<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:
    vector<int> bit;
    
    void update(int id, int val) {
        int n = bit.size();
        
        while (id < n) {
            bit[id] += val;
            id += (id & -id);
        }
    }

    int query(int id) {
        int ans = 0;
        
        while (id > 0) {
            ans += bit[id];
            id -= (id & -id);
        }
        
        return ans;
    }
    
    vector<int> countSmaller(vector<int>& nums) {
        int n = nums.size();

        vector<vector<int>> arr;

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

        //cout<<"arr built"<<endl;

        sort(arr.begin(), arr.end());

        bit.resize(n + 1, 0);

        vector<int> ans(n, 0);

        for (int i = 0;i < n;i ++) {
            int val = arr[i][0];
            int idx = arr[i][1];

            //cout<<val<<" "<<idx<<endl;

            update(idx + 1, 1);

            ans[idx] = (i + 1) - query(idx + 1);
        }

        return ans;
    }
};
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