Given a 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 rightmost bit have even values and sum of even numbers is always even .Now the rightmost bit can have value either 1 or 0 as we know even + odd = odd so when right most 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
C++
// 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
// 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 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#
// 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 // 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 ?> |
Output :
odd
This article is contributed by Ayush Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
