Java Program for Binary Search (Recursive and Iterative)

13 Jun, 2022
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:
Compare x with the middle element.If x matches with the middle element, we return the mid index.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.Else (x is smaller) recur for the left half.
Compare x with the middle element.
If x matches with the middle element, we return the mid index.
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.
Else (x is smaller) recur for the left half.
Example 1 
Java
// Java Program to Illustrate// Iterative Binary Search // Main class// BinarySearchclass 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);    }}
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 classesimport java.util.*; // Main classclass 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);    }}
Element found at index 3
Time Complexity: O(log n)
Auxiliary Space: O(1) 
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