Count of Pairs in Arrays a and b with Sum Comparison

Given two arrays a and b of the same size. Find the count of total pair in array a, such that the sum of elements of pair of array a is greater than the sum of elements of the corresponding pair of array b.

Example:

Input:
a={5,7,4,8,11}    
b={3,6,1,9,13}

Output: 6

Explanation: 
The pairs are (5,7)>(3,6), (5,7)>(3,6), (5,4)>(3,1), (5,8)>(3,9), (7,4)>(6,1), (4,8)>(1,9), (4,11)>(13,1)

Approach: It is only possible when we count the total number of pairs of difference array( a_i– b_i), Whose sum is positive.

Generate a difference array diff_i=a_i-b_{i .}
Sort the difference array .
Take a variable ans=0 which stores required output.
Iterate over the positive integer of the difference array, such that for every ith iteration find an index j such that a[j]>-a[i] and increment ans with (i-j), since all the elements before j when paired with a[i] gives -ve sum.
print ans.

Implementation in C++:

#include <bits/stdc++.h>
using namespace std;
int positivepair(vector<int> &diff)
{
	// Sorting the given array
	sort(diff.begin(), diff.end());

	// Variable to store the count of pairs
	int totalpair = 0;

	// Loop to iterate through the array
	for (int i = 0; i < diff.size(); i++)
	{
		//since we have to consider only positive number we ignor negative number
		if (diff[i] <= 0)
			continue;

		// Finding the index j such that all numbers before index j give -ve sum when paired with diff[i]
		int j = lower_bound(diff.begin(), diff.end(), -diff[i] + 1) - diff.begin();

		totalpair += i - j;
	}

	return totalpair;

}

//array to find the count of pair of a whose sum is greater than sum of crrosponding pair of b
int solve(vector<int> &a, vector<int> &b)
{
	//array to take diffrence of every element of a and b respectively.
	vector<int> diff;
	//loop to iterate through elements of both array and take the diffrence
	for (int i = 0; i < a.size(); i++)
	{
		diff.push_back(a[i] - b[i]);
	}

	return positivepair(diff);
}

//driver code
int main()
{
	vector<int> a = { 5, 7, 4, 8, 11 };

	vector<int> b = { 3, 6, 1, 9, 13 };

	cout << solve(a, b);
}

Output:

Time complexity: O(n * log(n))
Space complexity: O(n)