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Sum of digits of a number

In this article, we will discuss how to find the sum of digits of a given number. The task is to calculate the sum of all the individual digits present in the number, regardless of whether the number is positive or negative.

Problem Statement

Given a number, we need to calculate the sum of its digits. The input can be positive or negative.

Example:

  • Input: 123
  • Output: 6 (1 + 2 + 3 = 6)
  • Input: 562
  • Output: 13 (5 + 6 + 2 = 13)
  • Input: -562
  • Output: 13 (-5 + 6 + 2 = 13)

Algorithm

1. Start

2. Accept the number as input.

3. Initialize a variable called 'result' to store the sum of the digits.

4. If the number is negative, convert it to a positive number.

5. Use a while loop to iterate until the number becomes zero.

6. Inside the loop, extract the last digit of the number using the modulo operator and add it to the 'result' variable.

7. Divide the number by 10 to remove the last digit.

8. Repeat steps 6 and 7 until the number becomes zero.

9. Print the value of 'result', which represents the sum of the digits.

10. Stop

Pseudocode

sum_digits(number)
{
	result = 0
	if (number < 0)
	number = -number
	while (number != 0)
	{
	result += number % 10
	number /= 10
	}
	print result
}

main()
{
	sum_digits(123)
	sum_digits(562)
	sum_digits(-562)
}

Explanation

The given code calculates the sum of digits for a given number using the sum_digits function.

The sum_digits function takes a number as input and initializes a variable called 'result' to store the sum of the digits. If the number is negative, it converts it to a positive number by taking its absolute value.

It then uses a while loop to extract the last digit of the number using the modulo operator (%), adds it to the 'result', and divides the number by 10 to remove the last digit. This process continues until the number becomes zero.

Finally, the function prints the value of 'result', which represents the sum of the digits.

In the main function, several test cases are provided to demonstrate the functionality of the sum_digits function. The outputs for the given test cases are also shown.

Code Solution

Here given code implementation process.

//C Program
//Sum of digits of a number
#include <stdio.h>

// Method which are sum of given number 
// This method are add all digits and not take negative sign
void sum_digits(int number)
{
  int result = 0;
  if(number<0)
  {
    // Convert negative number to positive number
    number = - number;
  }
  while(number!=0)
  {
  
    // Adding last digit of a given number
    result+=number%10;
    
    // Remove last digit
    number/=10;
    
  }

  printf("%d\n",result );
}


int main()
{
  //Test Case
  sum_digits(123);
  sum_digits(562);

  //when passes negative number
  sum_digits(-562);

  return 0;
}

Output

6
13
13
/*
 C++ Program
 Sum of digits of a number
*/

#include<iostream>

using namespace std;

class MyNumber {
  public:

    // Method which are sum of given number 
    // This method are add all digits and not take negative sign
    void sum_digits(int number) {
      int result = 0;
      if (number < 0) {
        // Convert negative number to positive number
        number = -number;
      }
      while (number != 0) {
        // Adding last digit of a given number
        result += number % 10;
        // Remove last digit
        number /= 10;
      }
      cout << result << "\n";
    }
};
int main() {
  MyNumber obj ;
  //Test Case
  obj.sum_digits(123);
  obj.sum_digits(562);
  //When passes negative number
  obj.sum_digits(-562);
  return 0;
}

Output

6
13
13
/*
  Java Program
  Sum of digits of a number
*/

public class MyNumber {

  // Method which are sum of given number 
  // This method are add all digits and not take negative sign
  public void sum_digits(int number)
  {
    int result = 0;
   
    if(number<0)
    {
      // Convert negative number to positive number
      number = - number;
    }
    while(number!=0)
    {
    
      // Adding last digit of a given number
      result+=number%10;
      
      // Remove last digit
      number=number/10;
      
    }

    System.out.print(result+"\n" );
  }
  public static void main(String[] args) {
    
    MyNumber obj = new MyNumber();
    //Test Case
    obj.sum_digits(123);
    obj.sum_digits(562);

    //When passes negative number
    obj.sum_digits(-562);

  }


}

Output

6
13
13
/*
  C# Program
  Sum of digits of a number
*/
using System;
public class MyNumber {

	// Method which are sum of given number 
	// This method are add all digits and not take negative sign
	public void sum_digits(int number) {
		int result = 0;

		if (number < 0) {
			// Convert negative number to positive number
			number = -number;
		}
		while (number != 0) {

			// Adding last digit of a given number
			result += number % 10;

			// Remove last digit
			number = number / 10;

		}

		Console.Write(result + "\n");
	}
	public static void Main(String[] args) {

		MyNumber obj = new MyNumber();
		//Test Case
		obj.sum_digits(123);
		obj.sum_digits(562);

