<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 {
    vector<int> ans;
    void merge_sort(vector<pair<int, int>>& x) {

        if (x.size() > 1) {
            vector<pair<int, int>> a;
            for (int i = 0; i < x.size() / 2; ++i)
                a.emplace_back(x[i]);
            merge_sort(a);
            vector<pair<int, int>> b;
            for (int i = x.size() / 2; i < x.size(); ++i)
                b.emplace_back(x[i]);
            merge_sort(b);
            x.clear();

            int i = 0, j = 0;
            while (i < a.size() || j < b.size()) {
                if (i == a.size()) {
                    x.emplace_back(b[j]);
                    ans[x.back().second] += x.size() - 1 > a.size() + j
                                                ? x.size() - 1 - a.size() - j
                                                : 0;
                    ++j;
                } else if (j == b.size()) {
                    x.emplace_back(a[i]);
                    ans[x.back().second] +=
                        x.size() - 1 > i ? x.size() - 1 - i : 0;
                    ++i;
                } else {
                    if (a[i] <= b[j]) {
                        x.emplace_back(a[i]);
                        ans[x.back().second] +=
                            x.size() - 1 > i ? x.size() - 1 - i : 0;
                        ++i;
                    } else {
                        x.emplace_back(b[j]);
                        ans[x.back().second] +=
                            x.size() - 1 > a.size() + j
                                ? x.size() - 1 - a.size() - j
                                : 0;
                        ++j;
                    }
                }
            }
        }
    }

public:
    vector<int> countSmaller(vector<int>& nums) {
        vector<pair<int, int>> tbs;

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

        merge_sort(tbs);
        return ans;
    }
};
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