<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 nextGreatest(vector<long> & nums, long num){
        // index of just greatest than num
        int n = nums.size();
        int l = 0;
        int r = n-1;
        while(l <= r){
            int mid = l + (r-l)/2;
            if(nums[mid] > num){
                r = mid-1;
            }
            else{
                l = mid+1;
            }
        }
        return l;
    }

    int firstOccur(vector<long> & nums, long num){
        // index of first occ of num
        int n = nums.size();
        int l = 0;
        int r = n-1;
        while(l <= r){
            int mid = l + (r-l)/2;
            if(nums[mid] >= num){
                r = mid-1;
            }
            else{
                l = mid+1;
            }
        }
        return l;
    }

    int helper(vector<int> & nums, int left, int right, int lower, int upper){
        int cnt = 0;
        int mid = (left+right)/2;
        if(left == right) {
            return nums[left] >= lower && nums[left] <= upper;
        }
        vector<long> prefix;
        long sum = 0;
        for(int i = mid+1;i <= right;i++){
            // prefix sum of 2nd subarray
            sum += nums[i];
            prefix.push_back(sum);
        }
        sort(prefix.begin(), prefix.end());

        sum = 0; // suffix sum
        for(int i = mid;i >= left;i--){
            sum += (long)nums[i];
            cnt += nextGreatest(prefix, (long)upper-sum) - firstOccur(prefix, (long)lower-sum);
        }
        return helper(nums, left, mid, lower, upper) + helper(nums, mid+1, right, lower, upper)
        + cnt;
    }

    int countRangeSum(vector<int>& nums, int lower, int upper) {
        int n = nums.size();
        
        // O(n^2) is very easy and brute force, but that don't fit as per req.
        return helper(nums, 0, n-1, lower, upper);
    }
};
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