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# Find closest number in Sorted array

• Difficulty Level : Medium
• Last Updated : 27 Dec, 2022

Given an array of sorted integers. We need to find the closest value to the given number. Array may contain duplicate values and negative numbers.

Examples:

```Input : arr[] = {1, 2, 4, 5, 6, 6, 8, 9}
Target number = 11
Output : 9
9 is closest to 11 in given array

Input :arr[] = {2, 5, 6, 7, 8, 8, 9};
Target number = 4
Output : 5
5 is closest to 4 in given array

Input :arr[] = {2, 5, 6, 7, 8, 8, 9, 15, 19, 22, 32};
Target number = 34
Output : 32
32 is closest to 34 in given array```
Recommended Practice

A simple solution is to traverse through the given array and keep track of absolute difference of current element with every element. Finally return the element that has minimum absolute difference.

An efficient solution is to use Binary Search.

## C++

 `// CPP program to find element``// closest to given target using binary search.``#include ``using` `namespace` `std;` `int` `getClosest(``int``, ``int``, ``int``);` `// Returns element closest to target in arr[]``int` `findClosest(``int` `arr[], ``int` `n, ``int` `target)``{``    ``// Corner cases``  ``//left-side case``    ``if` `(target <= arr)``        ``return` `arr;``  ``//right-side case``    ``if` `(target >= arr[n - 1])``        ``return` `arr[n - 1];` `    ``// Doing binary search``    ``int` `i = 0, j = n, mid = 0;``    ``while` `(i < j) {``        ``mid = (i + j) / 2;` `        ``if` `(arr[mid] == target)``            ``return` `arr[mid];` `        ``/* If target is less than array element,``            ``then search in left */``        ``if` `(target < arr[mid]) {` `            ``// If target is greater than previous``            ``// to mid, return closest of two``            ``if` `(mid > 0 && target > arr[mid - 1])``                ``return` `getClosest(arr[mid - 1],``                                  ``arr[mid], target);  ``            ``j = mid;``        ``}``      ``/* Repeat for left half */` `        ``// If target is greater than mid``        ``else` `{``            ``if` `(mid < n - 1 && target < arr[mid + 1])``                ``return` `getClosest(arr[mid],``                                  ``arr[mid + 1], target);``            ``// update i``            ``i = mid + 1;``        ``}``    ``}` `    ``// Only single element left after search``    ``return` `arr[mid];``}` `// Method to compare which one is the more close.``// We find the closest by taking the difference``// between the target and both values. It assumes``// that val2 is greater than val1 and target lies``// between these two.``int` `getClosest(``int` `val1, ``int` `val2,``               ``int` `target)``{``    ``if` `(target - val1 >= val2 - target)``        ``return` `val2;``    ``else``        ``return` `val1;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 4, 5, 6, 6, 8, 8, 9 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `target = 11;``    ``cout << (findClosest(arr, n, target));``}` `// This code is contributed bu Smitha Dinesh Semwal`

## Java

 `// Java program to find element closest to given target using binary search.``import` `java.util.*;``import` `java.lang.*;``import` `java.io.*;` `class` `FindClosestNumber {``    ` `    ``// Returns element closest to target in arr[]``    ``public` `static` `int` `findClosest(``int` `arr[], ``int` `target)``    ``{``        ``int` `n = arr.length;` `        ``// Corner cases``        ``if` `(target <= arr[``0``])``            ``return` `arr[``0``];``        ``if` `(target >= arr[n - ``1``])``            ``return` `arr[n - ``1``];` `        ``// Doing binary search``        ``int` `i = ``0``, j = n, mid = ``0``;``        ``while` `(i < j) {``            ``mid = (i + j) / ``2``;` `            ``if` `(arr[mid] == target)``                ``return` `arr[mid];` `            ``/* If target is less than array element,``               ``then search in left */``            ``if` `(target < arr[mid]) {``       ` `                ``// If target is greater than previous``                ``// to mid, return closest of two``                ``if` `(mid > ``0` `&& target > arr[mid - ``1``])``                    ``return` `getClosest(arr[mid - ``1``],``                                  ``arr[mid], target);``                ` `                ``/* Repeat for left half */``                ``j = mid;             ``            ``}` `            ``// If target is greater than mid``            ``else` `{``                ``if` `(mid < n-``1` `&& target < arr[mid + ``1``])``                    ``return` `getClosest(arr[mid],``                          ``arr[mid + ``1``], target);               ``                ``i = mid + ``1``; ``// update i``            ``}``        ``}` `        ``// Only single element left after search``        ``return` `arr[mid];``    ``}` `    ``// Method to compare which one is the more close``    ``// We find the closest by taking the difference``    ``//  between the target and both values. It assumes``    ``// that val2 is greater than val1 and target lies``    ``// between these two.``    ``public` `static` `int` `getClosest(``int` `val1, ``int` `val2,``                                         ``int` `target)``    ``{``        ``if` `(target - val1 >= val2 - target)``            ``return` `val2;       ``        ``else``            ``return` `val1;       ``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``2``, ``4``, ``5``, ``6``, ``6``, ``8``, ``9` `};``        ``int` `target = ``11``;``        ``System.out.println(findClosest(arr, target));``    ``}``}`

## Python3

 `# Python3 program to find element``# closest to given target using binary search.` `# Returns element closest to target in arr[]``def` `findClosest(arr, n, target):` `    ``# Corner cases``    ``if` `(target <``=` `arr[``0``]):``        ``return` `arr[``0``]``    ``if` `(target >``=` `arr[n ``-` `1``]):``        ``return` `arr[n ``-` `1``]` `    ``# Doing binary search``    ``i ``=` `0``; j ``=` `n; mid ``=` `0``    ``while` `(i < j):``        ``mid ``=` `(i ``+` `j) ``/``/` `2` `        ``if` `(arr[mid] ``=``=` `target):``            ``return` `arr[mid]` `        ``# If target is less than array``        ``# element, then search in left``        ``if` `(target < arr[mid]) :` `            ``# If target is greater than previous``            ``# to mid, return closest of two``            ``if` `(mid > ``0` `and` `target > arr[mid ``-` `1``]):``                ``return` `getClosest(arr[mid ``-` `1``], arr[mid], target)` `            ``# Repeat for left half``            ``j ``=` `mid``        ` `        ``# If target is greater than mid``        ``else` `:``            ``if` `(mid < n ``-` `1` `and` `target < arr[mid ``+` `1``]):``                ``return` `getClosest(arr[mid], arr[mid ``+` `1``], target)``                ` `            ``# update i``            ``i ``=` `mid ``+` `1``        ` `    ``# Only single element left after search``    ``return` `arr[mid]`  `# Method to compare which one is the more close.``# We find the closest by taking the difference``# between the target and both values. It assumes``# that val2 is greater than val1 and target lies``# between these two.``def` `getClosest(val1, val2, target):` `    ``if` `(target ``-` `val1 >``=` `val2 ``-` `target):``        ``return` `val2``    ``else``:``        ``return` `val1` `# Driver code``arr ``=` `[``1``, ``2``, ``4``, ``5``, ``6``, ``6``, ``8``, ``9``]``n ``=` `len``(arr)``target ``=` `11``print``(findClosest(arr, n, target))` `# This code is contributed by Smitha Dinesh Semwal`

## C#

 `// C# program to find element``// closest to given target using binary search.