<p>Given an array of integers <code>nums</code> and an integer <code>k</code>, return <em>the total number of subarrays whose sum equals to</em> <code>k</code>.</p>

<p>A subarray is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,1,1], k = 2
<strong>Output:</strong> 2
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1,2,3], k = 3
<strong>Output:</strong> 2
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>-1000 &lt;= nums[i] &lt;= 1000</code></li>
	<li><code>-10<sup>7</sup> &lt;= k &lt;= 10<sup>7</sup></code></li>
</ul>

class Solution:
    def subarraySum(self, nums: List[int], k: int) -> int:
        prefix_sum = [0]
        for x in nums:
            prefix_sum.append(x + prefix_sum[-1])
        def check(sums, target, i, j) -> int:
            if i == j:
                return 0
            mid = (i + j) // 2
            ans = check(sums, target, i, mid) + check(sums, target, mid + 1, j)
            l, r = mid + 1, mid + 1
            for p in range(i, mid + 1):
                while l <= j and sums[l] - sums[p] < k: l += 1
                while r <= j and sums[r] - sums[p] <= k: r += 1
                ans += r - l
            sums[i:j + 1] = sorted(sums[i: j + 1])
            return ans
        return check(prefix_sum, k, 0, len(nums))
18500