<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 Solution {
public:
    


int merge(vector<int> &arr, int low, int mid, int high) {
    vector<int> temp; // temporary array
    int left = low;      // starting index of left half of arr
    int right = mid + 1;   // starting index of right half of arr

    //Modification 1: cnt variable to count the pairs:
    int cnt = 0;

    //storing elements in the temporary array in a sorted manner//

    while (left <= mid && right <= high) {
            if (arr[left] > 2LL * arr[right]) {
                cnt += (mid - left + 1);
                right++;
            } else {
                left++;
            }
        }

        // Reset indices to merge the sorted halves
        left = low;
        right = mid + 1;

        while (left <= mid && right <= high) {
            if (arr[left] <= arr[right]) {
                temp.push_back(arr[left]);
                left++;
            } else {
                temp.push_back(arr[right]);
                right++;
            }
        }
    // if elements on the left half are still left //

    while (left <= mid) {
        temp.push_back(arr[left]);
        left++;
    }

    //  if elements on the right half are still left //
    while (right <= high) {
        temp.push_back(arr[right]);
        right++;
    }

    // transfering all elements from temporary to arr //
    for (int i = low; i <= high; i++) {
        arr[i] = temp[i - low];
    }

    return cnt; // Modification 3
}

int mergeSort(vector<int> &arr, int low, int high) {
    int cnt = 0;
    if (low >= high) return cnt;
    int mid = (low + high) / 2 ;
    cnt += mergeSort(arr, low, mid);  // left half
    cnt += mergeSort(arr, mid + 1, high); // right half
    cnt += merge(arr, low, mid, high);  // merging sorted halves
    return cnt;
}

int reversePairs(vector<int>a) {

    // Count the number of pairs:
    return mergeSort(a, 0, a.size() - 1);
}




};
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