# Check in binary array the number represented by a subarray is odd or even

• Difficulty Level : Easy
• Last Updated : 13 Jul, 2022

Given an array such that all its terms is either 0 or 1.You need to tell the number represented by a subarray a[l..r] is odd or even

Examples :

Input : arr = {1, 1, 0, 1}
l = 1, r = 3
Output : odd
number represented by arr[l...r] is
101 which 5 in decimal form which is
odd

Input :  arr = {1, 1, 1, 1}
l = 0, r = 3
Output : odd

The important point to note here is all the odd numbers in binary form have 1 as their rightmost bit and all even numbers have 0 as their rightmost bit.
The reason is simple all other bits other than the rightmost bit have even values and the sum of even numbers is always even. Now the rightmost bit can have a value of either 1 or 0 as we know even + odd = odd so when the rightmost bit is 1 the number is odd and when it is 0 the number is even.
So to solve this problem we have to just check if a[r] is 0 or 1 and accordingly print odd or even

Implementation:

## C++

 // C++ program to find if a subarray// is even or odd.#includeusing namespace std; // prints if subarray is even or oddvoid checkEVENodd (int arr[], int n, int l, int r){    // if arr[r] = 1 print odd    if (arr[r] == 1)        cout << "odd" << endl;     // if arr[r] = 0 print even    else        cout << "even" << endl;} // driver codeint main(){    int arr[] = {1, 1, 0, 1};    int n = sizeof(arr)/sizeof(arr[0]);    checkEVENodd (arr, n, 1, 3);    return 0;}

## Java

 // java program to find if a subarray// is even or odd.import java.io.*; class GFG{    // prints if subarray is even or odd    static void checkEVENodd (int arr[], int n, int l, int r)    {        // if arr[r] = 1 print odd        if (arr[r] == 1)            System.out.println( "odd") ;             // if arr[r] = 0 print even        else            System.out.println ( "even") ;    }     // driver code    public static void main (String[] args)    {        int arr[] = {1, 1, 0, 1};        int n = arr.length;        checkEVENodd (arr, n, 1, 3);                      }} // This article is contributed by vt_m.

## Python3

 # Python3 program to find if a# subarray is even or odd. # Prints if subarray is even or odddef checkEVENodd (arr, n, l, r):     # if arr[r] = 1 print odd    if (arr[r] == 1):        print("odd")     # if arr[r] = 0 print even    else:        print("even") # Driver codearr = [1, 1, 0, 1]n = len(arr)checkEVENodd (arr, n, 1, 3) # This code is contributed by Anant Agarwal.

## C#

 // C# program to find if a subarray// is even or odd.using System;  class GFG {         // prints if subarray is even or odd    static void checkEVENodd (int []arr,                     int n, int l, int r)    {                 // if arr[r] = 1 print odd        if (arr[r] == 1)            Console.WriteLine( "odd") ;              // if arr[r] = 0 print even        else            Console.WriteLine( "even") ;    }      // driver code    public static void Main()    {                 int []arr = {1, 1, 0, 1};        int n = arr.Length;                 checkEVENodd (arr, n, 1, 3);    }}  // This article is contributed by Anant Agarwal.



## Javascript



Output

odd

Time Complexity : O(1)
Space Complexity : O(1)

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