Find the closest pair from two sorted arrays in c++

Suppose we have two sorted arrays and a number x, we have to find the pair whose sum is closest to x. And the pair has an element from each array. We have two arrays A1 [0..m-1] and A2 [0..n-1], and another value x. We have to find the pair A1[i] + A2[j] such that absolute value of (A1[i] + A2[j] – x) is minimum. So if A1 = [1, 4, 5, 7], and A2 = [10, 20, 30, 40], and x = 32, then output will be 1 and 30.

We will start from left of A1 and right from A2, then follow these steps to find such pair

Initialize diff, this will hold the difference between pair and x
Initialize two pointer left := 0 and right := n – 1
While left <= m and right >= 0, do
- if |A1[left] + A2[right] – sum| < diff, then
  - update diff and result
- if (A1[left] + A2[right]) < sum, then
  - increase left by 1
- otherwise
  - Decrease right by 1
Display the result

Example

#include<iostream>
#include<cmath>
using namespace std;

void findClosestPair(int A1[], int A2[], int m, int n, int x) {
   int diff = INT_MAX;
   int left_res, right_res;

   int left = 0, right = n-1;
   while (left<m && right>=0) {
      if (abs(A1[left] + A2[right] - x) < diff) {
         left_res = left;
         right_res = right;
         diff = abs(A1[left] + A2[right] - x);
      }

      if (A1[left] + A2[right] > x)
      right--;
      else
      left++;
   }

   cout << "The closest pair is [" << A1[left_res] << ", "<< A2[right_res] << "]";
}

int main() {
   int ar1[] = {1, 4, 5, 7};
   int ar2[] = {10, 20, 30, 40};

   int m = sizeof(ar1)/sizeof(ar1[0]);
   int n = sizeof(ar2)/sizeof(ar2[0]);

   int x = 32;
   findClosestPair(ar1, ar2, m, n, x);
}