re.search & eval clear solution for Safe Code by Sim0000 - python coding challenges

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re.search & eval solution in Clear category for Safe Code by Sim0000

import re

def safe_code(equation):
    eq = equation.replace('=', '==')
    f = bool(re.search('(^|[^0-9])##|#[0-9]', eq))
    for n in "0123456789"[f:]:
        if n not in eq and eval(eq.replace('#', n)): return int(n)
    return -1


print(safe_code('1-1=##'))
print(safe_code('111-11=1##'))
print(safe_code('#+#=#*#'))
print(safe_code('##=##'))
print(safe_code('##=5-5'))



if __name__ == '__main__':
    print("Example:")
    print(safe_code("-5#*-1=5#"))

    #These "asserts" using only for self-checking and not necessary for auto-testing
    assert safe_code("-5#*-1=5#") == 0
    assert safe_code("##*##=302#") == 5
    assert safe_code("19--45=5#") == -1
    assert safe_code("##--11=11") == -1
    assert safe_code("#9+3=22") == 1
    assert safe_code("11*#=##") == 2
    assert safe_code("#9+3=12") == -1
    print("Coding complete? Click 'Check' to earn cool rewards!")

Sept. 22, 2018

Comments:

Rapunzel85 on April 21, 2019, 10:02 p.m.
Thank you for this really good to understand and good to read solution! I'm not sure about the regular expression. Are patterns like -1#3*-1=1#3 allowed? There your expression doesn't work

Sim0000 on April 22, 2019, 1:04 a.m.
In this code, 0 must be excluded from the answer if certain conditions are met. The regular expression is used to check the condition. For -1#3*-1=1#3, this code returns 2. I need more parentheses. Following is correct regular expression. <pre class='brush: python'>f = bool(re.search('(^|[^0-9])(##|#[0-9])', eq)) </pre> This code returns 0 for the problem. Thank you @Rapunzel85

Escher

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