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# Find closest number in Sorted array

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[0])``        ``return` `arr[0];``  ``//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[0]);``    ``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[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()``    ``{``        ``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 by 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)

Approach 2: Using Two Pointers:

Another approach to solve this problem is to use two pointers technique, where we maintain two pointers left and right, and move them towards each other based on their absolute difference with target.

Initialize left = 0 and right = n-1, where n is the size of the array.
Loop while left < right
a. If the absolute difference between arr[left] and target is less than or equal to the absolute difference between arr[right] and target, move left pointer one step to the right, i.e. left++
b. Else, move right pointer one step to the left, i.e. right–
Return arr[left], which will be the element closest to the target.

## C++

 `#include ``using` `namespace` `std;` `int` `findClosest(``int` `arr[], ``int` `n, ``int` `target)``{``    ``int` `left = 0, right = n - 1;``    ``while` `(left < right) {``        ``if` `(``abs``(arr[left] - target)``            ``<= ``abs``(arr[right] - target)) {``            ``right--;``        ``}``        ``else` `{``            ``left++;``        ``}``    ``}``    ``return` `arr[left];``}` `int` `main()``{``    ``int` `arr[] = { 1, 2, 4, 5, 6, 6, 8, 8, 9 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `target = 11;``    ``cout << findClosest(arr, n, target);``    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `Main {``    ``public` `static` `int` `findClosest(``int``[] arr, ``int` `n,``                                  ``int` `target)``    ``{``        ``int` `left = ``0``, right = n - ``1``;``        ``while` `(left < right) {``            ``if` `(Math.abs(arr[left] - target)``                ``<= Math.abs(arr[right] - target)) {``                ``right--;``            ``}``            ``else` `{``                ``left++;``            ``}``        ``}``        ``return` `arr[left];``    ``}` `    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``1``, ``2``, ``4``, ``5``, ``6``, ``6``, ``8``, ``8``, ``9` `};``        ``int` `n = arr.length;``        ``int` `target = ``11``;``        ``System.out.println(findClosest(arr, n, target));``    ``}``}`

## Python3

 `import` `sys`  `def` `findClosest(arr, n, target):``    ``left, right ``=` `0``, n ``-` `1``    ``while` `left < right:``        ``if` `abs``(arr[left] ``-` `target) <``=` `abs``(arr[right] ``-` `target):``            ``right ``-``=` `1``        ``else``:``            ``left ``+``=` `1``    ``return` `arr[left]`  `if` `__name__ ``=``=` `"__main__"``:``    ``arr ``=` `[``1``, ``2``, ``4``, ``5``, ``6``, ``6``, ``8``, ``8``, ``9``]``    ``n ``=` `len``(arr)``    ``target ``=` `11``    ``print``(findClosest(arr, n, target))`

## C#

 `using` `System;` `public` `class` `Program {``    ``static` `int` `FindClosest(``int``[] arr, ``int` `n, ``int` `target)``    ``{``        ``int` `left = 0, right = n - 1;``        ``while` `(left < right) {``            ``if` `(Math.Abs(arr[left] - target)``                ``<= Math.Abs(arr[right] - target)) {``                ``right--;``            ``}``            ``else` `{``                ``left++;``            ``}``        ``}``        ``return` `arr[left];``    ``}` `    ``static` `void` `Main(``string``[] args)``    ``{``        ``int``[] arr = { 1, 2, 4, 5, 6, 6, 8, 8, 9 };``        ``int` `n = arr.Length;``        ``int` `target = 11;``        ``Console.WriteLine(FindClosest(arr, n, target));``    ``}``}`

## Javascript

 `function` `findClosest(arr, n, target) {``  ``let left = 0, right = n-1;``  ``while` `(left < right) {``    ``if` `(Math.abs(arr[left] - target) <= Math.abs(arr[right] - target)) {``      ``right--;``    ``} ``else` `{``      ``left++;``    ``}``  ``}``  ``return` `arr[left];``}` `const arr = [1, 2, 4, 5, 6, 6, 8, 8, 9];``const n = arr.length;``const target = 11;``console.log(findClosest(arr, n, target));``// This code is contributed by shivhack999`

Output

`9`

Time Complexity: O(N)
Auxiliary Space: O(1)

Approach 3: Recursive Approach

• “findClosestRecursive” function takes an array “arr” , right and left indices of the current search range and integer traget.
• “findClosestRecursive” function usese binary search approach to recursively search  array.
• “findClosestRecursive” function also calculates the middle index of the current search range and recursively searches the left and right halves of the array.
• Base case – It occurs when there is only one element in an array. In that case,  function will simply returns that element.

