Solution Explanation: Counting Pairs in Two Arrays

This problem asks us to count the number of pairs of indices (i, j) where i < j and nums1[i] + nums1[j] > nums2[i] + nums2[j]. The most efficient approach leverages sorting and a two-pointer technique.

1. Transformation:

The core inequality, nums1[i] + nums1[j] > nums2[i] + nums2[j], can be rewritten as:

nums1[i] - nums2[i] + nums1[j] - nums2[j] > 0

Let's define a new array diff where diff[i] = nums1[i] - nums2[i]. Now the inequality becomes:

diff[i] + diff[j] > 0

This simplification is crucial for the efficiency of the algorithm.

2. Sorting:

We sort the diff array. This allows us to efficiently find pairs that satisfy the condition using two pointers.

3. Two-Pointer Approach:

• Initialize two pointers: left (pointing to the beginning of the sorted diff array) and right (pointing to the end).

• Iterate: The core logic lies in the while loop (while (left < right)).

• Inner Loop: The inner while loop (while (left < right && diff[left] + diff[right] <= 0)) efficiently skips pairs that don't satisfy the condition. If the sum is less than or equal to 0, it means that any pair with diff[left] will also fail to satisfy the inequality with any element greater than diff[right] (because the array is sorted). Thus we increment left.

• Counting Pairs: Once we find a pair where diff[left] + diff[right] > 0, we know that all pairs (diff[left] with diff[left+1], diff[left+2], ..., diff[right]) satisfy the condition. The number of such pairs is right - left. We add this count to our result.

• Pointer Movement: After counting the pairs, we decrement right to consider pairs with a smaller element at the right pointer.

4. Time and Space Complexity:

• Time Complexity: O(n log n), dominated by the sorting step. The two-pointer traversal takes linear time, O(n).
• Space Complexity: O(n) in the worst case, used for storing the diff array. In-place sorting algorithms could reduce this space slightly depending on the implementation.

Code Examples (Python):

def countPairs(nums1: list[int], nums2: list[int]) -> int:
    diff = [nums1[i] - nums2[i] for i in range(len(nums1))]
    diff.sort()
    left, right = 0, len(diff) - 1
    count = 0
    while left < right:
        while left < right and diff[left] + diff[right] <= 0:
            left += 1
        count += right - left
        right -= 1
    return count
 

The code in other languages (Java, C++, Go, TypeScript, Rust, JavaScript) follows the same algorithmic structure, only differing in syntax and specific library functions used for sorting and array manipulation. The time and space complexity remain the same across all implementations.