Sliding window technique - Inside code

[@mubasakz7651]

Great explanation! Someone get this man an award!

[@recursion.]

alright I just gave him a turing award now piss off

[@biggn7441]

Your videos are impressive! Can you do all of the major technical interview patterns and the intuition behind them? Thanks 💯

[@insidecode]

Thanks a lot!

[@shaheerzaman620]

Nice video! Here is the solution for vowels # max number of vowels in a string of size k def max_vowels(s, k): vowels = ("a", "e", "i", "o", "u") maxvowels = 0 for ch in s[:k]: if ch in vowels: maxvowels += 1 total = maxvowels for i in range(len(s) - k): if s[i] in vowels: total -= 1 if s[i + k] in vowels: total += 1 maxvowels = max(maxvowels, total) return maxvowels

[@insidecode]

perfect! good job

[@DAOXINNo15]

Would you say in this optimal way to be O(n) then because of how the two if statements are linear and all right?

[@asunthakeded4967]

sliding window made very very easier. Now I will be able to solve 80% of the problem related to sliding easily without thinking much and by visualizing

[@chaos.n.dissonance]

Thank you. I've been struggling to wrap my mind around ChatGPT's explanation of sliding window technique for a few days now, this is so much simpler. def count_vowels_efficientcode(word:str, k:int) -> int: VOWELS = set('aeiou') max_vowels:int = 0 n:int = len(word) if n

[@sameerhasan7376]

Beautiful explanation! Thank you! Was struggling with this for quite some time.

[@ehapahmed1069]

your explanation is just simply brilliant ... I love it

[@tankstechtutorials5721]

great video. Had fun brushing up on this algorithm. function maxNumOfVowels(s, k) { let maxNum = 0; function isVowel(c) { return /[aeiou]/.test(c) ? 1 : 0; } let leftBound = 0; let rightBound = k-1; let firstSlice = s.substring(leftBound, k); for (let index = 0; index < firstSlice.length; index++) { const element = firstSlice[index]; maxNum += isVowel(element); } let currentCount = maxNum; while (rightBound < s.length - 1) { leftBound++; rightBound++; if(isVowel(s[leftBound-1])) { currentCount -= 1; } if (isVowel(s[rightBound])) { currentCount += 1; } maxNum = Math.max(currentCount, maxNum); } return maxNum; }

[@rnniloyiii9082]

The name should be caterpillar 🙂

[@pierrehebert9743]

When you first introduced the problem, we wanted to pick out the best set of five books, not what the price of the best set was. That means we were looking for argmax rather than max. so the code would be closer to: def best_set_for_price(items, k): if len(items) max): max = offset argmax = i + 1 return argmax As a bonus, note how we never need to calculate the initial sum.

[@insidecode]

Thanks for your interesting answer

[@Daniel_WR_Hart]

mind == blown

[@badermuteb4552]

nicely done. the best tutorial ever. please continue making new vide like this, here is my solution: is it correct? def mnv(s,k): v = "aeoiu" ms =0 for i in range(k): if s[i] in v: ms +=1 max_substring = ms for i in range(len(s) - k): if s[i] in v: ms -= 1 if s[i+k] in v: ms += 1 max_substring = max(max_substring, ms) return max_substring how would you solve it?????

[@sarthaksharma9322]

Beautiful visualisation! and explanation of time complexities and algorithm, love the video so much!!❤

[@algorithmdatastructures9244]

Best video on this topic all over youtube. 😍

[@megamehdi89]

Please continue making videos. I love them ❤️

[@insidecode]

Sure! thanks a lot

[@probabilitycodingisfunis1]

Amazing videos, content is explained so well with impressive animations!

[@insidecode]

Thank you!

[@ESLRepeat]

thanks man, you explained the thing really well. I had fun taking the class

[@iezioaudi22]

this was the best explanation I have seen! THANK YOUU!

[@Samad_27]

Plz make more videos no one can beat your level Absolutely Brilliant

[@insidecode]

Thanks!

[@preminfi4887]

Great explanation! Awesome visualisation, it is easy to understand the problem and the solution.

[@ysmerde2933]

You know you are gonna learn something when the English speaker has an foreign accent ty for the tuto dude you insane

[@anniamatthews6803]

great explanation! you made this very simple

[@value_investing_indonesia]

Absolute genius, please continue making videoss

[@insidecode]

Thanks! Sure

[@codelinx]

Great channel. Great illustrations and examples....🙋‍♂️

[@rodrigoelias1987]

here is a node solution for the exercise at the end of the video: import { strictEqual } from "assert"; const vowels = new Map([ ["a", "a"], ["e", "e"], ["i", "i"], ["o", "o"], ["u", "u"], ]); const isVowel = (v: string) => vowels.has(v); const countVowels = (string: string, span = 5) => { let vowelsCount = 0; let maxVowels = 0; for (let i = 0; i < string.length; i++) { if (i - span - 1 > -1) { if (isVowel(string[i - span - 1])) { vowelsCount--; } } if (isVowel(string[i])) { vowelsCount++; } if (vowelsCount > maxVowels) { maxVowels = vowelsCount; } } return maxVowels; }; const input = "bacacbefaobeacfe"; strictEqual( countVowels(input), 4, "The expected ammount of vowels was 4, got " + countVowels(input) );

[@sainithyamsani4062]

keep uploading this kind of videos. your videos are awesome!!!