## C++

 `#include ``using` `namespace` `std;` `int` `findClosestRecursive(``int` `arr[], ``int` `left, ``int` `right, ``int` `target) {``    ``// base case: when there is only one element in the array``    ``if` `(left == right) {``        ``return` `arr[left];``    ``}`` ` `    ``// calculate the middle index``    ``int` `mid = (left + right) / 2;`` ` `    ``// recursively search the left half of the array``    ``int` `leftClosest = findClosestRecursive(arr, left, mid, target);`` ` `    ``// recursively search the right half of the array``    ``int` `rightClosest = findClosestRecursive(arr, mid + 1, right, target);`` ` `    ``// compare the absolute differences of the closest elements in the left and right halves``    ``if` `(``abs``(leftClosest - target) <= ``abs``(rightClosest - target)) {``        ``return` `leftClosest;``    ``}``    ``else` `{``        ``return` `rightClosest;``    ``}``}` `int` `main() {``    ``int` `arr[] = {1, 2, 4, 5, 6, 6, 8, 8, 9};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `target = 11;``    ``cout << findClosestRecursive(arr, 0, n-1, target);``    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `Main {``  ``public` `static` `int` `findClosestRecursive(``int``[] arr,``                                         ``int` `left,``                                         ``int` `right,``                                         ``int` `target)``  ``{``    ` `    ``// base case: when there is only one element in the``    ``// array``    ``if` `(left == right) {``      ``return` `arr[left];``    ``}` `    ``// calculate the middle index``    ``int` `mid = (left + right) / ``2``;` `    ``// recursively search the left half of the array``    ``int` `leftClosest``      ``= findClosestRecursive(arr, left, mid, target);` `    ``// recursively search the right half of the array``    ``int` `rightClosest = findClosestRecursive(``      ``arr, mid + ``1``, right, target);` `    ``// compare the absolute differences of the closest``    ``// elements in the left and right halves``    ``if` `(Math.abs(leftClosest - target)``        ``<= Math.abs(rightClosest - target)) {``      ``return` `leftClosest;``    ``}``    ``else` `{``      ``return` `rightClosest;``    ``}``  ``}` `  ``public` `static` `void` `main(String[] args)``  ``{``    ``int``[] arr = { ``1``, ``2``, ``4``, ``5``, ``6``, ``6``, ``8``, ``8``, ``9` `};``    ``int` `n = arr.length;``    ``int` `target = ``11``;``    ``System.out.println(``      ``findClosestRecursive(arr, ``0``, n - ``1``, target));``  ``}``}`

## Javascript

 `function` `findClosestRecursive(arr, left, right, target) {``    ``// base case: when there is only one element in the array``    ``if` `(left == right) {``        ``return` `arr[left];``    ``}` `    ``// calculate the middle index``    ``let mid = Math.floor((left + right) / 2);` `    ``// recursively search the left half of the array``    ``let leftClosest = findClosestRecursive(arr, left, mid, target);` `    ``// recursively search the right half of the array``    ``let rightClosest = findClosestRecursive(arr, mid + 1, right, target);` `    ``// compare the absolute differences of the closest elements in the``    ``// left and right halves``    ``if` `(Math.abs(leftClosest - target) <= Math.abs(rightClosest - target)) {``        ``return` `leftClosest;``    ``} ``else` `{``        ``return` `rightClosest;``    ``}``}` `let arr = [1, 2, 4, 5, 6, 6, 8, 8, 9];``let n = arr.length;``let target = 11;``console.log(findClosestRecursive(arr, 0, n - 1, target));``// This code is contributed by user_dtewbxkn77n`

Output

`9`

Time Complexity: O(log n)
Auxiliary Space: O(log n)

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