Repeated String HackerRank Solution [Optimal Approach]

[@JavaAidTutorials]

If you find this tutorial helpful, please do not forget to like , comment, share and It would be great if you can leave your feedback about the tutorial, as I have put a lot of hard work to make things easy for you. Thanks ..!! 🙏🙏🙏

[@eyobghebremicael5791]

As the repeated string isn't important, we can calculate the count of a's easily as follows public static int counta(String a, int n){ int len = a.length(); int count=0; //empty string if (len==0){ return 0; } for (int j=0;j

[@jenish9854]

Osm bro...

[@randiaz95]

This implementation is slow for anything above a million

[@B3Band]

@@randiaz95 Check mine. Way faster since you don't actually check n letters of the string. Plus this one is lossy. If n > 2^31 the for loop can't handle it because you used an int. It shouldn't even compile.

[@raba650]

I use JavaScript as my choice for these challenges, but your explanation still allowed me to better understand the problem as we as come up with & understand a solution. Thank for the clearness.

[@JavaAidTutorials]

You're very welcome!

[@navinchainani4721]

Hi send this code in javascript email is [e-mail]

[@pritamhalder5723]

bhai please help me my email and Facebook hacking they are force fully asking money for me very urgent my account please suggest me please 🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

[@davidtemidayo8541]

Hello, thank you for this tutorial, i was able to write this using javaScript and it works function evaluateValue(s,n){ const sLength = s.length; const quotient = Math.floor(n / sLength); const r = n % sLength; const freqA = s.match(/a/g).length; const useRemainder= (r > 0) ? s.slice(0, r).match(/a/).length : 0 const a = quotient * freqA + useRemainder; return a; } evaluateValue("abc", 1000000000000);

[@TheManishYadav01]

This is best plateform of learning code, 👍👍

[@gamer-zy1uj]

Perfect👍 thanks 😊 you taught so calmly and your teaching way is literally nice .keep going on and please upload more videos on the questions which are asked by fang frequently in java itself🙏🙏🙏🙏🙏🙏 please sir🤗

[@NitishKumar-un9ev]

The most toughest thing in HackerEarth is understanding the problem/Question

[@pritamhalder5723]

Help me please

[@anil2009]

Guruji you are great!!

[@JavaAidTutorials]

😀

[@lasithadulshan7357]

Exellent teaching!! Grow up 💪🏽

[@jlecampana]

Fantastic way of solving it! Thank you.

[@JavaAidTutorials]

Happy to help!

[@jn2869]

Great teaching, you are awesome, keep up the good work

[@rocker31590]

Wow awesome 👏🏼

[@darrellwalker2149]

public static int AddString(string _str, int n) { string newStr = string.Empty; while(newStr.Length < n) { newStr += _str; } return newStr.Substring(0,n).ToCharArray().Where(x=>x.ToString() =="a").Count(); }

[@ronaldomaia]

const repeatedString = (s, n) => { const countA = s => s.split('').filter(char => char === 'a').length; return countA(s) * Math.floor(n / s.length) + countA(s.slice(0, n % s.length)); }

[@priyasurgical4599]

You are doing Great sir 😎

[@AkshaySangolli]

Excellent, but what if the a character is present in the middle of the string i.e. s = "bac"?

[@mayanksrivastava4452]

Great as Usual....

[@programminginc8539]

excellent approach

[@archana09424]

Wonderful explanation sir.Thank you for educating us.

[@harshmanola563]

long size = s.length(), repeated = n/size; long left = n - (size * repeated); int extra = 0; int count = 0; for(int i = 0; i < size; i++){ if(s.charAt(i) == 'a'){ ++count; } } for(int i = 0; i < left; i++){ if(s.charAt(i) == 'a'){ ++extra; } } repeated = (repeated * count) + extra; return repeated;

[@B3Band]

Lossy. Your for loop uses int, but it's expected to loop a long number of times.

[@hoddybhaba6704]

Nice explaination!

[@JavaAidTutorials]

Glad it was helpful!

[@faizandarwesh7867]

Quality content.

[@youshinde367]

Great As usual thank you so much

[@nareshle3159]

Hii sir am Naresh Can you please tell elobourately about coding in eclipse ide so that coding beginners also easily understand

[@daromarra5823]

Perfect! Thanks!

