# Search in an almost sorted array

• Difficulty Level : Medium
• Last Updated : 22 Jul, 2022

Given an array which is sorted, but after sorting some elements are moved to either of the adjacent positions, i.e., arr[i] may be present at arr[i+1] or arr[i-1]. Write an efficient function to search an element in this array. Basically the element arr[i] can only be swapped with either arr[i+1] or arr[i-1].
For example consider the array {2, 3, 10, 4, 40}, 4 is moved to next position and 10 is moved to previous position.
Example :

```Input: arr[] =  {10, 3, 40, 20, 50, 80, 70}, key = 40
Output: 2
Output is index of 40 in given array

Input: arr[] =  {10, 3, 40, 20, 50, 80, 70}, key = 90
Output: -1
-1 is returned to indicate element is not present```

A simple solution is to linearly search the given key in given array. Time complexity of this solution is O(n). We can modify binary search to do it in O(Logn) time.
The idea is to compare the key with middle 3 elements, if present then return the index. If not present, then compare the key with middle element to decide whether to go in left half or right half. Comparing with middle element is enough as all the elements after mid+2 must be greater than element mid and all elements before mid-2 must be smaller than mid element.
Following is the implementation of this approach.

## C++

 `// C++ program to find an element``// in an almost sorted array``#include ` `// A recursive binary search based function.``// It returns index of x in given array``// arr[l..r] is present, otherwise -1``int` `binarySearch(``int` `arr[], ``int` `l, ``int` `r, ``int` `x)``{``if` `(r >= l)``{``        ``int` `mid = l + (r - l) / 2;` `        ``// If the element is present at``        ``// one of the middle 3 positions``        ``if` `(arr[mid] == x)``            ``return` `mid;``        ``if` `(mid > l && arr[mid - 1] == x)``            ``return` `(mid - 1);``        ``if` `(mid < r && arr[mid + 1] == x)``            ``return` `(mid + 1);` `        ``// If element is smaller than mid, then``        ``// it can only be present in left subarray``        ``if` `(arr[mid] > x)``            ``return` `binarySearch(arr, l, mid - 2, x);` `        ``// Else the element can only be present``        ``// in right subarray``        ``return` `binarySearch(arr, mid + 2, r, x);``}` `// We reach here when element is not present in array``return` `-1;``}` `// Driver Code``int` `main(``void``)``{``int` `arr[] = {3, 2, 10, 4, 40};``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``int` `x = 4;``int` `result = binarySearch(arr, 0, n - 1, x);``(result == -1) ? ``printf``(``"Element is not present in array"``)``               ``: ``printf``(``"Element is present at index %d"``,``                         ``result);``return` `0;``}`

## Java

 `// Java program to find an element``// in an almost sorted array``class` `GFG``{``    ``// A recursive binary search based function.``    ``// It returns index of x in given array``    ``// arr[l..r] is present, otherwise -1``    ``int` `binarySearch(``int` `arr[], ``int` `l, ``int` `r, ``int` `x)``    ``{``        ``if` `(r >= l)``        ``{``            ``int` `mid = l + (r - l) / ``2``;` `            ``// If the element is present at``            ``// one of the middle 3 positions``            ``if` `(arr[mid] == x)``                ``return` `mid;``            ``if` `(mid > l && arr[mid - ``1``] == x)``                ``return` `(mid - ``1``);``            ``if` `(mid < r && arr[mid + ``1``] == x)``                ``return` `(mid + ``1``);` `            ``// If element is smaller than mid, then``            ``// it can only be present in left subarray``            ``if` `(arr[mid] > x)``                ``return` `binarySearch(arr, l, mid - ``2``, x);` `            ``// Else the element can only be present``            ``// in right subarray``            ``return` `binarySearch(arr, mid + ``2``, r, x);``        ``}` `        ``// We reach here when element is``        ``// not present in array``        ``return` `-``1``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String args[])``    ``{``        ``GFG ob = ``new` `GFG();``        ``int` `arr[] = {``3``, ``2``, ``10``, ``4``, ``40``};``        ``int` `n = arr.length;``        ``int` `x = ``4``;``        ``int` `result = ob.binarySearch(arr, ``0``, n - ``1``, x);``        ``if``(result == -``1``)``            ``System.out.println(``"Element is not present in array"``);``        ``else``            ``System.out.println(``"Element is present at index "` `+``                                ``result);``    ``}``}` `// This code is contributed by Rajat Mishra`