``using` `System;` `class` `GFG``{``    ` `    ``// Returns element closest``    ``// to target in arr[]``    ``public` `static` `int` `findClosest(``int` `[]arr,``                                  ``int` `target)``    ``{``        ``int` `n = arr.Length;` `        ``// Corner cases``        ``if` `(target <= arr)``            ``return` `arr;``        ``if` `(target >= arr[n - 1])``            ``return` `arr[n - 1];` `        ``// Doing binary search``        ``int` `i = 0, j = n, mid = 0;``        ``while` `(i < j)``        ``{``            ``mid = (i + j) / 2;` `            ``if` `(arr[mid] == target)``                ``return` `arr[mid];` `            ``/* If target is less``            ``than array element,``            ``then search in left */``            ``if` `(target < arr[mid])``            ``{``        ` `                ``// If target is greater``                ``// than previous to mid,``                ``// return closest of two``                ``if` `(mid > 0 && target > arr[mid - 1])``                    ``return` `getClosest(arr[mid - 1],``                                 ``arr[mid], target);``                ` `                ``/* Repeat for left half */``                ``j = mid;            ``            ``}` `            ``// If target is``            ``// greater than mid``            ``else``            ``{``                ``if` `(mid < n-1 && target < arr[mid + 1])``                    ``return` `getClosest(arr[mid],``                         ``arr[mid + 1], target);        ``                ``i = mid + 1; ``// update i``            ``}``        ``}` `        ``// Only single element``        ``// left after search``        ``return` `arr[mid];``    ``}` `    ``// Method to compare which one``    ``// is the more close We find the``    ``// closest by taking the difference``    ``// between the target and both``    ``// values. It assumes that val2 is``    ``// greater than val1 and target``    ``// lies between these two.``    ``public` `static` `int` `getClosest(``int` `val1, ``int` `val2,``                                 ``int` `target)``    ``{``        ``if` `(target - val1 >= val2 - target)``            ``return` `val2;    ``        ``else``            ``return` `val1;    ``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `[]arr = {1, 2, 4, 5,``                     ``6, 6, 8, 9};``        ``int` `target = 11;``        ``Console.WriteLine(findClosest(arr, target));``    ``}``}` `// This code is contributed by anuj_67.`

## PHP

 `= ``\$arr``[``\$n` `- 1])``        ``return` `\$arr``[``\$n` `- 1];` `    ``// Doing binary search``    ``\$i` `= 0;``    ``\$j` `= ``\$n``;``    ``\$mid` `= 0;``    ``while` `(``\$i` `< ``\$j``)``    ``{``        ``\$mid` `= (``\$i` `+ ``\$j``) / 2;` `        ``if` `(``\$arr``[``\$mid``] == ``\$target``)``            ``return` `\$arr``[``\$mid``];` `        ``/* If target is less than array element,``            ``then search in left */``        ``if` `(``\$target` `< ``\$arr``[``\$mid``])``        ``{` `            ``// If target is greater than previous``            ``// to mid, return closest of two``            ``if` `(``\$mid` `> 0 && ``\$target` `> ``\$arr``[``\$mid` `- 1])``                ``return` `getClosest(``\$arr``[``\$mid` `- 1],``                                  ``\$arr``[``\$mid``], ``\$target``);` `            ``/* Repeat for left half */``            ``\$j` `= ``\$mid``;``        ``}` `        ``// If target is greater than mid``        ``else``        ``{``            ``if` `(``\$mid` `< ``\$n` `- 1 &&``                ``\$target` `< ``\$arr``[``\$mid` `+ 1])``                ``return` `getClosest(``\$arr``[``\$mid``],``                                  ``\$arr``[``\$mid` `+ 1], ``\$target``);``            ``// update i``            ``\$i` `= ``\$mid` `+ 1;``        ``}``    ``}` `    ``// Only single element left after search``    ``return` `\$arr``[``\$mid``];``}` `// Method to compare which one is the more close.``// We find the closest by taking the difference``// between the target and both values. It assumes``// that val2 is greater than val1 and target lies``// between these two.``function` `getClosest(``\$val1``, ``\$val2``, ``\$target``)``{``    ``if` `(``\$target` `- ``\$val1` `>= ``\$val2` `- ``\$target``)``        ``return` `\$val2``;``    ``else``        ``return` `\$val1``;``}` `// Driver code``\$arr` `= ``array``( 1, 2, 4, 5, 6, 6, 8, 9 );``\$n` `= sizeof(``\$arr``);``\$target` `= 11;``echo` `(findClosest(``\$arr``, ``\$n``, ``\$target``));` `// This code is contributed bu Sachin.``?>`

## Javascript

 ``

Output:

`9`

Time Complexity: O(log  n) (Due to Binary Search)
Auxiliary Space: O(log n) (implicit stack is created due to recursion)

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