Double loop clear solution for The End of Other by bryukh - python coding challenges

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Double loop solution in Clear category for The End of Other by bryukh

def checkio(words):
    """
    You can iterate throught set.
    """
    for w1 in words:
        for w2 in words:
            if w1 != w2 and (w1.endswith(w2) or w2.endswith(w1)):
                return True
    return False

Feb. 20, 2014

Comments:

Lingson on Feb. 26, 2014, 4:33 p.m.
Ugh.. Not doing any programming for some time makes me forget that you can do that instead of using index. How stupid of me! Thanks for reminding me :)

chhyx2008 on Sept. 11, 2014, 8:39 a.m.
But, sets do not support indexing.

Cjkjvfnby on March 1, 2014, 9:11 p.m.
For one side it is clear and readable but for other is so uneducative for it algorithmic complexity. You can use combinations from itertools to make all possible pairs.

bryukh on March 1, 2014, 9:18 p.m.
Yep, but it give just constant part O(C*N^2)

Lijantropique on May 29, 2014, 2:14 a.m.
Hello, You can use product: <pre class='brush: python'>for words in (itertools.product(words_set,repeat=2)): </pre> and w1=words[0], w2=words[1]. Regards

bryukh on May 29, 2014, 2:18 a.m.
Yes, but why?

Lijantropique on May 29, 2014, 2:27 a.m.
Nothing special, just an alternative way of doing the same thing.

maurice.makaay on Nov. 3, 2014, 5:50 p.m.
It is not the same thing (which is logical, otherwise the Python team would not have created two functions for it I presume). The moment you use product() instead of permutations(), you are not only comparing pairs of words, but you are also comparing words against themselves. Permutations of [1,2] -> [1,2], [2,1] Products of [1,2] -> [1,1], [1,2], [2,1], [2,2]

Lijantropique on Nov. 4, 2014, 4:50 p.m.
I didn't said permutations were the same as product, I said product can be used when you are dealing with nested for loops. Check the following piece of code: <pre class='brush: python'>import itertools as it def test(seq): lst = [] for item1 in seq: for item2 in seq: lst.append((item1,item2)) return lst test_list = [1,2,3] print (list(it.product(test_list,repeat = 2))) print (test(test_list) </pre>

maurice.makaay on Nov. 4, 2014, 9:15 p.m.
Oh yes, if you were only talking about the two for loops, then you're correct. My apologies. I was looking at: <pre class='brush: python'>for w1 in words: for w2 in words: if w1 != w2 ... </pre> So including that if statement. When using product() in combination with the if statement, better use permutations(), so the w1 != w2 check can be omitted.

Lijantropique on Nov. 5, 2014, 9:04 p.m.
You're absolutely right. With the “if”, the product is not the best option but the permutations.

odwl on March 2, 2014, 6:26 p.m.
By sorting words by len, you can have just a triangle matrix path By sorting lexical order, you can even avoid a search

bryukh on March 2, 2014, 8:19 p.m.
This is "Clear" solution special for newbies -- simple and straight.

kamikaze on April 17, 2014, 1:07 p.m.
my is cleaner :))

bryukh on May 18, 2014, 9:30 a.m.
Why? Cleaner and faster are different words. :-)

kamikaze on May 21, 2014, 10:21 a.m.
"Clear" != "Speedy" categories :)

bryukh on May 21, 2014, 10:36 p.m.
I know. But you said about "cleaner".

sandarovich on May 13, 2014, 12:58 p.m.
Thank you for publication. endswith - is proper solution in this case. But as for me "if statement" may be much shorter, like this: if w1 != w2 and w1[::-1].find(w2[::-1]) == 0 {You reversed words and suffix and compared it) Educational: [start:end:stride], and string function find()

Cjkjvfnby on July 23, 2014, 6:51 a.m.
If will be much shorter if you reverse all worlds and remove duplicates before loops:)

bryukh on May 18, 2014, 9:29 a.m.
<blockquote> w1[::-1].find(w2[::-1]) </blockquote> I don't think this is better -- two redundant copies and " == 0" are not educational :)

jollyca on May 18, 2014, 9:11 a.m.
Isn't it faster if you make the double-loop over the set sorted in the descended order of the lengths? This way you iterate through smaller sets...

bryukh on May 18, 2014, 9:27 a.m.
I didn't try to write "the fastest" solution here.

RyanAlexander on May 21, 2014, 3:58 p.m.
I learn a better/more efficient way from almost every one of your solutions. Thanks for publishing these.

zver1zver2 on July 6, 2014, 10:13 a.m.
very good code

kozino on July 22, 2014, 3:44 p.m.
Hi. This is not necessary: (w1.endswith(w2) or w2.endswith(w1)). This will be enough: w1.endswith(w2).

bryukh on July 23, 2014, 12:03 a.m.
Yep, you are right.

Hayertjez on Nov. 18, 2015, 1:19 p.m.
Yes thats what I thought. Makes the program a little bit more efficient.

masero on Aug. 11, 2014, 9:47 p.m.
Hi, as endswith() accepts tuples, isn't better to write <pre class='brush: python'>for w in words: if w.endswith(tuple(x for x in words if x!=w)): </pre> instead of : <pre class='brush: python'>for w1 in words: for w2 in words: if w1 != w2 and (w1.endswith(w2) or w2.endswith(w1)): </pre> ?

bryukh on Aug. 12, 2014, 12:44 a.m.
Good point. Honestly, I didn't know about this feature of "endswith". Thx

Happiness on Feb. 8, 2015, 6:23 p.m.
Just another way to do the same: <pre class='brush: python'>new_set = words_set.difference({wd}) </pre>

