Check if sum of digits of a number exceeds the product of digits of that number

Given a positive integer N, the task is to check if the sum of the digits of N is strictly greater than the product of the digits of N or not. If found to be true, then print “Yes”. Otherwise, print “No”.

Examples:

Input: N = 1234
Output: No
Explanation:
The sum of the digits of N(= 1234) is = 1 + 2 + 3 + 4 = 10.
The product of the digits of N(= 1234) is 1*2*3*4 = 24.
As the sum of the digits is smaller than the product of the array. Therefore, print No.

Input: N = 1024
Output: Yes

Approach: Follow the steps below to solve the given problem:

• Initialize two variables, say sumOfDigit as 0 and prodOfDigit as 1 that stores the sum and the product of the digits of N.
• Iterate until N is greater than 0 and perform the following steps:
  • Find the last digit of N and store it in a variable, say rem.
  • Increment the value of sumOfDigit by rem.
  • Update the value of prodOfDigit as prodOfDigit*rem.
• After completing the above steps, if the value of sumOfDigit is greater than prodOfDigit then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

// Function to check if the sum of the digits of N is
// strictly greater than the product of the digits of N or
// not
void check(int n)
{
    // Stores the sum and the product of the digits of N
    int sumOfDigit = 0;
    int prodOfDigit = 1;
    while (n > 0) {
        // Stores the last digit if N
        int rem;
        rem = n % 10;
        // Increment the value of sumOfDigits
        sumOfDigit += rem;
        // Update the prodOfDigit
        prodOfDigit *= rem;
        // Divide N by 10
        n /= 10;
    }

    // Print the result
    if (sumOfDigit > prodOfDigit)
        cout << "Yes";
    else
        cout << "No";
}

// Driver Code
int main()
{
    int N = 1234;
    check(N);
    return 0;
}

C

// C program for the above approach
#include <stdio.h>

// Function to check if the sum of the digits of N is
// strictly greater than the product of the digits of N or
// not
void check(int n)
{
    // Stores the sum and the product of the digits of N
    int sumOfDigit = 0;
    int prodOfDigit = 1;
    while (n > 0) {
        // Stores the last digit if N
        int rem;
        rem = n % 10;
        // Increment the value of sumOfDigits
        sumOfDigit += rem;
        // Update the prodOfDigit
        prodOfDigit *= rem;
        // Divide N by 10
        n /= 10;
    }

    // Print the result
    if (sumOfDigit > prodOfDigit)
        printf("Yes");
    else
        printf("No");
}

// Driver Code
int main()
{
    int N = 1234;
    check(N);
    return 0;
}

// This code is contributed by Sania Kumari Gupta

Java

// Java program for the above approach
import java.io.*;
public class GFG{

// Function to check if the sum of the
// digits of N is strictly greater than
// the product of the digits of N or not
static void check(int n)
{

    // Stores the sum and the product of
    // the digits of N
    int sumOfDigit = 0;
    int prodOfDigit = 1;

    while (n > 0)
    {

        // Stores the last digit if N
        int rem;
        rem = n % 10;

        // Increment the value of
        // sumOfDigits
        sumOfDigit += rem;

        // Update the prodOfDigit
        prodOfDigit *= rem;

        // Divide N by 10
        n /= 10;
    }

    // Print the result
    if (sumOfDigit > prodOfDigit)
        System.out.println("Yes");
    else
        System.out.println("No");
}

// Driver code
public static void main(String[] args)
{
    int N = 1234;

    check(N);
}
}

// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach

# Function to check if the sum of the
# digits of N is strictly greater than
# the product of the digits of N or not
def check(n):

    # Stores the sum and the product of
    # the digits of N
    sumOfDigit = 0
    prodOfDigit = 1

    while n > 0:

        # Stores the last digit if N
        rem = n % 10

        # Increment the value of
        # sumOfDigits
        sumOfDigit += rem

        # Update the prodOfDigit
        prodOfDigit *= rem

        # Divide N by 10
        n = n // 10

    # Print the result
    if sumOfDigit > prodOfDigit:
        print("Yes")
    else:
        print("No")

# Driver Code
N = 1234

check(N)

# This code is contributed by jana_sayantan

C#

// C# program for the above approach
using System;

class GFG{

// Function to check if the sum of the
// digits of N is strictly greater than
// the product of the digits of N or not
static void check(int n)
{

    // Stores the sum and the product of
    // the digits of N
    int sumOfDigit = 0;
    int prodOfDigit = 1;

    while (n > 0)
    {

        // Stores the last digit if N
        int rem;
        rem = n % 10;

        // Increment the value of
        // sumOfDigits
        sumOfDigit += rem;

        // Update the prodOfDigit
        prodOfDigit *= rem;

        // Divide N by 10
        n /= 10;
    }

    // Print the result
    if (sumOfDigit > prodOfDigit)
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}


// Driver Code
public static void Main()
{
    int N = 1234;

    check(N);

}
}

// This code is contributed by code_hunt.

Javascript

<script>

// JavaScript program for the above approach

// Function to check if the sum of the
// digits of N is strictly greater than
// the product of the digits of N or not
function check(n)
{

    // Stores the sum and the product of
    // the digits of N
    let sumOfDigit = 0;
    let prodOfDigit = 1;

    while (n > 0)
    {

        // Stores the last digit if N
        let rem;
        rem = n % 10;

        // Increment the value of
        // sumOfDigits
        sumOfDigit += rem;

        // Update the prodOfDigit
        prodOfDigit *= rem;

        // Divide N by 10
        n = Math.floor(n / 10);
    }

    // Print the result
    if (sumOfDigit > prodOfDigit)
        document.write("Yes");
    else
        document.write("No");
}

// Driver Code

    let N = 1234;

    check(N);

</script>

Output:

No

Time Complexity: O(log₁₀N), as we are traversing the digits which will cost log₁₀N time.
Auxiliary Space: O(1), as we are not using any extra space.