<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>

using ll = long long;
using pii = pair<ll, int>;
#define fi first
#define se second

class Solution {
public:
    int reversePairs(vector<int>& nums) {
        vector<pii> v, vv;
        int n = nums.size();
        vector<int> bit(n + 1, 0);
        auto update = [&](int id) {
            while (id <= n) {
                bit[id]++;
                id += id & (-id);
            }
        };
        auto ps = [&](int id) {
            int res = 0;
            while (id > 0) {
                res += bit[id];
                id -= id & (-id);
            }
            return res;
        };
        auto qu = [&](int l, int r) {
            if (l > r) return 0;
            return ps(r) - ps(l - 1);
        };
        for (int i = 0; i < n; i++) {
            v.push_back({nums[i], i});
            vv.push_back({2LL * nums[i], i});
        }
        sort(v.begin(), v.end());
        sort(vv.begin(), vv.end());
        int ptr = 0;
        ll ans = 0;
        for (int i = 0; i < n; i++) {
            while (ptr < n && vv[ptr].fi < v[i].fi) {
                update(vv[ptr].se + 1);
                ptr++;
            }
            ans += qu(v[i].se + 2, n);
        }
        return ans;
    }
};
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