getting zeros from an integer using python recursion

user_17624417 0
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I want to extract all the zeros from a number and find the length of the number of zeros using Python recursion. I've tried the following code but it gives me an error. I can do that using for loop but I want to know how python recursion does this. For this code output should be 3.

def zeros(n):
    x = list(str(n))
    if x == []:
        return []
    else:
        head = x[0]
        tail = "0"
        return len([head]) + zeros(tail)


print(zeros(10010))

RecursionError: maximum recursion depth exceeded while calling a Python object

recursion python
3 Answers

because after first loop , head is always a scalar value (number) and tail is always 0 , so always the else part of your logic runs and there is no logic to stop the recursion for a specific condition, and eventually it hits the recursion limit with the error message you got.

you can count number of occurrence of zeros in a number this way easily:

print(str(10010).count('0'))

you don't need recursion to solve this problem really.

user_13674540

If, for some reason, you don't like the simplicity of using count as already suggested and REALLY want to use recursion, the following code works:

def zeros(n):
    n = str(n)
    if not len(n): 
        return 0
    elif n[0] == '0':
        return 1 + zeros(n[1:])
    else:
        return zeros(n[1:])
user_174846630

No need for strings, just arithmetic:

def zeros(n):
    if n == 0:
        return 1
    else:
        if n < 0:
            n = -n
        nb_zeros = 0
        while n > 0:
            n, u = divmod(n, 10)
            if u == 0:
                nb_zeros += 1
        return nb_zeros

# TESTING
for n in 0, 8, 100, 10010:
    print('{:6d} --> {}'.format(n, zeros(n)))

# OUTPUT
#      0 --> 1
#      8 --> 0
#    100 --> 2
#  10010 --> 3

Explanation: we repeatedly extract and remove the unit digit. For instance, if:

n == 1037

then:

n, u = divmod(n, 10)

results in:

n == 103, u == 7
user_30807230
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