[@insidecode]

Sure! Thanks a lot

[@kaushik.aryan04]

please make more videos like this on recursion medium hard questions

[@momal51]

Very nice explanation, thanks man !

[@ramseykarr6870]

Best explanation ever!

[@methylie]

Your lectures are good 😊.keep posting vedios

[@rishabhranjan5162]

I love these kind of explanations

[@TechDoctorMalayalam]

Great explanation and illustration

[@paulsingh11]

Sounds like the key word to use the Sliding Window is "contiguous" when dealing with an Array?

[@odell4709]

Made things very clear, thank you so much

[@biloliddinfarkhodov3956]

This is the best explanation I have ever watched. Thank you!

[@internetandcomputerprobe4426]

Really Informative

[@shreyaskulkarni4805]

Impressive Explanation ❤️

[@Fabreg01]

Great video!!. This technique can also be leveraged in rolling hash calculations. Solution to the sliding window to calculate max vowels count in a string class Solution: def maxVowels(self, s: str, k: int) -> int: if len(s) < k: return 0 total = 0 lookUpv = {'a':True, 'e':True, 'i':True, 'o':True, 'u':True} for c in s[:k]: if c in lookUpv: total += 1 maxTotal = total for i in range(len(s)-k): char_to_add = s[i+k] if char_to_add in lookUpv: total += 1 char_to_remove = s[i] if char_to_remove in lookUpv: total -= 1 maxTotal = max(maxTotal,total) return maxTotal

[@insidecode]

Yes, I'm planning a video on Rabin-Karp algorithm, which uses rolling hash

[@insidecode]

Hey, the video on Rabin-Karp is out

[@Fabreg01]

@@insidecode Great!!.Thanks

[@bobbrysonn]

Great explanation!

[@probabilitycodingisfunis1]

How do you make these animations?

[@insidecode]

I use PowerPoint

[@R_SinghRajput]

Crazy explanation 🔥😎

[@cecitorresmx]

Your video help me a lot! Thank you

[@aahringer]

Well done, thank you for sharing!

[@Syuzaki1301]

tried this in JS. Used some helper functions so I don't dilute the main function: // HELPER FUNCTIONS function isVowel(str) { let vowels = "aeiou"; return vowels.includes(str); } function countVowels(str) { let count = 0; for (let i = 0; i < str.length; i++) { if (isVowel(str[i])) count += 1; } return count; } // MAIN FUNCTION function maxVowels(s, k) { let total = countVowels(s.slice(0, 5)); let maxVowels = total; for (let i = 0; i < s.length - k; i++) { if (isVowel(s[i])) total -= 1; if (isVowel(s[i + k])) total += 1; if (maxVowels < total) maxVowels = total; } return maxVowels; }

[@rodrigoelias1987]

Nice content, dropping a like right now

[@ananyieie]

thankyou sir, great explanation☺

[@averrows]

great video! what software did you use to create this video?

[@insidecode]

Thanks! I used PowerPoint

[@Millicx]

Priceless video

[@GurtDovletov]

Very good explanation woow

[@noname_2108]

very helpful, thanks!

[@callmeitsh]

Nice man 👍 Thank u bro ♥️

[@arsh2544]

awesome video!

[@peacepeace9462]

i dont understand why the brute force for best_total_price has the "for i in range(len(prices)-k+1)" instead of "for i in range(len(prices)-k)". Could anyone explain this?

[@tapin130505]

because the last element should be included

[@peacepeace9462]

I figured out why. len(prices)-k+1 is the starting index of the last combination of the books. The calculations of brute force and sliding window have different index ranges.

[@chanukachathurangawk]

Thanks a lot.

[@DiogoSilva-xx8nz]

awesome channel

[@____r72]

this channel is so top

[@kurdi_x5842]

thaankyou it is so clear

[@hammadhassan9014]

you are genius

[@charann18]

Just awesome!!!!

[@SanjuKumar-hk8yy]

I like your videos because your content with animation and your hard work, love you dude❤️. But I am not gay.

[@prathagautam9872]

To the point!

[@subee128]

thanks

[@devmahad]

thanks :)

[@benhassineyoussef738]

"the size doesnt have an impact"

[@whiz-code]

I have come up with a solution but it seems like brute force, and if so, someone help me optimize it: static int maxVowelsNum(String s, int k) { var list = List.of('a', 'e', 'i', 'o', 'u'); var p1 = 0; var p2 = k + 1; var max = 0; var count = 0; while (p2 < s.length()) { var chars = s.substring(p1, p2); for (var ch : chars.toCharArray()) { if (list.contains(ch)) { count++; } } max = Math.max(max, count); p1++; p2++; count = 0; } return max; } Thank you.

[@ericlongteaches]

5:03 I have no clue what you just said here

[@ericlongteaches]

what accent is this?

[@ShivamKadam-bc4dn]

2.07

[@ShivamKadam-bc4dn]

2:07

[@nullptr.]

Thanks