<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>

class FenwickTree{
    public:
    int n;
    vector<int> bit;
    FenwickTree(int n){
        this->n=n;
        bit.assign(n+1,0);
    }
    int query(int ind){
        ind++;
        int sum=0;
        while(ind>0){
            sum+=bit[ind];
            ind-=(ind) &(-ind);
        }
        return sum;
    }
    void update(int ind,int val){
        ind++;
        while(ind<=n){
            bit[ind]+=val;
            ind+=(ind)&(-ind);
        }
    }
};
class Solution {
public:
    int reversePairs(vector<int>& nums) {
        vector<long long> all_nums; // To hold nums[i] and 2 * nums[j]

    // Store nums[i] and 2 * nums[j] for coordinate compression
    for (int num : nums) {
        all_nums.push_back(num);
        all_nums.push_back(2LL * num); // Store 2 * nums[j]
    }

    // Coordinate compression: Get unique values and assign ranks
    set<long long> unique_vals(all_nums.begin(), all_nums.end());
    map<long long, int> rank;
    int r = 1;
    for (long long val : unique_vals) {
        rank[val] = r++;
    }

    // Fenwick Tree for frequency counts
    FenwickTree fenwick(rank.size());

    int reverse_pairs = 0;

    // Iterate from right to left in the array
    for (int i = nums.size() - 1; i >= 0; --i) {
        // Query how many numbers are less than or equal to nums[i]
        reverse_pairs += fenwick.query(rank[nums[i]] - 1); // nums[i] > 2 * nums[j]

        // Update the Fenwick Tree with the rank of 2 * nums[i]
        fenwick.update(rank[2LL * nums[i]], 1); // Add 2 * nums[i] to Fenwick Tree
    }

    return reverse_pairs;
    }
};
107610