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 cab 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 C++ implementation of this approach.

## C++

// C++ program to find an element in an almost sorted // array #include <stdio.h> // 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 program to test above function 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 SearchAlmost { // 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 method public static void main(String args[]) { abc ob = new abc(); 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*/

Output:

Element is present at index 3

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

This article is contributed by **Abhishek**. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above