[@Shohan01]

What if the s = ‘bac’ how does that match formula works?

[@mohankadam4983]

Thank you so much Sir....It's very helpful

[@JavaAidTutorials]

Most welcome

[@michaeljackson6043]

I will calculate the number of a's of given string and then using n integer , i calculate the n/len(s) and n%len(s) result=(n/len(s)*no.of.a's) +how many chracters in n%len(s) substring

[@B3Band]

Similar to my posted solution (i posted the actual java). Substring can only take int (and n is long), but Since String can only hold 2^31, you know casting (n % length) to int will work.

[@B3Band]

public static long repeatedString(String s, long n) { int length = s.length(); //length of s long repeats = n / length; //if n is 10 and length is 3, repeats 3 times long append = n % length; //the extra substring at the end of the infinite string String suffix = s.substring(0, (int)append); //10 % 3 = 1, so append one letter to String //we know String can only hold 2^31, so //casting to an int is okay! long count = s.chars().filter(ch -> ch == 'a').count(); //this is how you count chars in Java 8+ //count the a chars in the original string long extraBit = suffix.chars().filter(ch -> ch == 'a').count(); //now count the extra chars from the substring long total = count * repeats + extraBit; //the total number of a chars. Return it! return total; }

[@justfun4976]

What is the use of sc.close()

[@simsonpeak3119]

Sir, does string should always start with a, what if the s=bca ,n=10 and 10%3 =1 and 10/3=3 and a occurrence will be 4 as per your code. But actual a occurrence is 3.

[@nareshkoude2133]

Correct !! it won't work...

[@sulemani2482]

first of all thanks sir, but I would like to know can we prepare ourself for .net core on macbook, I am facing an issue of connection string with database on mac after learning c# and sql

[@memesmacha61]

Loved it..!

[@manvirchoudhary107]

Thank you sir for the teaching. Would you tell us how we can come up with this kind of mathematical approach ? where we can learn ?. I am able to do simple brute force approach .Request from you please dont stop making videos on problem solving. Please do more videos on that,

[@mahmmadhusen6794]

Nice explanation

[@JavaAidTutorials]

Thank you ☺️

[@SarfrazRazaOfficial]

which software you are using for writing on the board...???

[@hareeshhari9293]

Hi thank you so much for the solution its really helpful I was little confused in partiallength calculation could you please explain

[@JavaAidTutorials]

You still have this doubt for partial length logic or its cleared now ?

[@pasumarthiashik1099]

@@JavaAidTutorials can u explain for loop

[@srinivaskaribandi1089]

@@pasumarthiashik1099 in python my way of sloving s=input() n=int(input()) list_a=[] lenth_of_stng=len(s) nums_add1=n//lenth_of_stng nums_add2=n%lenth_of_stng req1=nums_add1*s req2=s[:nums_add2] total=req1+""+req2 first_letter=total[0] count=0 for j in total: if first_letter==j: count+=1 print(count)

[@coffeemaker693]

you no need to concatenate the string, better split string into Individual character first. in C# as easy as : stringObj.ToArray(); in java as simple as : String s1="hello"; char[] ch=s1.toCharArray();

[@pasumarthiashik1099]

why have u taken acount and can u explain the for loop because string length in for loop is 3 or 10???

[@srinivaskaribandi1089]

My way of sloving problem like this s=input() n=int(input()) list_a=[] lenth_of_stng=len(s) nums_add1=n//lenth_of_stng nums_add2=n%lenth_of_stng req1=nums_add1*s req2=s[:nums_add2] total=req1+""+req2 first_letter=total[0] count=0 for j in total: if first_letter==j: count+=1 print(count)

[@MynameisJanRed]

Thank you

[@JavaAidTutorials]

You're welcome

[@shraddhapawar5674]

Can we edit main function given by hackerrank .

