2426 Number of Pairs Satisfying Inequality

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

• 0 <= i < j <= n - 1 and
• nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff. Return the number of pairs that satisfy the conditions.

Example 1:

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

Example 2:

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

Constraints:

• n == nums1.length == nums2.length
• 2 <= n <= 105
• -104 <= nums1[i], nums2[i] <= 104
• -104 <= diff <= 104

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class Solution:
    def numberOfPairs(self, nums1: List[int], nums2: List[int], diff: int) -> int:
        res = 0
        sorted_list = []
        for n1, n2 in zip(nums1, nums2):
            # nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff
            # (nums1[i] - nums2[i]) <= (nums1[j] - nums2[j]) + diff
            # find the number of i for each j
            res += bisect.bisect_right(sorted_list, n1 - n2 + diff)
            bisect.insort(sorted_list, n1 - n2)
        return res