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 = 0, do
  • if |A1[left] + A2[right] – sum|
  • update diff and result
• if (A1[left] + A2[right])
• increase left by 1

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);
}

Output

The closest pair is [1, 30]