<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, each of size <code>n</code>, and an integer <code>diff</code>. Find the number of <strong>pairs</strong> <code>(i, j)</code> such that:</p>

<ul>
	<li><code>0 &lt;= i &lt; j &lt;= n - 1</code> <strong>and</strong></li>
	<li><code>nums1[i] - nums1[j] &lt;= nums2[i] - nums2[j] + diff</code>.</li>
</ul>

<p>Return<em> the <strong>number of pairs</strong> that satisfy the conditions.</em></p>

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

<pre>
<strong>Input:</strong> nums1 = [3,2,5], nums2 = [2,2,1], diff = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong>
There are 3 pairs that satisfy the conditions:
1. i = 0, j = 1: 3 - 2 &lt;= 2 - 2 + 1. Since i &lt; j and 1 &lt;= 1, this pair satisfies the conditions.
2. i = 0, j = 2: 3 - 5 &lt;= 2 - 1 + 1. Since i &lt; j and -2 &lt;= 2, this pair satisfies the conditions.
3. i = 1, j = 2: 2 - 5 &lt;= 2 - 1 + 1. Since i &lt; j and -3 &lt;= 2, this pair satisfies the conditions.
Therefore, we return 3.
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [3,-1], nums2 = [-2,2], diff = -1
<strong>Output:</strong> 0
<strong>Explanation:</strong>
Since there does not exist any pair that satisfies the conditions, we return 0.
</pre>

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

<ul>
	<li><code>n == nums1.length == nums2.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums1[i], nums2[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= diff &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def numberOfPairs(self, nums1: List[int], nums2: List[int], diff: int) -> int:
        nums1 = [a - b for a, b in zip(nums1, nums2)]
        ans = [0]

        def merge(l, r):
            left, right = r,l
            left.sort()
            right.sort()

            a = [0] * len(left)
            i = 0
            j = 0

            while i < len(left) and j < len(right):
                if left[i] + diff >= right[j]:
                    j += 1
                else:
                    a[i] = j
                    i += 1

            if i < len(left):
                for n in range(i, len(a)):
                    a[n] = j

            ans[0] += sum(a)
            return l + r

        def mergesort(left, right):
            if left == right:
                return [nums1[left]]

            mid = left + (right - left) // 2
            l = mergesort(left, mid)
            r = mergesort(mid + 1, right)
            
            return merge(l, r)

        mergesort(0, len(nums1) - 1)
        return ans[0]

        
40200