Consider the following pseudocode for a reduce-and-conquer recursive SumDigit(num) algorithm that finds the sum of all d

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Consider the following pseudocode for areduce-and-conquer recursive SumDigit(num) algorithm that finds thesum of all digits in a number. Example calling SumDigit(1936)should return 19 (which is 6 + 3 + 9 +1)
integer SumDigit (num)
input: an integer number, num
output: sum of all the digits in num
if num is a one-digit number //base case
- then the sum of all the digits in num is, num
// example if num=7, then SumDigit(num), should return 7
otherwise //general case: num has multiple digits in it
- break num into 2 parts
- One part, part 1, has the rightmost digit (use modulo operation to dothat). Ex. 1936 % 10 -> 6
- The other part, part 2, has all the digits of num,except the rightmost digit. (use regular division to do this.Ex. 1936 / 10 -> 193
- Call the function SumDigit() itself on part 2, and getthe result of the sum of all the digits in part 2..
- Add the result from SumDigit(part 2) to part 1, and returnit
a. Write the corresponding C++ or Python code for thepseudocode. Submit only the code for the function
SumDigit() and not the whole program.
b. Using the modulo operation “%”, addition operation “+”, and division operation “/” as basic
operations, set up and solve a recurrence relation for the number of basic operations the algorithm
executes.
c. Modify the code to create a new recursive functionCountDigits(num) that counts the number of digits
in a number num. Example: a call to the functionCountDigits(1587) would return the value 4. Submit
only the code for the function CountDigits(num).
d. Using the modulo operation “%”, addition operation “+”, and division operation “/” as basic
operations, set up and solve a recurrence relation for thenumber of basic operations for
CountDigits(num).