Given a sorted array and a number x, find a pair in array whose sum is closest to x.
Examples:
Input: arr[] = {10, 22, 28, 29, 30, 40}, x = 54 Output: 22 and 30 Input: arr[] = {1, 3, 4, 7, 10}, x = 15 Output: 4 and 10
A simple solution is to consider every pair and keep track of closest pair (absolute difference between pair sum and x is minimum). Finally print the closest pair. Time complexity of this solution is O(n2)
An efficient solution can find the pair in O(n) time. The idea is similar to method 2 of this post. Following is detailed algorithm.
1) Initialize a variable diff as infinite (Diff is used to store the difference between pair and x). We need to find the minimum diff. 2) Initialize two index variables l and r in the given sorted array. (a) Initialize first to the leftmost index: l = 0 (b) Initialize second the rightmost index: r = n-1 3) Loop while l < r. (a) If abs(arr[l] + arr[r] - sum) < diff then update diff and result (b) Else if(arr[l] + arr[r] < sum ) then l++ (c) Else r--
Following is C++ implementation of above algorithm.
C++
// Simple C++ program to find the pair with sum closest to a given no. #include <iostream> #include <climits> #include <cstdlib> using namespace std; // Prints the pair with sum closest to x void printClosest(int arr[], int n, int x) { int res_l, res_r; // To store indexes of result pair // Initialize left and right indexes and difference between // pair sum and x int l = 0, r = n-1, diff = INT_MAX; // While there are elements between l and r while (r > l) { // Check if this pair is closer than the closest pair so far if (abs(arr[l] + arr[r] - x) < diff) { res_l = l; res_r = r; diff = abs(arr[l] + arr[r] - x); } // If this pair has more sum, move to smaller values. if (arr[l] + arr[r] > x) r--; else // Move to larger values l++; } cout <<" The closest pair is " << arr[res_l] << " and " << arr[res_r]; } // Driver program to test above functions int main() { int arr[] = {10, 22, 28, 29, 30, 40}, x = 54; int n = sizeof(arr)/sizeof(arr[0]); printClosest(arr, n, x); return 0; }
Java
// Java program to find pair with sum closest to x import java.io.*; import java.util.*; import java.lang.Math; class CloseSum { // Prints the pair with sum cloest to x static void printClosest(int arr[], int n, int x) { int res_l=0, res_r=0; // To store indexes of result pair // Initialize left and right indexes and difference between // pair sum and x int l = 0, r = n-1, diff = Integer.MAX_VALUE; // While there are elements between l and r while (r > l) { // Check if this pair is closer than the closest pair so far if (Math.abs(arr[l] + arr[r] - x) < diff) { res_l = l; res_r = r; diff = Math.abs(arr[l] + arr[r] - x); } // If this pair has more sum, move to smaller values. if (arr[l] + arr[r] > x) r--; else // Move to larger values l++; } System.out.println(" The closest pair is "+arr[res_l]+" and "+ arr[res_r]); } // Driver program to test above function public static void main(String[] args) { int arr[] = {10, 22, 28, 29, 30, 40}, x = 54; int n = arr.length; printClosest(arr, n, x); } } /*This code is contributed by Devesh Agrawal*/
Output:
The closest pair is 22 and 30
This article is contributed by Harsh. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
Recommended Posts:
- Count 1’s in a sorted binary array
- Find the closest pair from two sorted arrays
- Given an array A[] and a number x, check for pair in A[] with sum as x
- Find common elements in three sorted arrays
- Print All Distinct Elements of a given integer array
- Maximize array elements upto given number
- Longest Palindromic Substring using Palindromic Tree | Set 3
- Making zero array by decrementing pairs of adjacent
- k-th missing element in sorted array
- Euler Tour | Subtree Sum using Segment Tree
Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.