<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> tree;
    int n = 200001;
    int query(int i) {
        int res = 0;
        for (; i > 0; i -= (i & (-i)))
            res += tree[i];
        return res;
    }
    void add(int i, int val) {
        for (; i <= n; i += (i & (-i)))
            tree[i] += val;
    }

public:
    vector<int> countSmaller(vector<int>& nums) {
        tree.resize(n + 1);
        vector<int> ans(nums.size(), 0);
        for (int i = nums.size() - 1; i >= 0; i--) {
            int x = nums[i] + 10001;
            ans[i] = query(x - 1);
            add(x, 1);
        }
        return ans;
    }
};
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