<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>

const int N = 400007;
long long t[4 * N];
int a[10];

void update(int index, int tl, int tr, int pos, int val) {
    if (pos < tl || pos > tr) return;
    if (tl == tr) {
        t[index] += val;
        return;
    }
    int tm = (tl + tr) / 2;
    update(2 * index, tl, tm, pos, val);
    update(2 * index + 1, tm + 1, tr, pos, val);
    t[index] = t[2 * index] + t[2 * index + 1];
}

int query(int index, int tl, int tr, int l, int r) {
    if (r < tl || tr < l) return 0;
    if (tl >= l && tr <= r) return t[index];
    int tm = (tl + tr) / 2;
    return query(2 * index, tl, tm, l, r) + query(2 * index + 1, tm + 1, tr, l, r);
}

class Solution {
public:
    long long reversePairs(vector<int>& nums) {
        // memset(t, 0, sizeof(t));
        int n = nums.size();
        vector<long long> b;
        map<long long, int> mp;
        for (int i = 0; i < n; i++) {
            b.push_back(nums[i]);
            b.push_back(2 * 1LL * nums[i]);
        }
        sort(b.begin(), b.end());
        int id = 1;
        for (int i = 0; i < 2 * n; i++) {
            if (mp[b[i]]) continue;
            mp[b[i]] = id;
            id++;
        }
        long long ans = 0;
        for (int i = 0; i < n; i++) {
            ans += query(1, 0, 2 * n, mp[2 * 1LL * nums[i]] + 1, 2 * n);
            update(1, 0, 2 * n, mp[nums[i]], 1);
        }
        for (int i = 1; i <= 8 * n; i++) t[i] = 0; 
        return ans;
    }
};
88080