__ clear solution for The Hidden Word by Cjkjvfnby - python coding challenges

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__ solution in Clear category for The Hidden Word by Cjkjvfnby

from itertools import zip_longest

def find_start(text, word):
    """
    Find word stat position in text.
    Line numeration from starts 1.
    """
    start_pos = text.find(word)
    if start_pos >= 0:
        line_number = text.count('\n', 0, start_pos)
        line_start = text.rfind('\n', 0, start_pos)
        row_start = line_number + 1
        column_start = start_pos  - line_start
        return row_start, column_start

def checkio(text, word):
    text = text.replace(' ', '').lower()
    distance = len(word) - 1
    start =  find_start(text, word)
    if start:
       row_start, column_start = start
       return [row_start, column_start, row_start, column_start + distance]
    else:
        text =  '\n'.join(''.join(x) for x in zip_longest(*text.split('\n'), fillvalue=''))
        start =  find_start(text, word)
        column_start, row_start = start
        return [row_start, column_start, row_start + distance, column_start]

Feb. 7, 2014

Comments:

guido on Feb. 19, 2014, 2:49 a.m.
It's interesting that you prefer making several passes over the input string just to avoid splitting it in case you have a horizontal match. In my own solution I first broke the text up in lines and then searched each line for the word. In the extreme, your approach will find the horizontal match in less time, using less memory, but the input would have to be pretty darn huge for this to matter to anyone. (I'm not sure whether CheckIO prefers solutions that are asymptotically optimal for huge input, I'm just noticing that the code might have been easier to follow if you had taken a more traditional approach. :-) <pre class='brush: python'>return row_start, column_start </pre> It's a little jarring that there's no return statement here showing what it returns on failure. Good style would be to add "else: return None". <pre class='brush: python'>def checkio(text, word): text = text.replace(' ', '').lower() </pre> Shouldn't you lowercase the word to search as well? (TBH the puzzle description is unclear about this, and the example problems will pass even if you don't lowercase text. :-) <pre class='brush: python'>distance = len(word) - 1 start = find_start(text, word) if start: row_start, column_start = start return [row_start, column_start, row_start, column_start + distance] else: text = '\n'.join(''.join(x) for x in zip_longest(*text.split('\n'), fillvalue='')) </pre> Oh, wow. This could have used a comment. You're transposing the text by splitting it into lines, using a variant of zip(*a) to transpose a 2D array, then joining the lines again. This is very clever, although I worry that (even while enabling it) it defeats the purpose of the cleverness in find_start() to some extent -- there are a fair number of copies of the text being taken while doing all this. Have you done a careful analysis comparing your version to a more straightforward solution? <pre class='brush: python'> start = find_start(text, word) column_start, row_start = start return [row_start, column_start, row_start + distance, column_start] </pre> All in all points for cleverness and correctness, though perhaps not the most accessible or educational solution. Congrats! I'll just end by giving my <a href="http://www.checkio.org/mission/hidden-word/publications/guido/python-3/generator/share/3ca0a3bda46a56cb8ca54fe7de215809/">own version</a> so you have something to compare (and I'd be happy to admit defeat if you manage to benchmark or analyze this!). One unique property of the version below is that findit() is a generator that produces all solutions -- the for-loop in checkio() just abandons the generator as soon as the first solution is found.

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