Order-Agnostic Binary Search

Order-Agnostic Binary Search is a modified version of Binary Search algorithm. Here in this modified binary search comes with one more condition checking. The intuition behind this algorithm is what if the order of the sorted array is not given. So here check that the value of the first element is greater or smaller than the last element.

• If the first element is smaller than the last element-then if the search key value X is less than the middle of the interval then the end pointer will be changed to middle -1 otherwise start will be changed to middle + 1.
• If the first element is greater than the last element-then if the search key value X is less than the middle of the interval then the start pointer will move to the next element of the middle element otherwise the end pointer will move previous to the middle element.

In the end, if the search key value matches with the middle element then the element which was given to the search is found.

Implementation of Order-Agnostic Binary Search:

Let us see the implementation of Order-Agnostic Binary Search with the help of an example.

Given an array, arr[ ] of size N and an element X and the array is sorted in any order(ascending or descending), the task is to find whether the element x is present in the array or not. If yes, then print its index, else print -1.

Examples:

Input: arr[] = {40, 10, 5, 2, 1}, N=5, X=10
Output: 1
Explanation:

The array is sorted in descending order and the element is present at index 1.

Input: arr[] = {1}, N=1, X=10
Output: -1

Approach: The brute force idea would be to linearly traverse the array and check whether the element is present in the array or not. Optimization to this algorithm would be to use binary search if the order of sorting of the array was known – ascending/descending. A variation of binary search can be used i.e, Order-Agnostic Binary search as stated below:

Follow the steps below to solve the problem using Order-Agnostic Binary Search:

• Initialize a boolean variable isAsc as true if arr[start] is less than arr[end] else set it as false.
• Traverse over a while loop till start is less than equal to end and perform the following steps:
  • Initialize the variable middle as the average of start and end.
  • If arr[middle] equals X, then return the value of middle as the answer,
  • If the array is in ascending order then perform the following steps:
    • If arr[middle] is less than X, then set the value of start as  middle+1 else set the value of end as middle-1.
  • Else, if arr[middle] is greater than X, then set the value of end as middle-1 else set the value of start as middle+1.
• After performing the above steps, return the value of -1 as the answer as the element is not found.

Iterative implementation of Order-Agnostic Binary Search.

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// An iterative binary search function.
int binarySearch(int arr[], int start, int end, int x)
{

    // Checking the sorted order of the given array
    bool isAsc = arr[start] < arr[end];
    while (start <= end) {
        int middle = start + (end - start) / 2;

        // Check if x is present at mid
        if (arr[middle] == x)
            return middle;

        // Ascending order
        if (isAsc == true) {

            // If x greater, ignore left half
            if (arr[middle] < x)
                start = middle + 1;

            // If x smaller, ignore right half
            else
                end = middle - 1;
        }

        // Descending order
        else {

            // If x smaller, ignore left half
            if (arr[middle] > x)
                start = middle + 1;

            // If x greater, ignore right half
            else
                end = middle - 1;
        }
    }

    // Element is not present
    return -1;
}

// Driver Code
int main()
{
    int arr[] = { 40, 10, 5, 2, 1 };
    int x = 10;
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << binarySearch(arr, 0, n - 1, x);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG {

// An iterative binary search function.
static int binarySearch(int arr[], int start, int end, int x)
{

    // Checking the sorted order of the given array
    boolean isAsc = arr[start] < arr[end];
    while (start <= end) {
        int middle = start + (end - start) / 2;

        // Check if x is present at mid
        if (arr[middle] == x)
            return middle;

        // Ascending order
        if (isAsc == true) {

            // If x greater, ignore left half
            if (arr[middle] < x)
                start = middle + 1;

            // If x smaller, ignore right half
            else
                end = middle - 1;
        }

        // Descending order
        else {

            // If x smaller, ignore left half
            if (arr[middle] > x)
                start = middle + 1;

            // If x greater, ignore right half
            else
                end = middle - 1;
        }
    }

    // Element is not present
    return -1;
}

    // Driver Code
    public static void main(String[] args)
    {
    int arr[] = { 40, 10, 5, 2, 1 };
    int x = 10;
    int n = arr.length;
    System.out.println(binarySearch(arr, 0, n - 1, x));
    }
}

// This code is contributed by sanjoy_62.

