Welcome back, Problem solving is the most important part of a programmers life cause each and every day they work on solving issues so they need to be good in problem solving, So today we are going to Create a Program for Sum of the digits of a given number, We are going to implement this in Golang Programming Language if anyone wants the solutions in other languages as well please don’t hesitate to leave a comment, we are going trying to solve this problem via multiple approaches, approaches mentioned below:

1. Sum of the digits of a given number with input as string not using modular and divide
2. Sum of the digits of a given number using modular and divide
3. Sum of the digits of a given number using recursion
4. Sum of the digits of a given number using tail recursion

EXAMPLES

Input: n = 12345
Output: 15

Input: n = 1234
Output: 10

1. Given a number, find the sum of its digits.

When the number of digits of that number exceeds 1019 , we can’t take that number as an integer since the range of long long int doesn’t satisfy the given number. So take input as a string, run a loop from start to the length of the string and increase the sum with that character(in this case it is numeric)

Follow the below steps to solve the problem:

• Declare a variable sum equal to zero
• Run a loop from zero to the length of the input string
• Add the value of each character into the sum, by converting the character into it’s integer value
• Return sum

Below is the implementation of the above approach:

package main

import (
    "fmt"
    _ "strconv"
)



// function to get the sum of digits
func doSum(n string) int {
    // step one to get length of the digits of given number convert it to string
    sum := 0
    for i := range n {
        // Since ascii value of
        // numbers starts from 48
        // so we subtract it from sum
        sum += int(n[i]) - 48
    }
  
    // SECOND APPROACH IF FUNCTION INPUT IS INT
    // nn := strconv.Itoa(n)
    // for i := range nn {
    // 	// Since ascii value of
    // 	// numbers starts from 48
    // 	// so we subtract it from sum
    // 	sum = sum + int(nn[i]) - 48
    // }
  
    return sum
}


func main() {
    n := "12345"
    fmt.Println(doSum(n)) 
}

Output:

15

2. Sum of the digits of a given number using modular and divide:

Follow the below steps to solve the problem:

• Get the number
• Declare a variable to store the sum and set it to 0
• Repeat the next two steps till the number is not 0
• Get the rightmost digit of the number with help of the remainder ‘%’ operator by dividing it by 10 and adding it to the sum.
• Divide the number by 10 with help of ‘/’ operator to remove the rightmost digit.
• Print or return the sum

Below is the implementation of the above approach:

 
package main

import (
    "fmt"
)


// function to get sum of digits
func doSum(n int) int {
    sum := 0
    for n != 0 {
        sum += n % 10
        n = int(n / 10)
    }
    return sum
}

func main() {
    n := 12345
    fmt.Println(doSum(n)) 
}

Output:

15

3. Sum of the digits of a given number using recursion

Follow the below steps to solve the problem:

• Get the number
• Get the remainder and pass the next remaining digits
• Get the rightmost digit of the number with help of the remainder ‘%’ operator by dividing it by 10 and adding it to the sum.
• Divide the number by 10 with help of the ‘/’ operator to remove the rightmost digit.
• Check the base case with n = 0
• Print or return the sum

Below is the implementation of the above approach:

package main

import (
    "fmt"
)


// function to get sum of digits
func doSum(n int) int {
    if n == 0 {
        return 0
    }
    return n%10 + doSum(n/10)
}

func main() {
    n := 12345
    fmt.Println(doSum(n)) 
}

Output:

15

4. Sum of the digits of a given number using tail recursion

Follow the below steps to solve the problem:

• Add another variable “Val” to the function and initialize it to ( Val = 0 )
• On every call to the function add the mod value (n%10) to the variable as “(n%10)+val” which is the last digit in n. Along with passing the variable n as n/10. 
• So on the First call, it will have the last digit. As we are passing n/10 as n, It follows until n is reduced to a single digit. 
• n<10 is the base case so When n < 10, then add the n to the variable as it is the last digit and return the val which will have the sum of digits

Below is the implementation of the above approach:

package main

import (
    "fmt"
)


// function to get sum of digits
func tailRecursion(n int, v int) int {
  if n < 10 {
    v += n
    return v
  }

  return tailRecursion(int(n/10), (n%10)+v)
}

func main() {
  n := 12345
  fmt.Println(tailRecursion(n,0)) 
}

Output:

15

FULL CODE:

package main

import (
    "fmt"
)

// function to get the sum of digits
func simpleSum(n string) int {
    sum := 0
    for i := range n {
        // Since ascii value of
      // numbers starts from 48
      // so we subtract it from sum
    sum += int(n[i]) - 48
    }
    return sum
}

// function to get the sum of digits
func sumUsingModular(n int) int {
    sum := 0
    for n != 0 {
        sum += n % 10
        n = int(n / 10)
    }
    return sum
}

// function to get the sum of digits
func sumUsingRecursion(n int) int {
    if n == 0 {
        return 0
    }
    return n%10 + sumUsingRecursion(n/10)
}

// function to get sum of digits
func tailRecursion(n int, v int) int {
  if n < 10 {
    v += n
    return v
  }
  return tailRecursion(int(n/10), (n%10)+v)
}

func main() {
  n := 12345
  sn := "12345"
  simpleSum := simpleSum(sn)
  sumUsingModular := sumUsingModular(n)
  sumUsingRecursion := sumUsingRecursion(n)
  tailSum := tailRecursion(n,0)
  fmt.Printf("simple sum: %d\nmodular sum: %d\nrecursive sum: %d\ntail sum: %d\n",simpleSum, sumUsingModular, sumUsingRecursion,tailSum)
}

code link:
If you found code/algorithm incorrect or found better approach then please comment.