Number of Pairs Satisfying Inequality
<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>
Binary Indexed Tree; Segment Tree; Array; Binary Search; Divide and Conquer; Ordered Set; Merge Sort
class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, delta):
        while x <= self.n:
            self.c[x] += delta
            x += BinaryIndexedTree.lowbit(x)

    def query(self, x):
        s = 0
        while x:
            s += self.c[x]
            x -= BinaryIndexedTree.lowbit(x)
        return s


class Solution:
    def numberOfPairs(self, nums1: List[int], nums2: List[int], diff: int) -> int:
        tree = BinaryIndexedTree(10**5)
        ans = 0
        for a, b in zip(nums1, nums2):
            v = a - b
            ans += tree.query(v + diff + 40000)
            tree.update(v + 40000, 1)
        return ans
