Java Program for Binary Search (Recursive and Iterative)

So as we all know binary search is one of the searching algorithms that is most frequently applied while dealing with data structures where the eccentric goal is not to traverse the whole array. Here array must be sorted as we check the middle element and ignore the half of the array which is of no use as per the number system. We basically ignore half of the elements just after one comparison. So do we keep on iterating till the element is found or land upon a conclusion that element is not present n the array.

Algorithms:

1. Compare x with the middle element.
2. If x matches with the middle element, we return the mid index.
3. Else If x is greater than the mid element, then x can only lie in the right half subarray after the mid element. So we recur for the right half.
4. Else (x is smaller) recur for the left half.

Example 1

Java

// Java Program to Illustrate
// Iterative Binary Search

// Main class
// BinarySearch
class GFG {

    // Method 1
    // Returns index of x
    // if it is present in arr[],
    // else return -1
    int binarySearch(int arr[], int x)
    {

        int l = 0, r = arr.length - 1;

        // Checking element in whole array
        while (l <= r) {
            int m = l + (r - l) / 2;

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

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

            // If x is smaller,
            // element is on left side
            // so ignore right half
            else
                r = m - 1;
        }

        // If we reach here,
        // element is not present
        return -1;
    }

    // Method 2
    // Main driver method
    public static void main(String args[])
    {

        GFG ob = new GFG();

        // Input array
        int arr[] = { 2, 3, 4, 10, 40 };
        // Length of array
        int n = arr.length;
        // Element to be checked if present or not
        int x = 10;

        // Calling the method 1 and
        // storing result
        int result = ob.binarySearch(arr, x);

        // Element present
        if (result == -1)

            // Print statement
            System.out.println("Element not present");

        // Element not present
        else

            // Print statement
            System.out.println("Element found at index "
                               + result);
    }
}

Output

Element found at index 3

Time Complexity: O(log n)

Auxiliary Space: O(1)

Example 2 

Java

// Java Program to Illustrate Recursive Binary Search

// Importing required classes
import java.util.*;

// Main class
class GFG {

    // Method 1
    // Recursive binary search
    // Returns index of x if it is present
    // in arr[l..r], else return -1
    int binarySearch(int arr[], int l, int r, int x)
    {
        // Restrict the boundary of right index
        // and the left index to prevent
        // overflow of indices
        if (r >= l && l <= arr.length - 1) {

            int mid = l + (r - l) / 2;

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

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

            // Else the element can only be present
            // in right subarray
            return binarySearch(arr, mid + 1, r, x);
        }

        // We reach here when element is not present in
        // array
        return -1;
    }

    // Method 2
    // Main driver method
    public static void main(String args[])
    {

        // Creating object of above class
        GFG ob = new GFG();

        // Custom input array
        int arr[] = { 2, 3, 4, 10, 40 };

        // Length of array
        int n = arr.length;

        // Custom element to be checked
        // whether present or not
        int x = 10;

        // Calling above method
        int result = ob.binarySearch(arr, 0, n - 1, x);

        // Element present
        if (result == -1)

            // Print statement
            System.out.println("Element not present");

        // Element not present
        else

            // Print statement
            System.out.println("Element found at index "
                               + result);
    }
}

Output

Element found at index 3

Time Complexity: O(log n)

Auxiliary Space: O(1)