Given a sorted array and a number x, find a pair in an 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
Naive Approach:- A simple solution is to consider every pair and keep track of the closest pair (the absolute difference between pair sum and x is minimum). Finally, print the closest pair. The time complexity of this solution is O(n2)
Implementation:-
C++
// Simple C++ program to find the pair with sum closest to a given no. #include <bits/stdc++.h> 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 //variable to store current minimum difference int temp = INT_MAX; //iterating over array for(int i=0;i<n-1;i++) { for(int j=i+1;j<n;j++) { //if found more closest pair if(abs(arr[i]+arr[j]-x)<temp) { res_l=i; res_r=j; temp=abs(arr[i]+arr[j]-x); } } } 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; } // Code By shubhamrajput6156 |
Java
// Simple Java program to find the pair // with sum closest to a given no. import java.util.*; class GFG { // Function to Prints the pair with sum closest to x public static void printClosest(int[] arr, int n, int x) { int res_l = 0, res_r = 0; // To store indexes of result pair // variable to store current minimum difference int temp = Integer.MAX_VALUE; // iterating over array for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { // if found more closest pair if (Math.abs(arr[i] + arr[j] - x) < temp) { res_l = i; res_r = j; temp = Math.abs(arr[i] + arr[j] - x); } } } System.out.println("The closest pair is " + arr[res_l] + " and " + arr[res_r]); } // Driver program to test above functions public static void main(String[] args) { int[] arr = { 10, 22, 28, 29, 30, 40 }; int x = 54; int n = arr.length; printClosest(arr, n, x); } } // This code is Contributed by Prasad Kandekar(prasad264) |
Python3
# Python3 program to find the pair # with sum # closest to a given no. import sys # Prints the pair with sum closest to x def printClosest(arr, n, x): res_l = res_r = 0 # To store indexes of result pair temp = sys.maxsize # variable to store current minimum difference for i in range(n-1): for j in range(i+1, n): # if found more closest pair if abs(arr[i] + arr[j] - x) < temp: res_l = i res_r = j temp = abs(arr[i] + arr[j] - x) print("The closest pair is", arr[res_l], "and", arr[res_r]) #driver code arr = [10, 22, 28, 29, 30, 40] x = 54n = len(arr) printClosest(arr, n, x) |
C#
// Simple C# program to find the pair with sum closest to a given no. using System; public class GFG { // Prints the pair with sum closest to x static void PrintClosest(int[] arr, int n, int x) { // To store indexes of result pair int res_l = 0, res_r = 0; // variable to store current minimum difference int temp = int.MaxValue; // iterating over array for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { // if found more closest pair if (Math.Abs(arr[i] + arr[j] - x) < temp) { res_l = i; res_r = j; temp = Math.Abs(arr[i] + arr[j] - x); } } } Console.WriteLine("The closest pair is " + arr[res_l] + " and " + arr[res_r]); } // Driver program to test above functions static public void Main(string[] args) { int[] arr = { 10, 22, 28, 29, 30, 40 }; int x = 54; int n = arr.Length; PrintClosest(arr, n, x); } } // This Code is Contributed by Prasad Kandekar(prasad264) |
Javascript
// Simple JavaScript program to find the pair with sum closest to a given number. // Function to find the pair with sum closest to x function printClosest(arr, n, x) { let res_l, res_r; // To store indexes of result pair // Variable to store current minimum difference let temp = Number.MAX_SAFE_INTEGER; // Iterating over array for (let i = 0; i < n - 1; i++) { for (let j = i + 1; j < n; j++) { // If found more closest pair if (Math.abs(arr[i] + arr[j] - x) < temp) { res_l = i; res_r = j; temp = Math.abs(arr[i] + arr[j] - x); } } } console.log("The closest pair is " + arr[res_l] + " and " + arr[res_r]); } // Driver program to test above function let arr = [10, 22, 28, 29, 30, 40]; let x = 54; let n = arr.length; printClosest(arr, n, x); // This code is contributed By Prajwal Kandekar |
Output
The closest pair is 22 and 30
Time Complexity:- O(N^2)
Auxiliary Space:- O(1)
Binary Search Approach:- The more efficient solution than the above approach is to use Binary Search because the given array is in a sorted format.
Step-by-step algorithm for implementing the above approach:
- Initialize variables:
- l and r to point to the first and last elements of the array, respectively.
- res_l and res_r to store the indexes of the closest pair.
- minDiff to store the current minimum difference.
- Iterate over the array using a loop:
- Set e to the current element.
- While left is less than or equal to right:
- Set mid to the middle element of the subarray.
- If arr[mid] + e is equal to x,
- set res_l to i, res_r to mid, and minDiff to 0. Break out of the loop.
- If abs(arr[mid] + e – x) is less than minDiff,
- set minDiff to abs(arr[mid] + e – x) and res_l to i
- and res_r to mid.
- If arr[mid] + e is less than x, set left to mid + 1.
- Otherwise, set right to mid – 1.
- Set left and right to point to the first and last elements of the remaining subarray, respectively.
- Print the pair with the values of arr[res_l] and arr[res_r].
