Sum of Digits code - returns array of 0s

Question

I am currently writing a python code that tries to find out the number of four digit numbers that add up to a specific value. My code is below:

def findSum ():
   target = 0;
   checkno = 1000;
   comboSum = [];
   sumArray = [];
   while target<=36 :
     while checkno<10000 :
        digitSum = int(str(checkno)[0]) + int(str(checkno)[1]) + int(str(checkno)[2]) + int(str(checkno)[3]);
        if digitSum == target :
            comboSum.append(checkno);
        checkno = checkno + 1;
     sumArray.append(len(comboSum));
     target = target + 1;
   print (sumArray);

findSum();

However, when I put this through the python interpreter, it returns an array of 36 "0"s. I am not quite sure why this is the case, when I increase the target every time, and then loops back through.

Does anyone know why this is the case?

Thanks!

Answer 1

You don't reset checkno to 1000 after you increase target and loop.

So first iteration on target, you get the correct answer 0. Second iteration, where target is 1 , your checkno is already 10000 , so the inner loop doesn't execute.

You need to move your initialisation of checkno and comboSum inside the outer loop.

Answer 2

You can replace your while loops by for variable in range() , like:

def findSum():
    # target = 0
    # checkno = 1000
    # comboSum = []
    sumArray = []
    for target in range(36):
        comboSum = []
        for checkno in range(1000, 10000):
            digitSum = int(str(checkno)[0]) +\
                       int(str(checkno)[1]) +\
                       int(str(checkno)[2]) +\
                       int(str(checkno)[3])
            if digitSum == target:
                comboSum.append(checkno)
            # checkno = checkno + 1
        sumArray.append(len(comboSum))
        # target = target + 1

    print(sumArray)

findSum()

It would be a more Pythonic way to do.

As highlighted by some other comments:

• never end lines by semicolon
• usually try to stick to PEP8 rules (whitespaces aroud operators for instance)

Answer 3

If you just want to find the count of how many four digit numbers sum to your target you can simply use divmod to get the digits and sum all the times the sum of the digits is equal to your target number:

def sum_digits(n, target):
    n, sm = divmod(n, 10)
    while n and sm <= target:
        n, rem = divmod(n, 10)
        sm += rem
    return sm == target



def find_sum(targ):
    for n in range(1000, 10000):
        yield sum_digits(n, targ)


print(sum(findSum(36)))

You are checking from 0-target which is wrong based on your description, find out the number of four digit numbers that add up to a specific value you should only be checking the target number.