# Java Program for Binary Search (Recursive and Iterative)

• Difficulty Level : Medium
• Last Updated : 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:

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)

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