Maximum Vowels in a Substring (LeetCode 1456) | Full solution with animations | Sliding Window

[@jatin_9911]

your explanation is on next level 🔥🔥...
and your final thoughts is really helpful
thank u so much..

[@udayrajvadeghar8555]

thank you sir : )

[@zainulabdin-h7r]

thanks for the tutorial teacher

[@rupeshpatil6957]

Thanks Brother .. Subscribed!!

[@mohammadsalim2397]

Thanks, bro, but I didn't understand why you started the second loop with 'i=k' (we only need to increment 'i' once). Let's say, for example, k=3; at that point, we don't check the values 0 and 1. Also, why should the condition be 'i

[@jritzeku]

First loop populates our first window which is 'k' length. In second loop he is resuming loop from where we left off in first loop so we start at 'k'. I think we could have done this all in one loop tho.

[@nikoo28]

I didn’t understand your question.

[@ankitgupta9272]

Please make a playlist, so i can solve all these type of question at first and then watch your video if i got stuck

[@saikiran1426]

May be Elements of the array are contiguous .but string is also a char array,so,i think elements in the string also contiguous

[@AngelOfDeath943]

Initially I used string to store aeiou then use string contains method to check if vowels are there or not , so iam guessing this check takes longer time than looking in Hashset

[@nikoo28]

Hashset is the fastest

[@SubhajitDas-mt7sn]

cpp
class Solution {
public:
int maxVowels(string s, int k) {
unordered_set vowel = {'a', 'e', 'i', 'o', 'u'};
int Max_vowel = 0;
int Window_vowel = 0;
for (int i = 0; i < k; i++) {
if (vowel.count(s[i])) {
Window_vowel++;
}
}
Max_vowel = Window_vowel;
for (int i = k; i < s.size(); i++) {
if (vowel.count(s[i - k])) {
Window_vowel--;
}
if (vowel.count(s[i])) {
Window_vowel++;
}
Max_vowel = max(Max_vowel, Window_vowel);
}
return Max_vowel;
}
};

[@ChefBouny]

Your sound feels like chocolate 🍫

[@nikoo28]

hahah....

[@danishshaikh9205]

You are doing soo great Please never stop 🤌🏻✨✨