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IRC log for #friendlyjava, 2019-08-24

##friendlyjava on freenode

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Time S Nick Message
00:01 Jantz_ joined ##friendlyjava
00:37 Jantz joined ##friendlyjava
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06:00 nanoz joined ##friendlyjava
06:00 nanoz aditsu, a single for loop is fine
06:01 nanoz https://pastebin.com/MM9e3Vjs
06:44 nanoz english is not my first language
06:44 nanoz Jantz
10:43 nanozz joined ##friendlyjava
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15:23 nanozz hello aditsu pdurbin
15:45 Jantz joined ##friendlyjava
16:15 aditsu hi
16:18 aditsu nanozz: I looked at your paste, it is very incomplete and extremely inefficient
16:22 nanozz aditsu, thanks for reviewing
16:22 nanozz how else i can make it efficent any thoughts
16:24 aditsu first of all, you don't need a StringBuilder
16:25 nanozz stringbuilder is used to add/remove and also match the regex for count increase
16:26 aditsu regex?????
16:26 aditsu sounds like you're going full Rube Goldberg
16:26 nanozz yes , build the string match with regex does it start with 1 and end with 1 or you can use startWith and endWith
16:29 aditsu this is your code: https://www.youtube.com/watch?v=nORRgU8sGdE
16:34 nanozz :P
16:35 nanozz please review and let me know :)
16:35 aditsu let's start with some basic things: how can you check if a substring of your string from i to j starts and ends with 1?
16:36 nanozz psubstring.startWith == psubstring.endWith == 1
16:39 aditsu that's bad (inefficient) for a couple of reasons: 1) creating a substring, 2) startsWith and endsWith work with strings, and there's some overhead to check string lengths, set up a 1-iteration loop, etc
16:40 aditsu how many characters from your string do you actually need to check?
16:40 nanozz for example if you take characters as follow
16:40 nanozz 1234
16:41 nanozz its substrings are
16:41 nanozz 1,2,3,4,123,12,234,23,34,1234
16:41 aditsu why are you talking about that?
16:42 nanozz an example so we can discuss on this for reference of our talk
16:43 aditsu well, you have the string, i and j, how many characters do you need to check to find out if the substring from i to j starts and ends with 1?
16:44 nanozz 2 characters
16:44 nanozz the first and last
16:45 aditsu good
16:45 aditsu and what are the positions of those characters in the original string?
16:46 nanozz 0 and n-1
16:46 aditsu no, those would be the positions in the substring
16:50 nanozz :?
16:51 aditsu well, ok, let's take your example, string is "1234", i=1, j=2
16:52 aditsu which characters do you need to check?
16:53 nanozz ith character and the last character
16:53 aditsu you're only half right..
16:54 aditsu what is the substring in this case?
16:54 nanozz 1234 is the substring sequence
16:54 aditsu no, "1234" is the original string
16:54 nanozz yes
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