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 <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 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.
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
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