<p>Given an integer array <code>nums</code> and two integers <code>lower</code> and <code>upper</code>, return <em>the number of range sums that lie in</em> <code>[lower, upper]</code> <em>inclusive</em>.</p>

<p>Range sum <code>S(i, j)</code> is defined as the sum of the elements in <code>nums</code> between indices <code>i</code> and <code>j</code> inclusive, where <code>i &lt;= j</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [-2,5,-1], lower = -2, upper = 2
<strong>Output:</strong> 3
<strong>Explanation:</strong> The three ranges are: [0,0], [2,2], and [0,2] and their respective sums are: -2, -1, 2.
</pre>

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

<pre>
<strong>Input:</strong> nums = [0], lower = 0, upper = 0
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
	<li><code>-10<sup>5</sup> &lt;= lower &lt;= upper &lt;= 10<sup>5</sup></code></li>
	<li>The answer is <strong>guaranteed</strong> to fit in a <strong>32-bit</strong> integer.</li>
</ul>

class Solution {

public:

    int countRangeSum(vector<int>& nums, int lower, int upper) {

        int n = nums.size();

        vector<long long> prefix(n + 1, 0);

        

        // Step 1: Calculate prefix sums

        for (int i = 0; i < n; ++i) {

            prefix[i + 1] = prefix[i] + nums[i];

        }

        

        // Step 2: Use modified merge sort to count valid range sums

        return mergeSortAndCount(prefix, 0, n + 1, lower, upper);

    }

    

private:

    int mergeSortAndCount(vector<long long>& prefix, int left, int right, int lower, int upper) {

        if (right - left <= 1) {

            return 0;

        }

        

        int mid = left + (right - left) / 2;

        int count = mergeSortAndCount(prefix, left, mid, lower, upper) +

                    mergeSortAndCount(prefix, mid, right, lower, upper);

        

        // Step 3: Count valid range sums for the current split

        int j = mid, k = mid, l = mid;

        vector<long long> temp;

        for (int i = left; i < mid; ++i) {

            while (k < right && prefix[k] - prefix[i] < lower) ++k;

            while (j < right && prefix[j] - prefix[i] <= upper) ++j;

            count += (j - k);

        }

        // Step 4: Merge the two halves

        int r = mid;

        for (int i = left; i < mid; ++i) {

            while (r < right && prefix[r] < prefix[i]) {

                temp.push_back(prefix[r++]);

            }

            temp.push_back(prefix[i]);

        }

        

        while (r < right) {

            temp.push_back(prefix[r++]);

        }

        

        // Copy the merged result back to the original array

        for (int i = left; i < right; ++i) {

            prefix[i] = temp[i - left];

        }

        

        return count;

    }

};
    
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