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Find 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
- if |A1[left] + A2[right] – sum| < diff, then
- 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); }
Output
The closest pair is [1, 30]
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