Python

# Python program for the above approach

# An iterative binary search function.
def binarySearch(arr, start, end, x):

    # Checking the sorted order of the given array
    isAsc = arr[start] < arr[end]

    while (start <= end):
        middle = start + (end - start) // 2

        # Check if x is present at mid
        if (arr[middle] == x):
            return middle

        # Ascending order
        if (isAsc == True):

            # If x greater, ignore left half
            if (arr[middle] < x):
                start = middle + 1

            # If x smaller, ignore right half
            else:
                end = middle - 1

        # Descending order
        else:

            # If x smaller, ignore left half
            if (arr[middle] > x):
                start = middle + 1

            # If x greater, ignore right half
            else:
                end = middle - 1

    # Element is not present
    return -1

# Driver Code
arr = [ 40, 10, 5, 2, 1 ]
x = 10
n = len(arr)
print(binarySearch(arr, 0, n - 1, x))

# This code is ciontributed by Samim Hossain Mondal.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG {

// An iterative binary search function.
static int binarySearch(int[] arr, int start, int end, int x)
{

    // Checking the sorted order of the given array
    bool isAsc = arr[start] < arr[end];
    while (start <= end) {
        int middle = start + (end - start) / 2;

        // Check if x is present at mid
        if (arr[middle] == x)
            return middle;

        // Ascending order
        if (isAsc == true) {

            // If x greater, ignore left half
            if (arr[middle] < x)
                start = middle + 1;

            // If x smaller, ignore right half
            else
                end = middle - 1;
        }

        // Descending order
        else {

            // If x smaller, ignore left half
            if (arr[middle] > x)
                start = middle + 1;

            // If x greater, ignore right half
            else
                end = middle - 1;
        }
    }

    // Element is not present
    return -1;
}

    // Driver Code
    public static void Main(String[] args) {

    int[] arr = { 40, 10, 5, 2, 1 };
    int x = 10;
    int n = arr.Length;
    Console.Write(binarySearch(arr, 0, n - 1, x));

    }
}

// This code is contributed by code_hunt.

Javascript

<script>

        // JavaScript Program to implement
        // the above approach

        // An iterative binary search function.
        function binarySearch(arr, start, end, x) {

            // Checking the sorted order of the given array
            let isAsc = arr[start] < arr[end];
            while (start <= end) {
                let middle = start + Math.floor((end - start) / 2);

                // Check if x is present at mid
                if (arr[middle] == x)
                    return middle;

                // Ascending order
                if (isAsc == true) {

                    // If x greater, ignore left half
                    if (arr[middle] < x)
                        start = middle + 1;

                    // If x smaller, ignore right half
                    else
                        end = middle - 1;
                }

                // Descending order
                else {

                    // If x smaller, ignore left half
                    if (arr[middle] > x)
                        start = middle + 1;

                    // If x greater, ignore right half
                    else
                        end = middle - 1;
                }
            }

            // Element is not present
            return -1;
        }

        // Driver Code
        let arr = [40, 10, 5, 2, 1];
        let x = 10;
        let n = arr.length;
        document.write(binarySearch(arr, 0, n - 1, x));

    // This code is contributed by Potta Lokesh
    </script>

Output:

1

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

Recursive implementation of Order-Agnostic Binary Search :

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// A recursive binary search function.
// It returns location of x in given
// array arr[l..r] is present,
// otherwise -1
int binarySearch(int arr[], int start,
                 int end, int x)
{
    bool isAsc = arr[start] < arr[end];
    if (end >= start) {
        int middle = start + (end - start) / 2;

        // If the element is present
        // at the middle itself
        if (arr[middle] == x)
            return middle;

        if (isAsc == true) {

            // If element is smaller than mid,
            // then it can only be
            // present in left subarray
            if (arr[middle] > x)
                return binarySearch(
                    arr, start,
                    middle - 1, x);

            // Else the element can only be present
            // in right subarray
            return binarySearch(arr, middle + 1,
                                end, x);
        }
        else {
            if (arr[middle] < x)
                return binarySearch(arr, start,
                                    middle - 1, x);

            // Else the element can only be present
            // in left subarray
            return binarySearch(arr, middle + 1,
                                end, x);
        }
    }

    // Element not found
    return -1;
}

// Driver Code
int main(void)
{
    int arr[] = { 40, 10, 5, 2, 1 };
    int x = 10;
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << binarySearch(arr, 0, n - 1, x);

    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG {

  // A recursive binary search function.
  // It returns location of x in given
  // array arr[l..r] is present,
  // otherwise -1
  static int binarySearch(int arr[], int start, int end, int x) {
    boolean isAsc = arr[start] < arr[end];
    if (end >= start) {
      int middle = start + (end - start) / 2;

