Check if product of digits of a number at even and odd places is equal

Given an integer N, the task is to check whether the product of digits at even and odd places of a number are equal. If they are equal, print Yes otherwise print No.

Examples:

Input: N = 2841
Output: Yes
Product of digits at odd places = 2 * 4 = 8
Product of digits at even places = 8 * 1 = 8

Input: N = 4324
Output: No
Product of digits at odd places = 4 * 2 = 8
Product of digits at even places = 3 * 4 = 12

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Find product of digits at even places and store it in prodEven.
Find product of digits at odd places and store it in prodOdd.
If prodEven = prodOdd then print Yes otherwise print No.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function that returns true if the product 
// of even positioned digits is equal to 
// the product of odd positioned digits in n 
bool productEqual(int n) 
{ 
  
    // If n is a single digit number 
    if (n < 10) 
        return false; 
    int prodOdd = 1, prodEven = 1; 
  
    while (n > 0) { 
  
        // Take two consecutive digits 
        // at a time 
        // First digit 
        int digit = n % 10; 
        prodOdd *= digit; 
        n /= 10; 
  
        // If n becomes 0 then 
        // there's no more digit 
        if (n == 0) 
            break; 
  
        // Second digit 
        digit = n % 10; 
        prodEven *= digit; 
        n /= 10; 
    } 
  
    // If the products are equal 
    if (prodEven == prodOdd) 
        return true; 
  
    // If products are not equal 
    return false; 
} 
  
// Driver code 
int main() 
{ 
    int n = 4324; 
  
    if (productEqual(n)) 
        cout << "Yes"; 
    else
        cout << "No"; 
  
    return 0; 
} 

Java

// Java implementation of the approach  
  
class GFG  
{  
      
    // Function that returns true  
    // if the product of even positioned  
    // digits is equal to the product of  
    // odd positioned digits in n  
    static boolean productEqual(int n)  
    {  
      
        // If n is a single digit number  
        if (n < 10)  
            return false;  
        int prodOdd = 1, prodEven = 1;  
      
        while (n > 0)  
        {  
      
            // Take two consecutive digits  
            // at a time  
            // First digit  
            int digit = n % 10;  
            prodOdd *= digit;  
            n /= 10;  
      
            // If n becomes 0 then  
            // there's no more digit  
            if (n == 0)  
                break;  
      
            // Second digit  
            digit = n % 10;  
            prodEven *= digit;  
            n /= 10;  
        }  
      
        // If the products are equal  
        if (prodEven == prodOdd)  
            return true;  
      
        // If products are not equal  
        return false;  
    }  
      
    // Driver code  
    public static void main(String args[])  
    {  
        int n = 4324;  
      
        if (productEqual(n))  
            System.out.println("Yes");  
        else
            System.out.println("No");  
    }  
    // This code is contributed by Ryuga 
}  

Python3

# Python implementation of the approach 
  
# Function that returns true if the product 
# of even positioned digits is equal to 
# the product of odd positioned digits in n 
def productEqual(n): 
    if n < 10: 
        return False
    prodOdd = 1; prodEven = 1
  
    # Take two consecutive digits 
    # at a time 
    # First digit 
    while n > 0: 
        digit = n % 10
        prodOdd *= digit 
        n = n//10
  
        # If n becomes 0 then 
        # there's no more digit 
        if n == 0: 
            break; 
        digit = n % 10
        prodEven *= digit 
        n = n//10
  
    # If the products are equal 
    if prodOdd == prodEven: 
        return True
  
    # If the products are not equal 
    return False
  
# Driver code 
n = 4324
if productEqual(n): 
    print("Yes") 
else: 
    print("No") 
  
# This code is contributed by Shrikant13 

C#

// C# implementation of the approach 
using System; 
  
class GFG 
{ 
      
// Function that returns true  
// if the product of even positioned 
// digits is equal to the product of 
// odd positioned digits in n 
static bool productEqual(int n) 
{ 
  
    // If n is a single digit number 
    if (n < 10) 
        return false; 
    int prodOdd = 1, prodEven = 1; 
  
    while (n > 0) 
    { 
  
        // Take two consecutive digits 
        // at a time 
        // First digit 
        int digit = n % 10; 
        prodOdd *= digit; 
        n /= 10; 
  
        // If n becomes 0 then 
        // there's no more digit 
        if (n == 0) 
            break; 
  
        // Second digit 
        digit = n % 10; 
        prodEven *= digit; 
        n /= 10; 
    } 
  
    // If the products are equal 
    if (prodEven == prodOdd) 
        return true; 
  
    // If products are not equal 
    return false; 
} 
  
// Driver code 
static void Main() 
{ 
    int n = 4324; 
  
    if (productEqual(n)) 
        Console.WriteLine("Yes"); 
    else
        Console.WriteLine("No"); 
} 
} 
  
// This code is contributed by mits 

PHP

0)
{

// Take two consecutive digits
// at a time

// First digit
$digit = $n % 10;
$prodOdd *= $digit;
$n /= 10;

// If n becomes 0 then
// there’s no more digit
if ($n == 0)
break;

// Second digit
$digit = $n % 10;
$prodEven *= $digit;
$n /= 10;
}

// If the products are equal
if ($prodEven == $prodOdd)
return true;

// If products are not equal
return false;
}

// Driver code
$n = 4324;
if (productEqual(!$n))
echo “Yes”;
else
echo “No”;

// This code is contributed by jit_t
?>

Output:

No

My Personal Notes

Suggest a Topic

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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