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

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++ program to find if a subarray
// is even or odd.
#include<bits/stdc++.h>
using namespace std;

// prints if subarray is even or odd
void 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 code
int main()
{
    int arr[] = {1, 1, 0, 1};
    int n = sizeof(arr)/sizeof(arr[0]);
    checkEVENodd (arr, n, 1, 3);
    return 0;
}

// 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 program to find if a 
# subarray is even or odd.

# Prints if subarray is even or odd
def 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 code
arr = [1, 1, 0, 1]
n = len(arr)
checkEVENodd (arr, n, 1, 3)

# This code is contributed by Anant Agarwal.

// 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.

<?php
// PHP program to find if a subarray
// is even or odd.

// prints if subarray is even or odd
function checkEVENodd ($arr, $n, $l, $r)
{
    // if arr[r] = 1 print odd
    if ($arr[$r] == 1)
        echo "odd", "\n";

    // if arr[r] = 0 print even
    else
        echo "even", "\n";
}

// Driver code
$arr = array(1, 1, 0, 1);
$n = sizeof($arr);
checkEVENodd ($arr, $n, 1, 3);

// This code is Contributed by Ajit
?>

<script>

    // Javascript program to find 
    // if a subarray is even or odd.
    
    // prints if subarray is even or odd
    function checkEVENodd (arr, n, l, r)
    {
          
        // if arr[r] = 1 print odd
        if (arr[r] == 1)
            document.write("odd") ;
       
        // if arr[r] = 0 print even
        else
            document.write("even") ;
    }
    
    let arr = [1, 1, 0, 1];
    let n = arr.length;

    checkEVENodd (arr, n, 1, 3);

</script>

odd

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

This article is contributed by Ayush Jha.