<p>Given an integer array <code>nums</code>, return <em>the number of <strong>reverse pairs</strong> in the array</em>.</p>

<p>A <strong>reverse pair</strong> is a pair <code>(i, j)</code> where:</p>

<ul>
	<li><code>0 &lt;= i &lt; j &lt; nums.length</code> and</li>
	<li><code>nums[i] &gt; 2 * nums[j]</code>.</li>
</ul>

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

<pre>
<strong>Input:</strong> nums = [1,3,2,3,1]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The reverse pairs are:
(1, 4) --&gt; nums[1] = 3, nums[4] = 1, 3 &gt; 2 * 1
(3, 4) --&gt; nums[3] = 3, nums[4] = 1, 3 &gt; 2 * 1
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,4,3,5,1]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The reverse pairs are:
(1, 4) --&gt; nums[1] = 4, nums[4] = 1, 4 &gt; 2 * 1
(2, 4) --&gt; nums[2] = 3, nums[4] = 1, 3 &gt; 2 * 1
(3, 4) --&gt; nums[3] = 5, nums[4] = 1, 5 &gt; 2 * 1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

//<<<<<<CP TEMPLATE<<<<<<<<
//<<<<<<<<<<<BIT TEMPLATE
//bit for prefix sum query
//for higher range maxi and mini me segment tree use kro
#define ll long long
#define all(x) x.begin(), x.end()
class Solution {
public:
    const int N = 1e5 + 10;
    vector<int> bit;
    void update(int i, int x) {
        for (; i < N; i += (i & -i))
            bit[i] += x;
    }
    int getSum(int i) {
        int sum = 0;
        for (; i > 0; i -= (i & -i))
            sum += bit[i];
        return sum;
    }
    int reversePairs(vector<int>& nums) {
        int n = nums.size();
        bit.resize(N, 0); 
        vector<ll> res;
        for (int num : nums) {
            res.push_back(num);
            res.push_back(2LL * num); 
        }
        sort(all(res));
        res.erase(unique(all(res)), res.end());
        //step 2<<<<<<<<<<compression>>>>>>>>>>>>>>>>.
        auto getIndex = [&](ll x) {
            return lower_bound(res.begin(), res.end(), x) - res.begin() + 1;
        };
        int cnt = 0;
        // Traverse nums from the end to count reverse pairs
        for (int i = n - 1; i >= 0; --i) {
            cnt += getSum(getIndex(nums[i]) - 1); 
            update(getIndex(2LL * nums[i]), 1);
        }
        return cnt;
    }
};

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