<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:
        n=len(nums1)
        nums=[nums1[i]-nums2[i] for i in range(n)]
        def merge(l,mid,r):
            i,j,count=l,mid+1,0
            temp=[]
            while i<=mid and j<=r:
                if nums[i]<=nums[j]+diff:
                    count+=r-j+1
                    i+=1
                else:
                    j+=1
            i,j=l,mid+1
            while i<=mid and j<=r:
                if nums[i]<nums[j]:
                    temp.append(nums[i])
                    i+=1
                else:
                    temp.append(nums[j])
                    j+=1
            while i<=mid:
                temp.append(nums[i])
                i+=1
            while j<=r:
                temp.append(nums[j])
                j+=1
            nums[l:r+1]=temp
            return count
        def merge_sort(l,r):
            count=0
            if l<r:
                mid=(l+r)//2
                count+=merge_sort(l,mid)
                count+=merge_sort(mid+1,r)
                count+=merge(l,mid,r)
            return count
        return merge_sort(0,n-1)
                
40900