## Python3

 `# Python 3 program to find an element``# in an almost sorted array` `# A recursive binary search based function.``# It returns index of x in given array arr[l..r]``# is present, otherwise -1``def` `binarySearch(arr, l, r, x):` `    ``if` `(r >``=` `l):``        ` `        ``mid ``=` `int``(l ``+` `(r ``-` `l) ``/` `2``)``        ` `        ``# If the element is present at one``        ``# of the middle 3 positions``        ``if` `(arr[mid] ``=``=` `x): ``return` `mid``        ``if` `(mid > l ``and` `arr[mid ``-` `1``] ``=``=` `x):``            ``return` `(mid ``-` `1``)``        ``if` `(mid < r ``and` `arr[mid ``+` `1``] ``=``=` `x):``            ``return` `(mid ``+` `1``)``            ` `        ``# If element is smaller than mid, then``        ``# it can only be present in left subarray``        ``if` `(arr[mid] > x):``            ``return` `binarySearch(arr, l, mid ``-` `2``, x)``        ` `        ``# Else the element can only``        ``# be present in right subarray``        ``return` `binarySearch(arr, mid ``+` `2``, r, x)` `    ``# We reach here when element``    ``# is not present in array``    ``return` `-``1` `# Driver Code``arr ``=` `[``3``, ``2``, ``10``, ``4``, ``40``]``n ``=` `len``(arr)``x ``=` `4``result ``=` `binarySearch(arr, ``0``, n ``-` `1``, x)``if` `(result ``=``=` `-``1``):``    ``print``(``"Element is not present in array"``)``else``:``    ``print``(``"Element is present at index"``, result)` `# This code is contributed by Smitha Dinesh Semwal.`

## C#

 `// C# program to find an element``// in an almost sorted array``using` `System;` `class` `GFG``{``    ``// A recursive binary search based function.``    ``// It returns index of x in given array``    ``// arr[l..r] is present, otherwise -1``    ``int` `binarySearch(``int` `[]arr, ``int` `l, ``int` `r, ``int` `x)``    ``{``        ``if` `(r >= l)``        ``{``            ``int` `mid = l + (r - l) / 2;` `            ``// If the element is present at``            ``// one of the middle 3 positions``            ``if` `(arr[mid] == x)``                ``return` `mid;``            ``if` `(mid > l && arr[mid - 1] == x)``                ``return` `(mid - 1);``            ``if` `(mid < r && arr[mid + 1] == x)``                ``return` `(mid + 1);` `            ``// If element is smaller than mid, then``            ``// it can only be present in left subarray``            ``if` `(arr[mid] > x)``                ``return` `binarySearch(arr, l, mid - 2, x);` `            ``// Else the element can only be present``            ``// in right subarray``            ``return` `binarySearch(arr, mid + 2, r, x);``        ``}` `        ``// We reach here when element is``        ``// not present in array``        ``return` `-1;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``GFG ob = ``new` `GFG();``        ``int` `[]arr = {3, 2, 10, 4, 40};``        ``int` `n = arr.Length;``        ``int` `x = 4;``        ``int` `result = ob.binarySearch(arr, 0, n - 1, x);``        ``if``(result == -1)``            ``Console.Write(``"Element is not present in array"``);``        ``else``            ``Console.Write(``"Element is present at index "` `+``                           ``result);``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

 `= ``\$l``)``    ``{``        ``\$mid` `= ``\$l` `+ (``\$r` `- ``\$l``) / 2;` `        ``// If the element is present at``        ``// one of the middle 3 positions``        ``if` `(``\$arr``[``\$mid``] == ``\$x``)``            ``return` `\$mid``;``        ``if` `(``\$mid` `> ``\$l` `&& ``\$arr``[``\$mid` `- 1] == ``\$x``)``            ``return` `(``\$mid` `- 1);``        ``if` `(``\$mid` `< ``\$r` `&& ``\$arr``[``\$mid` `+ 1] == ``\$x``)``            ``return` `(``\$mid` `+ 1);` `        ``// If element is smaller than mid, then``        ``// it can only be present in left subarray``        ``if` `(``\$arr``[``\$mid``] > ``\$x``)``            ``return` `binarySearch(``\$arr``, ``\$l``,``                           ``\$mid` `- 2, ``\$x``);` `        ``// Else the element can only be present``        ``// in right subarray``        ``return` `binarySearch(``\$arr``, ``\$mid` `+ 2,``                                    ``\$r``, ``\$x``);``}` `// We reach here when element``// is not present in array``return` `-1;``}` `// Driver Code``\$arr` `= ``array``(3, 2, 10, 4, 40);``\$n` `= sizeof(``\$arr``);``\$x` `= 4;``\$result` `= binarySearch(``\$arr``, 0, ``\$n` `- 1, ``\$x``);``if``(``\$result` `== -1)``    ``echo``(``"Element is not present in array"``);``else``    ``echo``(``"Element is present at index \$result"``);` `//This code is contributed by nitin mittal``?>`

## Javascript

 ``

Output :

`Element is present at index 3`

Time complexity of the above function is O(Logn).