[@mohamedfarhan5894]

I have tried python3, and my code works well. But it won't work for very large n values (more than 8-digit) .please check and give suggestions . def char_occ(s,n,char): k = s*n return k[0:n].count(char) examples >char_occ('abcac',10,'a') >4 >char_occ('aba',10,'a') >7

[@orchidnewbie]

Thank you for the video. What is the Time Complexity on this solution, max s.Length

[@yashwanthchigari14]

It's going to be linear: O(n)

[@coffeemaker693]

C# version , my solution static void Main(string[] args) { string s = "aba"; Console.WriteLine(RepeatString(10, s)); Console.ReadKey(); } private static int RepeatString(int n, string s) { string result = string.Empty; var characters = s.ToArray(); var countA = 0; for (int i = 0, index = 0; i < n; i++, index++) { if (index == characters.Length) index = 0; if (characters[index] == 'a') countA++; result += string.Concat(characters[index]); } Console.WriteLine(result); return countA; }

[@prasuguna]

If it's bca then this formula fails?

[@babr5401]

why you usen't stream ?

[@anandkatam8150]

Sir please make hacker rank python preparation kit solutions also please sir

[@arjishat]

java7 , java8, java15 er modde partokko ta ki??

[@NateFreestyle]

HackerRank classifies this as easy? I'm not sure I agree with that.

[@hariprasad1727]

How We Need To Set Environment In Web IDE For Spring Boot Applications

[@genericbinary558]

what about "baac" and n=13 🙂

[@kuldeepsinghrajput2600]

why u used long for storing values?

[@NateFreestyle]

To take into account very large input values.

[@ashutosh3387]

Is there any complete course of dsa in java

[@JavaAidTutorials]

Currently we do not have any courses yet, but lot of subscribers asked this question, so planning to create in future and inform you all. :)

[@samiksharamteke2855]

long repeatedString(string s, long n) { long freq = n /s.length(); long remains = n % s.length(); int count = 0; for(int i=0;i

[@shenth27]

Quite a bit complicated solution. I'd simply loop n number of times, maintains another index for the s, which I resets every time index >=s.length(). Then in the loop just increment the counter whenever you find 'a'. But I guess not an optimal solution if n is huge number though.

[@JavaAidTutorials]

yes , if n is huge, it will not pass all TCs

[@B3Band]

"simply" loop n times, when n can be 10^12 hahahaha Very slow solution. Check out mine, which only checks the original String, not the infinite string.

[@IdontWatchIsee]

I didnt understand how you translate the algorithm in code :((( what can i do to understand this part better?? Please help

[@JavaAidTutorials]

Do write lots of code. Learn more about your chosen programming language what are the utils , functionality etc it offers and do practice. you will see the difference. practice is the only key which can quickly help you to translate algorithm into code.

[@IdontWatchIsee]

@@JavaAidTutorials thank you very much for taking the time to answer ❤️

[@abhisekpanda3701]

import java.util.Scanner; public class recur { public static void main(String[] args) { Scanner sc=new Scanner(System.in); System.out.println("Enter the word"); String word=sc.next(); System.out.println("Enter the number"); int num=sc.nextInt(); long l=(num/(word.length())); long l1=num%(word.length()); String s=sort(word,l,l1); System.out.println(s); } static String sort(String word,long l , long l1) { String str=""; char ch1=' '; for(int i=0;i

[@hariprasad1727]

WEB IDE I am Not able To Understand

[@salilkumarghosh9861]

Sir can I solve these problems using C after watching all videos?

[@JavaAidTutorials]

We always try to teach generic algorithm which is independent of any specific language. so you can use any language to implement the same .

[@tauqeerahmed1366]

If someone wants to see this in swift let c = "baa" let n = 10 func checkNoOfCharactersIn(souceString : String , character : Character) -> Int { var count = 0 for char in souceString { if character == char { count += 1 } } return count } func getSubstringToString(sourceString : String,to : Int) -> String { return String(sourceString[.. Int { var answer = 0 let lengthOfOneTimeString = sourceString.count let repeatTimese = n / lengthOfOneTimeString let reminderLimit = n % lengthOfOneTimeString answer = checkNoOfCharactersIn(souceString: sourceString, character: "a") answer = answer * repeatTimese answer = answer + checkNoOfCharactersIn(souceString: getSubstringToString(sourceString: sourceString, to: reminderLimit), character: "a") return answer } var count = numberOfTimeAInNLetters(sourceString: c, n: n)

[@PrivacyRocks]

Your heavy India/pakistani accent makes it impossible to understand you. I’m looking for someone else