C++
// C++ program to find the pair with sum closest // to a given no using above approach. #include <bits/stdc++.h> using namespace std; // Function to prints the pair with sum closest to x void closestPair(int arr[], int n, int x) { // Initialize variables pointing to the // first and last elements of the array int l = 0, r = n - 1; // To store indexes of result pair int res_l, res_r; // variable to store current minimum difference int minDiff = INT_MAX; // Iterate over the array using for (int i = 0; i < n; i++) { int e = arr[i]; // Use binary search to find the element // 'elem' in the array such that 'e+elem' // is closest to 'x'. int left = i + 1, right = n - 1; while (left <= right) { int mid = (left + right) / 2; if (arr[mid] + e == x) { res_l = i; res_r = mid; minDiff = 0; break; } // Check if this pair is closer than the // closest pair so far if (abs(arr[mid] + e - x) < minDiff) { minDiff = abs(arr[mid] + e - x); res_l = i; res_r = mid; } if (arr[mid] + e < x) { left = mid + 1; } else { right = mid - 1; } } } // Print the pair 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}; int x = 54; int n = sizeof(arr) / sizeof(arr[0]); // Function Call closestPair(arr, n, x); return 0; } // This Code is Contributed by Prasad Kandekar(prasad264) |
Java
import java.util.*; public class Main { public static void main(String[] args) { // Initialize the array int[] arr = { 10, 22, 28, 29, 30, 40 }; // Initialize the target sum int x = 54; // Get the length of the array int n = arr.length; // Call the closestPair function closestPair(arr, n, x); } // Function to find the pair with sum closest to a given // no using above approach public static void closestPair(int[] arr, int n, int x) { // Initialize variables pointing to the first and // last elements of the array int l = 0, r = n - 1; // To store indexes of result pair int res_l = 0, res_r = 0; // variable to store current minimum difference int minDiff = Integer.MAX_VALUE; // Iterate over the array for (int i = 0; i < n; i++) { int e = arr[i]; // Use binary search to find the element 'elem' // in the array such that 'e+elem' is closest to // 'x'. int left = i + 1, right = n - 1; while (left <= right) { int mid = (left + right) / 2; if (arr[mid] + e == x) { res_l = i; res_r = mid; minDiff = 0; break; } // Check if this pair is closer than the // closest pair so far if (Math.abs(arr[mid] + e - x) < minDiff) { minDiff = Math.abs(arr[mid] + e - x); res_l = i; res_r = mid; } if (arr[mid] + e < x) { left = mid + 1; } else { right = mid - 1; } } } // Print the pair System.out.println("The closest pair is " + arr[res_l] + " and " + arr[res_r]); } } |
Python3
# Python program to find the pair with sum closest # to a given no using above approach. import sys # Function to prints the pair with sum closest to x def closestPair(arr, n, x): # Initialize variables pointing to the # first and last elements of the array l, r = 0, n - 1 # To store indexes of result pair res_l, res_r = 0, 0 # variable to store current minimum difference minDiff = sys.maxsize # Iterate over the array using for i in range(n): e = arr[i] # Use binary search to find the element # 'elem' in the array such that 'e+elem' # is closest to 'x'. left, right = i + 1, n - 1 while left <= right: mid = (left + right) // 2 if arr[mid] + e == x: res_l = i res_r = mid minDiff = 0 break # Check if this pair is closer than the # closest pair so far if abs(arr[mid] + e - x) < minDiff: minDiff = abs(arr[mid] + e - x) res_l = i res_r = mid if arr[mid] + e < x: left = mid + 1 else: right = mid - 1 # Print the pair print("The closest pair is", arr[res_l], "and", arr[res_r]) # Driver program to test above functions arr = [10, 22, 28, 29, 30, 40] x = 54n = len(arr) # Function Call closestPair(arr, n, x) # This Code is Contributed by prasad264 |
C#
using System; class Program { // Function to prints the pair with sum closest to x static void closestPair(int[] arr, int n, int x) { // Initialize variables pointing to the // first and last elements of the array int l = 0, r = n - 1; // To store indexes of result pair int res_l = -1, res_r = -1; // variable to store current minimum difference int minDiff = int.MaxValue; // Iterate over the array using for (int i = 0; i < n; i++) { int e = arr[i]; // Use binary search to find the element // 'elem' in the array such that 'e+elem' // is closest to 'x'. int left = i + 1, right = n - 1; while (left <= right) { int mid = (left + right) / 2; if (arr[mid] + e == x) { res_l = i; res_r = mid; minDiff = 0; break; } // Check if this pair is closer than the // closest pair so far if (Math.Abs(arr[mid] + e - x) < minDiff) { minDiff = Math.Abs(arr[mid] + e - x); res_l = i; res_r = mid; } if (arr[mid] + e < x) { left = mid + 1; } else { right = mid - 1; } } } // Print the pair Console.WriteLine($"The closest pair is {arr[res_l]} and {arr[res_r]}"); } // Driver program to test above functions static void Main(string[] args) { int[] arr = { 10, 22, 28, 29, 30, 40 }; int x = 54; int n = arr.Length; // Function Call closestPair(arr, n, x); } } |
Javascript
