Java Program for Binary Search (Recursive and Iterative)

We basically ignore half of the elements just after one comparison.

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

// Java implementation of recursive Binary Search 
class BinarySearch { 
    // 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) 
    { 
        if (r >= l) { 
            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; 
    } 
  
    // Driver method to test above 
    public static void main(String args[]) 
    { 
        BinarySearch ob = new BinarySearch(); 
        int arr[] = { 2, 3, 4, 10, 40 }; 
        int n = arr.length; 
        int x = 10; 
        int result = ob.binarySearch(arr, 0, n - 1, x); 
        if (result == -1) 
            System.out.println("Element not present"); 
        else
            System.out.println("Element found at index " + result); 
    } 
} 
/* This code is contributed by Rajat Mishra */

Output:

Element found at index 3

// Java implementation of iterative Binary Search 
class BinarySearch { 
    // 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; 
        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, ignore right half 
            else
                r = m - 1; 
        } 
  
        // if we reach here, then element was not present 
        return -1; 
    } 
  
    // Driver method to test above 
    public static void main(String args[]) 
    { 
        BinarySearch ob = new BinarySearch(); 
        int arr[] = { 2, 3, 4, 10, 40 }; 
        int n = arr.length; 
        int x = 10; 
        int result = ob.binarySearch(arr, x); 
        if (result == -1) 
            System.out.println("Element not present"); 
        else
            System.out.println("Element found at index " + result); 
    } 
} 

Output:

Element found at index 3

Please refer complete article on Binary Search for more details!

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