HackerRank - Repeated String | Full Solution with Simplified Examples and Visuals | Study Algorithms

[@shruthibudidha5508]

great and clear explanation sir!!Thank you so much sir Not only this sir,each and every video was awesome and easily understandable

[@kumarisuruchisuman8919]

No one can explain like you ☺

[@C__janhaviKadam]

crisp and clear ! thankss

[@CostaKazistov]

Really clear explanation. Definitely helped to understand this interesting HackerRank problem and the optimal solution. Many thanks!!

[@pradeeps9633]

bro you're just amazing👌👌.we want this type of teaching.

[@Lion-st5g3s378]

10 out of 10 explanation 🌼️, tysm!

[@MrMclit]

Great explanations, thank you

[@pratyakshamaheshwari8269]

Thank you Sir, you taught it very well.

[@TheAxelgr]

I love you man!

[@rohitsaka]

Can you Please make it on Stock Span Problem

[@nikoo28]

Reach out to me on my email in channel description with the problem link.

[@vaitheeswaranr2115]

StringLeftLength=n%strLength; Will be helpful?

[@prativa810]

why is the time complexity O(1)? You have run the two loops in the code

[@arjunmsit]

I guess it's because the loop is run for size of input string and not on N. I'm this case it's 5 which can be dropped to 0(1)

[@_JAGATHEESANV]

Why can't we use this stringLeftLength=n%strLength;

[@nikoo28]

you can

[@spencersedano]

how is this an easy question

[@niharikamurulidhara7017]

Sir....I tried the below code.. #include #include using namespace std; int main() { long long int n,count=0; int length,i; string sub_str; getline(cin,sub_str); cin>>n; length = sub_str.length(); if(length==1) { if(sub_str[i]=='a') { count=n; } else { count=0; } } // finding number of a's in substring else if(length>1) { for(i=0;i

[@nikoo28]

Try looking at the test cases I have in my code and debug your code with those cases. Maybe you can catch the problem

[@_httpantwon]

It still wasn't clear

[@nikoo28]

Which part did you face a problem at? I can try to elaborate

[@k16e]

Shouldn't you `Math.floor(n / sLength)` the number of reps?