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++
// 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 ?> |
Javascript
<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.
