
        merge(nums, l, r, ans);

        return;
    }

    vector<int> countSmaller(vector<int>& nums) 
    {
        vector<int> ans(nums.size(), 0);
        vector<pair<int, int>> dupNum;

        for(int i = 0; i < nums.size(); i++) dupNum.push_back({nums[i], i});

        mergeSort(dupNum, 0, nums.size() - 1, ans);

        return ans;    
    }
};
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