      // If the element is present
      // at the middle itself
      if (arr[middle] == x)
        return middle;

      if (isAsc == true) {

        // If element is smaller than mid,
        // then it can only be
        // present in left subarray
        if (arr[middle] > x)
          return binarySearch(arr, start, middle - 1, x);

        // Else the element can only be present
        // in right subarray
        return binarySearch(arr, middle + 1, end, x);
      } else {
        if (arr[middle] < x)
          return binarySearch(arr, start, middle - 1, x);

        // Else the element can only be present
        // in left subarray
        return binarySearch(arr, middle + 1, end, x);
      }
    }

    // Element not found
    return -1;
  }

  // Driver Code
  public static void main(String[] args) {
    int arr[] = { 40, 10, 5, 2, 1 };
    int x = 10;
    int n = arr.length;
    System.out.print(binarySearch(arr, 0, n - 1, x));

  }
}

// This code is contributed by Rajput-Ji

Python3

# Python program for the above approach

# A recursive binary search function.
# It returns location of x in given
# array arr[l..r] is present,
# otherwise -1
def binarySearch(arr, start,
                 end, x):
    isAsc = arr[start] < arr[end]
    if (end >= start):
        middle = (int)(start + (end - start) / 2)

        # If the element is present
        # at the middle itself
        if (arr[middle] == x):
            return middle

        if (isAsc == True):

            # If element is smaller than mid,
            # then it can only be
            # present in left subarray
            if (arr[middle] > x):
                return binarySearch(
                    arr, start,
                    middle - 1, x)

            # Else the element can only be present
            # in right subarray
            return binarySearch(arr, middle + 1,
                                end, x)

        else:
            if (arr[middle] < x):
                return binarySearch(arr, start,
                                    middle - 1, x)

            # Else the element can only be present
            # in left subarray
            return binarySearch(arr, middle + 1,
                                end, x)

    # Element not found
    return -1

# Driver Code

arr = [40, 10, 5, 2, 1]
x = 10
n = len(arr)
print(binarySearch(arr, 0, n - 1, x))

# This code is contributed by Taranpreet

C#

// C# program for the above approach
using System;

public class GFG {

  // A recursive binary search function.
  // It returns location of x in given
  // array arr[l..r] is present,
  // otherwise -1
  static int binarySearch(int []arr, int start, int end, int x) {
    bool isAsc = arr[start] < arr[end];
    if (end >= start) {
      int middle = start + (end - start) / 2;

      // If the element is present
      // at the middle itself
      if (arr[middle] == x)
        return middle;

      if (isAsc == true) {

        // If element is smaller than mid,
        // then it can only be
        // present in left subarray
        if (arr[middle] > x)
          return binarySearch(arr, start, middle - 1, x);

        // Else the element can only be present
        // in right subarray
        return binarySearch(arr, middle + 1, end, x);
      } else {
        if (arr[middle] < x)
          return binarySearch(arr, start, middle - 1, x);

        // Else the element can only be present
        // in left subarray
        return binarySearch(arr, middle + 1, end, x);
      }
    }

    // Element not found
    return -1;
  }

  // Driver Code
  public static void Main(String[] args) {
    int []arr = { 40, 10, 5, 2, 1 };
    int x = 10;
    int n = arr.Length;
    Console.Write(binarySearch(arr, 0, n - 1, x));

  }
}

// This code contributed by Rajput-Ji

Javascript

<script>
// javascript program for the above approach

    // A recursive binary search function.
    // It returns location of x in given
    // array arr[l..r] is present,
    // otherwise -1
    function binarySearch(arr , start , end , x) {
        let isAsc = arr[start] < arr[end];
        if (end >= start) {
            var middle = start + parseInt((end - start) / 2);

            // If the element is present
            // at the middle itself
            if (arr[middle] == x)
                return middle;

            if (isAsc == true) {

                // If element is smaller than mid,
                // then it can only be
                // present in left subarray
                if (arr[middle] > x)
                    return binarySearch(arr, start, middle - 1, x);

                // Else the element can only be present
                // in right subarray
                return binarySearch(arr, middle + 1, end, x);
            } else {
                if (arr[middle] < x)
                    return binarySearch(arr, start, middle - 1, x);

                // Else the element can only be present
                // in left subarray
                return binarySearch(arr, middle + 1, end, x);
            }
        }

        // Element not found
        return -1;
    }

    // Driver Code
        var arr = [ 40, 10, 5, 2, 1 ];
        var x = 10;
        var n = arr.length;
        document.write(binarySearch(arr, 0, n - 1, x));

// This code is contributed by Rajput-Ji
</script>

Output:

1

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