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
Explanation: Product = 2 * 4 * 3 * 5 = 120, 120 is even.

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:

1. Product of two even numbers is even.
2. Product of two odd numbers is odd.
3. 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;
}

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
}

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

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

Javascript

<script>
// JavaScript 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 (let 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

    let arr = [ 2, 4, 3, 5 ];
    let n = arr.length;
    if (isProductEven(arr, n))
        document.write("Even");
    else
        document.write("Odd");

// This code is contributed by Surbhi Tyagi.

</script>

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
?>

Output

Even

Time Complexity: O(n).
Auxiliary Space: O(1) 

Approach 2: Using Memoization:

We can use memoization to optimize the function if the same function call with the same arguments is expected to occur multiple times. Here is an example implementation of the function that uses memoization:

C++

#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;
}

Java

// Java program to check whether product of
// 'n' numbers is even or odd
import java.util.*;

public class Main {
    // memoization table to store previously
    // computed results
    static HashMap<String, Boolean> memo = new HashMap<>();

    // function to check whether product of
    // 'n' numbers is even or odd
    static boolean isProductEven(int arr[], int n) {
        // convert array to string for memoization
        String key = Arrays.toString(arr);

        // check if result is already computed
        if (memo.containsKey(key)) {
            return memo.get(key);
        }

        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) {
                memo.put(key, true);
                return true;
            }
        }

        // product is an odd number
        memo.put(key, false);
        return false;
    }

    // Driver program to test above
    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");
        }
    }
}

Python3

# memoization table to store previously computed results
memo = {}

# function to check whether product of 'n' numbers is even or odd
def isProductEven(arr, n):
    # convert array to string for memoization
    key = str(arr)

    # check if result is already computed
    if key in memo:
        return memo[key]

    for i in range(n):
        # if a single even number is found, then
        # final product will be an even number
        if arr[i] % 2 == 0:
            memo[key] = True
            return True

    # product is an odd number
    memo[key] = False
    return False

# Driver program to test above
arr = [2, 4, 3, 5]
n = len(arr)
if isProductEven(arr, n):
    print("Even")
else:
    print("Odd")

C#

using System;
using System.Collections.Generic;

public class MainClass
{

  // memoization table to store previously
  // computed results
  static Dictionary<string, bool> memo = new Dictionary<string, bool>();

  // function to check whether product of
  // 'n' numbers is even or odd
  static bool IsProductEven(int[] arr, int n)
  {

    // convert array to string for memoization
    string key = string.Join(",", arr);

    // check if result is already computed
    if (memo.ContainsKey(key)) {
      return memo[key];
    }

    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) {
        memo[key] = true;
        return true;
      }
    }

    // product is an odd number
    memo[key] = false;
    return false;
  }

  // Driver program to test above
  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");
    }
  }
}

Javascript

// JS equivalent
// memoization table to store previously computed results
let memo = {};

// function to check whether product of 'n' numbers is even or odd
function isProductEven(arr, n) {
  // convert array to string for memoization
  let key = arr.toString();

  // check if result is already computed
  if (key in memo) {
    return memo[key];
  }

  for (let i = 0; i < n; i++) {
    // if a single even number is found, then
    // final product will be an even number
    if (arr[i] % 2 === 0) {
      memo[key] = true;
      return true;
    }
  }

  // product is an odd number
  memo[key] = false;
  return false;
}

// Driver program to test above
let arr = [2, 4, 3, 5];
let n = arr.length;
if (isProductEven(arr, n)) {
  console.log("Even");
} else {
  console.log("Odd");
}

Output

Even

Time Complexity: O(N).
Auxiliary Space: O(1), a constant amount of extra space to store the intermediate result

Approach 3:

Here’s another approach to check whether the product of ‘n’ numbers is even or odd:

• Traverse the given array of ‘n’ numbers and count the number of even elements.
• If the count of even elements is greater than or equal to 1, then the product of ‘n’ numbers will be even. Otherwise, it will be odd.
  Here’s the implementation of this approach in C++:

C++

import java.util.*;

public class SpecialPrime {

    // Function to check if a number is prime
    public static boolean isPrime(int x) {
        if (x < 2) {
            return false;
        }
        for (int i = 2; i <= Math.sqrt(x); i++) {
            if (x % i == 0) {
                return false;
            }
        }
        return true;
    }

    // Function to check if a number is a special prime
    public static boolean isSpecialPrime(int x) {
        int s = 0;
        for (int i = 2; i <= x; i++) {
            if (isPrime(i)) {
                s += i;
            }
        }
        return isPrime(s);
    }

    // Function to count special primes up to 'n' and determine if there are at least 'k' of them
    public static String countSpecialPrimes(int n, int k) {
        int count = 0;
        for (int i = 2; i <= n; i++) {
            if (isSpecialPrime(i)) {
                count++;
            }
            if (count >= k) {
                return "YES";
            }
        }
        return "NO";
    }

    public static void main(String[] args) {
        int n = 27;
        int k = 2;

        // Call the function to count special primes and output the result
        System.out.println(countSpecialPrimes(n, k));
    }
}

Java

import java.util.Arrays;

public class GFG {

    // function to check whether the product of 'n' numbers
   // is even or odd
    private static boolean isProductEven(int[] arr, int n) {
        int evenCount = 0;
        for (int i = 0; i < n; i++) {
            if (arr[i] % 2 == 0) {
                evenCount++;
            }
        }

        if (evenCount >= 1) {
            return true;
        } else {
            return false;
        }
    }

    // Driver program to test above
    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");
    }
}

Python3

# Function to check whether the product of 'n' numbers is even or odd
def isProductEven(arr):
    even_count = 0
    for i in range(len(arr)):
        if arr[i] % 2 == 0:
            even_count += 1

    if even_count >= 1:
        return True
    else:
        return False

# Driver program to test above function
if __name__ == "__main__":
    arr = [2, 4, 3, 5]
    if isProductEven(arr):
        print("Even")
    else:
        print("Odd")

C#

using System;

public class GFG {
    // Function to check whether the product of 'n' numbers
    // is even or odd
    static bool IsProductEven(int[] arr, int n)
    {
        int evenCount = 0;
        for (int i = 0; i < n; i++) {
            if (arr[i] % 2 == 0) {
                evenCount++;
            }
        }

        if (evenCount >= 1) {
            return true;
        }
        else {
            return false;
        }
    }

    // Driver program to test above
    public static void Main(string[] args)
    {
        int[] arr = { 2, 4, 3, 5 };
        int n = arr.Length;
        if (IsProductEven(arr, n))
            Console.WriteLine("Even");
        else
            Console.WriteLine("Odd");
    }
}

Javascript

//javascript code for the above approach.
function isPrime(x) {
    if (x < 2) {
        return false;
    }
    for (let i = 2; i <= Math.sqrt(x); i++) {
        if (x % i === 0) {
            return false;
        }
    }
    return true;
}

function isSpecialPrime(x) {
    let s = 0;
    for (let i = 2; i <= x; i++) {
        if (isPrime(i)) {
            s += i;
        }
    }
    return isPrime(s);
}

function countSpecialPrimes(n, k) {
    let count = 0;
    for (let i = 2; i <= n; i++) {
        if (isSpecialPrime(i)) {
            count++;
        }
        if (count >= k) {
            return "YES";
        }
    }
    return "NO";
}

const n = 27;
const k = 2;

// Call the function to count special primes and output the result
console.log(countSpecialPrimes(n, k));

Output

Even

Time Complexity: O(n).
Auxiliary Space: O(1)