// JavaScript program to find the pair with sum closest // to a given no using above approach. // Function to prints the pair with sum closest to x function closestPair(arr, n, x) { // Initialize variables pointing to the // first and last elements of the array let l = 0, r = n - 1; // To store indexes of result pair let res_l, res_r; // variable to store current minimum difference let minDiff = Number.MAX_SAFE_INTEGER; // Iterate over the array using for (let i = 0; i < n; i++) { let e = arr[i]; // Use binary search to find the element // 'elem' in the array such that 'e+elem' // is closest to 'x'. let left = i + 1, right = n - 1; while (left <= right) { let mid = Math.floor((left + right) / 2); if (arr[mid] + e == x) { res_l = i; res_r = mid; minDiff = 0; break; } // Check if this pair is closer than the // closest pair so far if (Math.abs(arr[mid] + e - x) < minDiff) { minDiff = Math.abs(arr[mid] + e - x); res_l = i; res_r = mid; } if (arr[mid] + e < x) { left = mid + 1; } else { right = mid - 1; } } } // Print the pair console.log(`The closest pair is ${arr[res_l]} and ${arr[res_r]}`); } // Driver program to test above functions let arr = [10, 22, 28, 29, 30, 40]; let x = 54; let n = arr.length; // Function Call closestPair(arr, n, x); |
Output
The closest pair is 22 and 30
Complexity Analysis:
Time Complexity: O(n log n), because we are using a binary search algorithm to search for the pair, and for each element, we are performing a binary search, which has a time complexity of O(logn). Hence, the total time complexity of the approach becomes O(n log n).
Auxiliary Space: O(1), because we are not using any extra space to store the elements of the array or the result. The only extra space used is for storing some variables, which is constant and does not depend on the size of the input.
Efficient Approach:- An efficient solution can find the pair in O(n) time. The idea is similar to method 1 of this post. The following is a 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) If(arr[l] + arr[r] < sum ) then l++
(c) Else r--
Following is the implementation of the above algorithm.
C++14
// Simple C++ program to find the pair with sum closest to a given no. #include <bits/stdc++.h> 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; } // Code By Mayur Patil |
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 closest 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*/ |
Python3
# Python3 program to find the pair # with sum # closest to a given no. # A sufficiently large value greater # than any # element in the input array MAX_VAL = 1000000000 #Prints the pair with sum closest to x def printClosest(arr, n, x): # To store indexes of result pair res_l, res_r = 0, 0 #Initialize left and right indexes # and difference between # pair sum and x l, r, diff = 0, n-1, MAX_VAL # 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 arr[l] + arr[r] > x: # If this pair has more sum, move to # smaller values. r -= 1 else: # Move to larger values l += 1 print('The closest pair is {} and {}' .format(arr[res_l], arr[res_r])) # Driver code to test above if __name__ == "__main__": arr = [10, 22, 28, 29, 30, 40] n = len(arr) x=54 printClosest(arr, n, x) # This code is contributed by Tuhin Patra |
C#
// C# program to find pair with sum closest to x using System; class GFG { // Prints the pair with sum closest to x static void printClosest(int []arr, int n, int x) { // To store indexes of result pair int res_l = 0, res_r = 0; // Initialize left and right indexes and // difference between pair sum and x int l = 0, r = n-1, diff = int.MaxValue; // 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++; } Console.Write(" The closest pair is " + arr[res_l] + " and " + arr[res_r]); } // Driver program to test above function public static void Main() { int []arr = {10, 22, 28, 29, 30, 40}; int x = 54; int n = arr.Length; printClosest(arr, n, x); } } // This code is contributed by nitin mittal. |
PHP
<?php // Simple PHP program to find the // pair with sum closest to a // given no. // Prints the pair with // sum closest to x function printClosest($arr, $n, $x) { // To store indexes // of result pair $res_l; $res_r; // Initialize left and right // indexes and difference between // pair sum and x $l = 0; $r = $n - 1; $diff = PHP_INT_MAX; // While there are elements // between l and r while ($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--; // Move to larger values else $l++; } echo " The closest pair is " , $arr[$res_l] ," and " , $arr[$res_r]; } // Driver Code $arr = array(10, 22, 28, 29, 30, 40); $x = 54; $n = count($arr); printClosest($arr, $n, $x); // This code is contributed by anuj_67. ?> |
Javascript
<script> // JavaScript program to find pair // with sum closest to x // Prints the pair with sum closest to x function printClosest(arr,n,x) { // To store indexes of result pair let res_l=0, res_r=0; // Initialize left and right indexes // and difference between // pair sum and x let l = 0, r = n-1, diff = Number.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++; } document.write( " The closest pair is "+arr[res_l]+" and "+ arr[res_r] ); } // Driver program to test above function let arr = [10, 22, 28, 29, 30, 40], x = 54; let n = arr.length; printClosest(arr, n, x); // This code is contributed by sravan kumar </script> |
Output
The closest pair is 22 and 30
Time Complexity: O(n), where n is the length of an Array.
Auxiliary Space: O(1)
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