<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:
    
    struct seg_tree {
        
        vector<int> tree;
        seg_tree(int n) {
            this->tree = vector<int>(4 * 100002, 0);
        }
        
        void update(int index, int left, int right, int l, int r) {
            
            if (l > r) return;
            if (left >= l && right <= r) {
                tree[index] += 1;
            } else {
                int mid = (right - left) / 2 + left;
                update(2 * index, left, mid, l, min(mid, r));
                update(2 * index + 1, mid + 1, right, max(mid + 1, l), r);
                tree[index] = tree[2 * index] + tree[2 * index + 1];
            }
            
        }
        
        int query(int index, int left, int right, int l, int r) {
            
            if (l > r) return 0;
            if (left >= l && right <= r) {
                return tree[index];
            } else {
                int mid = (right - left) / 2 + left;
                return query(2 * index, left, mid, l, min(mid, r)) + query(2 * index + 1, mid + 1, right, max(mid + 1, l), r);
            }
            
        }
        
        
        
    };
    
    vector<int> countSmaller(vector<int>& nums) {
        
        int n = nums.size();
        seg_tree st = seg_tree(5);
        for (int i = 0; i < n; i++) nums[i] += 10000;
        
        vector<int> res(n, 0);
        for (int i = n - 1; i >= 0; i--) {
            res[i] = st.query(1, 0, 100002, 0, nums[i] - 1);
            st.update(1, 0, 100002, nums[i], nums[i]);
        }
        
        return res;
        
        
    }
};
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