Blukano on Sept. 30, 2014, 7 a.m.
This is a nice solution - I didn't know about endswith, so my solution was a bit more complex. Also, what happens if the set contains two identical words? Then by definition one would end with the other (i.e. return True), but your solution would return False. Obviously that wasn't a condition that was checked, but I didn't realise that, so I converted to a list and iterated by index! Your solution is a little cleaner :)

cle99y on Jan. 9, 2015, 11:57 a.m.
a set can include identical data, try it :)

pynur on Oct. 16, 2014, 10:53 a.m.
why you use "w1.endswith(w2) or w2.endswith(w1)"? You think it's be faster then we use only 'w1.endswith(w2)'? On one side we return True on earlier iteration what we iterate. But other side in every iterate we evaluate the two expression. What you think about it? Whatever LIKE for clear solution.

bryukh on Oct. 16, 2014, 11:40 a.m.
I think in average it's the same. Honestly I don't remember why I did that way :-)

netgamer on Nov. 14, 2014, 6:54 p.m.
First time did i see this function (endswith)

viktor.pecheniuk on Nov. 26, 2014, 3:28 p.m.
cool:)

andreylarionov on Dec. 6, 2014, 1:31 p.m.
Cool! But, I think that OR condition is not needed: <pre class='brush: python'>if w1 != w2 and w1.endswith(w2): return True </pre> Correct me if I'm wrong. I'm a very newbie in Python.

yang.wang04 on May 27, 2016, 3:49 a.m.
Have the same doubts

hendrikrichard.scholten_e44275 on Dec. 20, 2014, 6:21 p.m.
Nice solution short and easy

marshall.zheng on Dec. 25, 2014, 7:26 a.m.
good and clear

cle99y on Jan. 9, 2015, 11:56 a.m.
better than slicing (DOH !)

pbeesley on Jan. 20, 2015, 4:51 p.m.
I got hung up on the indentation of the "return False". Why doesn't the indented "return false" override the "return True" inside the nest?

bryukh on Jan. 21, 2015, 11:29 a.m.
It's not "override", these returns are place in various "blocks"

pbeesley on Jan. 21, 2015, 5:36 p.m.
So the indentations are not so much a hierarchy (where the left most block(s) trump the return value of the more indented blocks)? So the False is assigned the return value by virtue of no "True" existing?

bryukh on Jan. 21, 2015, 6:16 p.m.
Yep, if inside "for"s we didn't return True, then after upper "for" we return False, because there no "correct" words.

TierprotB. on March 5, 2015, 1:02 a.m.
Only one minor thing - according to task in principle you can receive doubled word so this case should be counted too, it`s more, in general, correct to remove word while cycling and place it to the set after. If i understand task correctly ofcourse.

OlegKulik on March 30, 2015, 3:31 p.m.
You don't have to check that both w1.endswith(w2) OR w2.endswith(w1) as both conditions are duplicated at some point of the loops intersection so w1.endswith(w2) would be enough (it's covering all combinations)

kimonel99 on April 29, 2015, 8:38 p.m.
Wrong. <pre class='brush: python'>>>>checkio({"hello","hello"}) >>>False //wrong </pre>

bryukh on April 30, 2015, 5:29 p.m.
{"hello","hello"} -- is it possible? ;-)

bryukh on April 30, 2015, 5:30 p.m.
Yep.

iwill on May 15, 2015, 1:56 p.m.
I should have guessed there was a built-in function for this.

melizzaryan on May 26, 2015, 4:54 p.m.
<pre class='brush: python'>for word in words_set: for string in words_set: if string.endswith(word,1): return True </pre> puts the beginning of the range to check at the second letter so that you don't have to have the w1!= w2 statement Not sure why I didn't need to check if w1 ends with w2 OR if w2 ends with w1?

gnucian on June 21, 2015, 3:42 a.m.
Aha! The .endswith() method!

LokiOfFire on June 21, 2015, 5:16 a.m.
.endswith ...... Now I know. LOL. Think I was manually doing endswith. <pre class='brush: python'>word_list = list(words_set) found = False for i in range(0, len(word_list)): for j in range(0, len(word_list)): result = word_list[i].rfind(word_list[j]) if result > 0: if word_list[i][:result] + word_list[j] == word_list[i]: found = True return found </pre>

slon_ok on Nov. 27, 2015, 9:43 p.m.
When I look at this solution, I feel ashamed of my solution :(

zakroishikov on Dec. 1, 2015, 7:14 a.m.
Hi, here's a note: Russian translation says to check if one element of the set end with another OR there are any complete matches. So your solution does not fit the task in translation because of the first part of <pre class='brush: python'>if w1 != w2 and (w1.endswith(w2) or w2.endswith(w1)): </pre> Someone must fix Russian translation to avoid misunderstanding by someone :)

ploadyfree on Jan. 22, 2016, 8:44 a.m.
you could check only w1.endswith(w2), it would be faster

kastianoza on May 7, 2016, 11:27 p.m.
No point in writing : w1.endswith(w2) or w2.endswith(w1), one of them would be enough because when iterating through pairs of words you get both pair (w1, w2) and (w2,w1)

ljy95135 on May 25, 2016, 7:34 a.m.
w1.endswith(w2) or w2.endswith(w1) using <pre class='brush: python'>w1.endswith(w2) </pre> is also ok i think.

Shmebula on May 8, 2017, 4:20 p.m.
This is the solution I came up with and here I was feeling lazy and dumb for something so trivial. :p

qiang.zu.7 on March 5, 2018, 3:37 p.m.
The solution does not cover the case {'hello', 'aa', 'aa'}.

bryukh on March 6, 2018, 7:55 a.m.
Do you know what `set` data type is?

viktor.chyrkin on Oct. 26, 2021, 10:56 p.m.
you used your own hints to solve this. right?

Different kind of sets

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