Check whether product of ‘n’ numbers is even or odd
Given an array arr[] containing n numbers. The problem is to check whether the product of the given n numbers is even or odd.
Examples:
Input : arr[] = {2, 4, 3, 5}
Output : Even
Product = 2 * 4 * 3 * 5 = 120
Input : arr[] = {3, 9, 7, 1}
Output : Odd
A simple solution is to first find the product, then check if the product is even or odd. This solution causes overflow for large arrays.
A better solution is based on following mathematical calculation facts:
- Product of two even numbers is even.
- Product of two odd numbers is odd.
- Product of one even and one odd number is even.
Based on the above facts, if a single even number occurs then the entire product of n numbers will be even else odd.
C++
// C++ implementation to check whether product of // 'n' numbers is even or odd #include <bits/stdc++.h> using namespace std; // function to check whether product of // 'n' numbers is even or odd bool isProductEven(int arr[], int n) { for (int i = 0; i < n; i++) // if a single even number is found, then // final product will be an even number if ((arr[i] & 1) == 0) return true; // product is an odd number return false; } // Driver program to test above int main() { int arr[] = { 2, 4, 3, 5 }; int n = sizeof(arr) / sizeof(arr[0]); if (isProductEven(arr, n)) cout << "Even"; else cout << "Odd"; return 0; } |
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Java
// Java implementation to check whether product of // 'n' numbers is even or odd public class GFG { // function to check whether product of // 'n' numbers is even or odd static boolean isProductEven(int arr[], int n) { for (int i = 0; i < n; i++) // if a single even number is found, then // final product will be an even number if ((arr[i] & 1) == 0) return true; // product is an odd number return false; } // Driver code public static void main(String args[]) { int arr[] = { 2, 4, 3, 5 }; int n = arr.length ; if (isProductEven(arr, n)) System.out.println("Even"); else System.out.println("Odd") ; } // This Code is contributed by ANKITRAI1 } |
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Python3
# Python3 implementation to # check whether product of 'n' # numbers is even or odd # function to check whether # product of 'n' numbers is # even or odd def isProductEven(arr, n): for i in range(0, n): # if a single even number is # found, then final product # will be an even number if ((arr[i] & 1) == 0): return True # product is an odd number return False # Driver Code arr = [ 2, 4, 3, 5 ] n = len(arr) if (isProductEven(arr, n)): print("Even") else: print("Odd") # This code is contributed # by ihritik |
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C#
// C# implementation to check // whether product of 'n' // numbers is even or odd using System; class GFG { // function to check whether // product of 'n' numbers // is even or odd static bool isProductEven(int []arr, int n) { for (int i = 0; i < n; i++) // if a single even number is // found, then final product // will be an even number if ((arr[i] & 1) == 0) return true; // product is an odd number return false; } // Driver code public static void Main() { int []arr = { 2, 4, 3, 5 }; int n = arr.Length; if (isProductEven(arr, n)) Console.WriteLine("Even"); else Console.WriteLine("Odd") ; } } // This code is contributed by ihritik |
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PHP
<?php // PHP implementation to check // whether product of 'n' numbers // is even or odd // function to check whether // product of 'n' numbers is // even or odd function isProductEven($arr, $n) { for ($i = 0; $i < $n; $i++) // if a single even number is // found, then final product // will be an even number if (($arr[$i] & 1) == 0) return true; // product is an odd number return false; } // Driver code $arr = array( 2, 4, 3, 5 ); $n = sizeof($arr); if (isProductEven($arr, $n)) echo "Even"; else echo "Odd"; // This code is contributed by ihritik ?> |
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Output:
Even
Time Complexity: O(n).
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