Number of Pairs Satisfying Inequality

Problem Description

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
• nums1[i] - nums1[j] <= nums2[i] - nums2[j] + diff.

Return the number of pairs that satisfy the conditions.

Key Insights

• The problem can be transformed into counting the number of valid pairs based on a derived inequality.
• The derived condition can be simplified to finding pairs where one value is less than or equal to another, allowing for efficient counting.
• Utilizing sorting and binary search can significantly optimize the solution compared to a naive O(n^2) approach.

Space and Time Complexity

Time Complexity: O(n log n)
Space Complexity: O(n)

Solution

To solve the problem, we can transform the conditions into a more manageable form. We can define a new array transformed where each element is calculated as:

transformed[i] = nums1[i] - nums2[i] + diff.

The goal then becomes counting how many pairs (i, j) exist such that transformed[j] >= transformed[i] for i < j. We can achieve this by sorting the transformed array and using binary search to count valid pairs efficiently.

1. Create a new array, transformed, based on the original arrays and the diff.
2. Sort this transformed array.
3. For each element in the sorted array, use binary search to find how many elements that come after the current element are greater than or equal to it.

This approach reduces the number of comparisons needed and allows us to efficiently count valid pairs.

Code Solutions