# Check whether bitwise OR of N numbers is Even or Odd

Given an array arr[] containing N numbers. The task is to check whether the bitwise-OR of the given N numbers is even or odd.

Examples:

```Input : arr[] = { 2, 12, 20, 36, 38 }
Output : Even Bit-wise OR

Input : arr[] = { 3, 9, 12, 13, 15 }
Output : Odd Bit-wise OR
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A Simple Solution is to first find the OR of the given N numbers, then check if this OR is even or odd.

Time Complexity: O(N)

A Better Solution is based on bit manipulation and Mathematical facts.

• Bitwise OR of any two even numbers is an even number.
• Bitwise OR of any two odd number is an odd number.
• Bitwise OR of an even and an odd number is an odd number.

Based on the above facts, it can be deduced that if at least one odd number is present in the array then the bitwise OR of whole array will be odd otherwise even.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to check whether ` `// bitwise OR of n numbers is even or odd ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check if bitwise OR ` `// of n numbers is even or odd ` `bool` `check(``int` `arr[], ``int` `n) ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// if at least one odd number is found, ` `        ``// then the bitwise OR of all numbers will be odd ` `        ``if` `(arr[i] & 1) ` `            ``return` `true``; ` `    ``} ` ` `  `    ``// Bitwise OR is an odd number ` `    ``return` `false``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 3, 9, 12, 13, 15 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``if` `(check(arr, n)) ` `        ``cout << ``"Odd Bit-wise OR"``; ` `    ``else` `        ``cout << ``"Even Bit-wise OR"``; ` `    ``return` `0; ` `}  `

## Java

 `// Java implementation to check whether  ` `// bitwise OR of n numbers is even or odd  ` `class` `GFG  ` `{ ` ` `  `// Function to check if bitwise OR  ` `// of n numbers is even or odd  ` `static` `boolean` `check(``int` `arr[], ``int` `n)  ` `{ ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``// if at least one odd number is  ` `        ``// found, then the bitwise OR of  ` `        ``// all numbers will be odd  ` `        ``if` `(arr[i] % ``2` `== ``1``) ` `        ``{ ` `            ``return` `true``; ` `        ``} ` `    ``} ` ` `  `    ``// Bitwise OR is an odd number  ` `    ``return` `false``; ` `} ` ` `  `// Driver Code  ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `arr[] = {``3``, ``9``, ``12``, ``13``, ``15``}; ` `    ``int` `n = arr.length; ` ` `  `    ``if` `(check(arr, n)) ` `    ``{ ` `        ``System.out.println(``"Odd Bit-wise OR"``); ` `    ``}  ` `    ``else`  `    ``{ ` `        ``System.out.println(``"Even Bit-wise OR"``); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed ` `// by 29AjayKumar `

## Python3

 `# Python3 implementation to check whether  ` `# bitwise OR of n numbers is even or odd  ` ` `  `# Function to check if bitwise OR  ` `# of n numbers is even or odd  ` `def` `check(arr,n): ` `    ``for` `i ``in` `range``(n): ` ` `  `        ``# if at least one odd number is found,  ` `        ``# then the bitwise OR of all numbers  ` `        ``# will be odd  ` `        ``if` `arr[i] & ``1``: ` `            ``return` `True` ` `  `    ``# Bitwise OR is an odd number  ` `    ``return` `False` ` `  `# Driver code ` `if` `__name__``=``=``'__main__'``: ` `    ``arr ``=` `[``3``, ``9``, ``12``, ``13``, ``15``] ` `    ``n ``=` `len``(arr) ` `    ``if` `check(arr,n): ` `        ``print``(``"Odd Bit-wise OR"``) ` `    ``else``: ` `        ``print``(``"Even Bit-wise OR"``)  ` ` `  `# This code is contributed by  ` `# Shrikant13 `

## C#

 `// C# implementation to check whether  ` `// bitwise OR of n numbers is even or odd  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` ` `  `// Function to check if bitwise OR  ` `// of n numbers is even or odd  ` `static` `bool` `check(``int` `[]arr, ``int` `n)  ` `{ ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `    ``{ ` `        ``// if at least one odd number is  ` `        ``// found, then the bitwise OR of  ` `        ``// all numbers will be odd  ` `        ``if` `(arr[i] % 2 == 1) ` `        ``{ ` `            ``return` `true``; ` `        ``} ` `    ``} ` ` `  `    ``// Bitwise OR is an odd number  ` `    ``return` `false``; ` `} ` ` `  `// Driver Code  ` `public` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = {3, 9, 12, 13, 15}; ` `    ``int` `n = arr.Length; ` ` `  `    ``if` `(check(arr, n)) ` `    ``{ ` `        ``Console.WriteLine(``"Odd Bit-wise OR"``); ` `    ``}  ` `    ``else` `    ``{ ` `        ``Console.WriteLine(``"Even Bit-wise OR"``); ` `    ``} ` `} ` `} ` ` `  `// This code is contributed  ` `// by 29AjayKumar `

## PHP

 ` `

Output:

```Odd Bit-wise OR
```

Time Complexity: O(N) in worst case.

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Improved By : shrikanth13, jit_t, 29AjayKumar