Hmmm... not easy clear solution for Surjection Strings by pakicetus - python coding challenges

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Hmmm... not easy solution in Clear category for Surjection Strings by pakicetus

def isometric_strings(str1: str, str2: str) -> bool:
    dic = {}
    for i in range(0, len(str1)):
        if str1[i] in dic and str2[i] != dic[str1[i]]: return False
        dic[str1[i]] = str2[i]
    return True

July 29, 2020

Comments:

veky on Aug. 11, 2022, 6:09 a.m.
It's not easy when you're not using the tools you have. :-] Look: You intend to go through str1 and str2 simultaneously. That's what `zip` is for. <pre class='brush: python'>m = {} for x, y in zip(str1, str2, strict=True): if x in m and y != m[x]: return False m[x] = y return True </pre> Now, you check whether m[x] exists and is different from y. You can do it easier with .get: <pre class='brush: python'>if y != m.get(x, y): return False </pre> However, since you also need to set it to y if it doesn't exist, you can rewrite the body as <pre class='brush: python'>for x, y in zip(str1, str2, strict=True): if y != m.setdefault(x, y): return False </pre> But then, the whole loop becomes a standard bounded universal quantification, expressible with `all`: <pre class='brush: python'>return all(y == m.setdefault(x, y) for x, y in zip(str1, str2)) </pre> You still think it's not easy? :-D

Strings Theory

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