		//When passes negative number
		obj.sum_digits(-562);

	}


}

Output

6
13
13
# Python 3 Program
# Sum of digits of a number
class MyNumber :
  # This method are add all digits and not take negative sign

  # Method which are sum of given number 
  def sum_digits(self, number) :
    result = 0
    if (number < 0) :
      # Convert negative number to positive number
      number = -number
    
    while (number != 0) :
      # Adding last digit of a given number
      result += number % 10
      # Remove last digit
      number = int(number / 10)
    
    print(result)
  

def main() :
  obj = MyNumber()
  #Test Case
  obj.sum_digits(123)
  obj.sum_digits(562)
  #When passes negative number
  obj.sum_digits(-562)


if __name__ == "__main__":
  main()

Output

6
13
13
# Ruby Program 

# Sum of digits of a number
class MyNumber 
	# This method are add all digits and not take negative sign

	# Method which are sum of given number 
	def sum_digits(number) 
		result = 0
		if (number < 0) 
			# Convert negative number to positive number
			number = -number
		end
		while (number != 0) 
			# Adding last digit of a given number
			result += number % 10
			# Remove last digit
			number = number / 10
		end
		print(result ,"\n")
	end
end
def main() 
	obj = MyNumber.new()
	#Test Case
	obj.sum_digits(123)
	obj.sum_digits(562)
	#When passes negative number
	obj.sum_digits(-562)
end
main()

Output

6
13
13

/*
 Scala Program
 Sum of digits of a number
*/
class MyNumber {
	// Method which are sum of given number 
	// This method are add all digits and not take negative sign
	def sum_digits(value: Int): Unit = {
		var result: Int = 0;
        var number: Int = value;
		if (number < 0) {
			// Convert negative number to positive number
			number = -number;
		}
		while (number != 0) {
			// Adding last digit of a given number
			result += number % 10;
			// Remove last digit
			number = number / 10;
		}
		print(result);
        print("\n");
	}
}
object Main {
	def main(args: Array[String]): Unit = {
		var obj: MyNumber = new MyNumber();
		//Test Case
		obj.sum_digits(123);obj.sum_digits(562);
		//When passes negative number
		obj.sum_digits(-562);
	}
}

Output

6
13
13
/*
  Swift 4 Program
  Sum of digits of a number
*/
class MyNumber {
	// Method which are sum of given number 
	// This method are add all digits and not take negative sign
	func sum_digits(_ value: Int) {
		var result: Int = 0;
      	var number: Int = value;
		if (number < 0) {
			// Convert negative number to positive number
			number = -number;
		}
		while (number != 0) {
			// Adding last digit of a given number
			result += number % 10;
			// Remove last digit
			number = number / 10;
		}
		print(result);
	}
}
func main() {
	let obj: MyNumber = MyNumber();
	//Test Case
	obj.sum_digits(123);
	obj.sum_digits(562);
	//When passes negative number
	obj.sum_digits(-562);
}
main();

Output

6
13
13
<?php
/*
  Php Program
  Sum of digits of a number
*/
class MyNumber {
	// Method which are sum of given number 
	// This method are add all digits and not take negative sign

	public 	function sum_digits($number) {
		$result = 0;
		if ($number < 0) {
			// Convert negative number to positive number
			$number = -$number;
		}
		while ($number != 0) {
			// Adding last digit of a given number
			$result += $number % 10;
			// Remove last digit
			$number = intval($number / 10);
		}
		echo ($result."\n");
	}
}
function main() {
	$obj = new MyNumber();
	//Test Case

	$obj->sum_digits(123);
	$obj->sum_digits(562);
	//When passes negative number

	$obj->sum_digits(-562);
}
main();
?>

Output

6
13
13
/*
 Node Js Program
 Sum of digits of a number
*/
class MyNumber {
	// Method which are sum of given number 
	// This method are add all digits and not take negative sign
	sum_digits(number) {
		var result = 0;
		if (number < 0) {
			// Convert negative number to positive number
			number = -number;
		}
		while (number != 0) {
			// Adding last digit of a given number
			result += number % 10;
			// Remove last digit
			number = parseInt(number / 10);
		}
		process.stdout.write(result + "\n");
	}
}

function main(args) {
	var obj = new MyNumber();
	//Test Case
	obj.sum_digits(123);
	obj.sum_digits(562);
	//When passes negative number
	obj.sum_digits(-562)
}
main();

Output

6
13
13

Time Complexity

The time complexity of the given code is O(log n), where n is the absolute value of the input number. This is because the while loop iterates the number of digits times, and the number of digits in a number is proportional to log n.

Last updated on by Kalkicode




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