Question

How to check if a number is cyclops (number with odd number of digits and zero in the center) in Python

Answer 1

import math 

def isCyclopsNumber(n) :
    if (n == 0) :
        return True
    m = n % 10
    count = 0
    
    while (m != 0) :
        count += 1
        n = math.floor(n / 10)
        m = n % 10
    
    n = math.floor(n / 10)
    m = n % 10
    
    while (m != 0) :
        count -= 1
        n = math.floor(n / 10)
        m = n % 10
    
    return n == 0 and count == 0

print(("yes" if isCyclopsNumber(209) else "no"))
print(("yes" if isCyclopsNumber(18037) else "no"))
print(("yes" if isCyclopsNumber(5604) else "no"))



'''
run:

yes
yes
no

'''

 

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Answer 2

def isCyclopsNumber(n) :
    if (n == 0) :
        return True
    m = n % 10
    count = 0
    
    while (m != 0) :
        count += 1
        n = n // 10
        m = n % 10
    
    n = n // 10
    m = n % 10
    
    while (m != 0) :
        count -= 1
        n = n // 10
        m = n % 10
    
    return n == 0 and count == 0

print(("yes" if isCyclopsNumber(209) else "no"))
print(("yes" if isCyclopsNumber(18037) else "no"))
print(("yes" if isCyclopsNumber(5604) else "no"))



'''
run:

yes
yes
no

'''

 

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Answer 3

def isCyclopsNumber(num) :
    s = str(num)
    
    if not len(s) % 2 :
        return False
    
    if not s.count('0') == 1: 
        return False
    
    mid_index = len(s) // 2
    if s[mid_index] == '0' :
        return True
    
    return False

print(("yes" if isCyclopsNumber(209) else "no"))
print(("yes" if isCyclopsNumber(18037) else "no"))
print(("yes" if isCyclopsNumber(5604) else "no"))




'''
run:

yes
yes
no

'''

 

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