CODEWITHSUNDEEP
python-3.x
Swap01
Swap01

Reputation: 31

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit

I don't understand why this code gives me 'Error' and not 2. I am trying to do this without recursion. An example with recursion would also be appreciated. 

def addDigits(num):
   num = str(num)
   if(len(num) > 1):
   for i in range(0, len(num)-1):
        for j in range(1, len(num)):
            if(len(str(int(num[i]) + int(num[j]))) > 1):
                x = len(str(int(num[i]) + int(num[j])))
                for i2 in range(0, x-1):
                    for j2 in range(1, x):
                        if(x == 1):
                            return(int(num[i]) + int(num[j]))
                        else: 
                            return('Error')
            else:
                return('Error2')
    else:
        return(num)
print(addDigits(38))

Upvotes: 0

Views: 1688

Answers (3)

Aviroxi
Aviroxi

Reputation: 115

Here is one way to do it with loops,i am from java so bear with me with semicolons.

def addDigits(num):

        while(num >= 10):
            int temp = 0;
            while(num > 0):
                temp = temp + num % 10;
                num = num/10;
            
            num = temp;
        
        return num;
addDigits(38)

this should give you the correct answer which is 2

Upvotes: 0

Sridhar Murali
Sridhar Murali

Reputation: 390

Like you mentioned, recursive is the best way to do it. If you ever write a code with more than 2 loops, just think again whether it is absolutely necessary.

def add_digits(x):
    if x/10>1:
        x=add_digits(sum([int(i) for i in str(x)]))
    else:
        return x
    if x/10<1:
        return x
add_digits(123456)

output:

3

The above code will help you sum the digits until its a single digit number.

Unlike C or C++, python is not a super fast language. Avoiding loops will make a significant difference.

Upvotes: 1

Minghao
Minghao

Reputation: 31

I think if you output x and you will find x=2 rather than x=1, so the program run execute the else: return('Error').

Upvotes: 0

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