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 = 1234Output:NoExplanation:

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 = 1024Output: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**.

- Find the last digit of
- 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;` `}` |

## Java

`// Java program for the above approach` `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 coode 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);` ` ` `}` ` ` ` ` `// Prlet 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(log_{10}N)**Auxiliary Space:** O(1)

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