First clear solution for Largest Rectangle in a Histogram by Moff - python coding challenges

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First solution in Clear category for Largest Rectangle in a Histogram by Moff

def largest_histogram(h):
    result = min(h) * len(h)
    for w in range(1, len(h)):
        for i in range(len(h) - w + 1):
            result = max(result, min(h[i:i + w]) * w)
    return result

Nov. 16, 2016

Comments:

znatny_kulish on Jan. 4, 2017, 8:26 a.m.
I used the same logic :)

ositowang on June 2, 2018, 1:19 a.m.
I have a question. if you are going to update the result..why do we need initialize the result like that

serg3y1999 on Sept. 13, 2018, 10:10 a.m.
because you use result here: max(result, min(h[i:i + w]) * w) to find max in result,which was initialized

Goran7T on Nov. 26, 2019, 2:21 p.m.
Also, it is useful if len(h)=1 and code does not enter the for loop.

wadada220104 on April 7, 2022, 6:08 a.m.
I don't think so when using range(len(h))

serg3y1999 on Sept. 13, 2018, 10:20 a.m.
can you explain pls how it works : result = max(result, min(h[i:i + w]) * w) ?this key for max

Moff on Dec. 16, 2018, 8:29 a.m.
It's short way of `if` statement: ``` if value > result: result = value ``` equals ``` result = max(result, value) ```

piter239 on April 25, 2020, 7:33 p.m.
Nice and clear! I hoped to start with brute-force and then find better solution, but it seems to be the best approach. I used itertools.combinations - if you'd care to take a look <a href="https://py.checkio.org/mission/largest-histogram/publications/piter239/python-3/brute-force-maxmin-histogramstartstop-for-all-combinations-in-rangelen-1/">https://py.checkio.org/mission/largest-histogram/publications/piter239/python-3/brute-force-maxmin-histogramstartstop-for-all-combinations-in-rangelen-1/</a>

Darkalkemy on Aug. 22, 2020, 8:51 p.m.
This is exactly the solution I was looking for... but I implemented it in in a very ugly way.

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