How to check if one string ends with another, or the other way around?

Question

Given two strings, return True if either of the strings appears at the very end of the other string, ignoring upper/lower case differences (in other words, the computation should not be "case sensitive").

Examples / Tests:

>>> end_other('Hiabc', 'abc') True >>> end_other('AbC', 'HiaBc') True >>> end_other('abc', 'abXabc') True

My Code:

def end_other(s1, s2): s1 = s1.upper() s2 = s2.upper() if s1[2:6] == s2: return True elif s2[2:6] == s1: return True elif s2 == s1: return True else: return False

What I expect is wrong.

(NB: this is a code practice from CodingBat

DSM · Accepted Answer · 2012-11-17 19:16:57Z

Any reason you can't use the built-in functions?

def end_other(s1, s2): s1 = s1.upper() s2 = s2.upper() return s1.endswith(s2) or s2.endswith(s1)

Your code with the arbitrary slices doesn't make a lot of sense.

Ashwini Chaudhary · Accepted Answer · 2012-11-17 19:21:06Z

something like this, using str.rfind():

In [114]: def end(s1,s2): s1=s1.lower() s2=s2.lower() if s1 in s2 and s2.rfind(s1)+len(s1) == len(s2): return True elif s2 in s1 and s1.rfind(s2)+len(s2) == len(s1): return True return False .....: In [115]: end('Hiabc','abc') Out[115]: True In [116]: end('abC','HiaBc') Out[116]: True In [117]: end('abc','abxabc') Out[117]: True In [118]: end('abc','bc') Out[118]: True In [119]: end('ab','ab12') Out[119]: False

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