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Transform to A[i] = nums1[i] - nums2[i] and count pairs i < j with A[i] <= A[j] + diff (equivalently A[i] - A[j] <= diff). This is a classic counting-within-bound problem for which divide-and-conquer (merge-sort counting) or a Fenwick tree with coordinate compression yields an O(n log n) solution.
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Leetcode 2426. Number of Pairs Satisfying Inequality
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