NYJ1qCr Use Python 3:
s, n = input().strip(), int(input().strip()) print(s.count("a") * (n // len(s)) + s[:n % len(s)].count("a"))
satyasri12511001 print(s.count("a") * (n // len(s)) + s[:n % len(s)].count("a")) please any one explain me this
BMichaels It counts the number of "a" in a full string, and in the last, potentially partial string, and calculates the total amount of "a" based on that.
"a" count of full string * number of string repeats + "a" count of last string.
satyasri12511001 thanks for your help bro
vishalbty Iterating the whole string and the sliced string is not necessary. Here's how I did it by dividing our string into two parts and iterating each part only once.
yashpalsinghdeo1 here is problem solution in Python java c++ and c programming. https://solution.programmingoneonone.com/2020/07/hackerrank-repeated-string-solution.html
Baji_Shaik Thank you
hemanthkumar12 Thank you
jayanthsaikiran [deleted]addhawk [deleted]prasadbiradar037 could u explain the code
pranita_kadlag can you explain the code
kcsals This code works pretty well even though it can be written a lot shorter. I'll help explain since he never did. The first two if statements are trying to avoid really long series if it just has n = 100000000 and s=a. Not needed if you coded in a better way, but it works. Now starting with the "l" variable. That counts the number of "a's" in a given series, so [a,b,c,a] would return 2. math.floor part makes it so it divides by the integral and leaves the remainder out. I recommend just doing n//len(s) though. It's faster. l*c counts all the a's for the integral side, but now you have to take into account the remainder side. That is what res += .... does. the n%len(s) gives the remainder side. Puting it in a position of s[:x] makes it count only the list up to the remainder number. Then s[].count('a') counts the a's in that remainer. The res += just adds the a's from the remainder side to the integral side that you did in the line above.
jayanthsaikiran I tried to explain the code. But people are going against me as I reveiled the solution. Some of them are saying that the code needs improvement.😂😂
Anyways thanks for the explanation.
ejayt I think it's important to note the difference between '/' and '//' in python 3. '/' will return a decimal number, whereas '//' uses truncate division (floor division) returning a whole number.
cbr26504 whats the difference between '//' and math.floor ? I understand it as the same but my code fails for 12/23 cases with '//' and works fine with math.floor which i got from @jayanthsaikiran above
miaohuang1976 I got the same problem, just make sure you add parenthesis for n//lens, this is the version that works for me:
def repeatedString(s, n):
return s.count('a') *( n//(len(s))) + s[:(n%(len(s)))].count('a')Meanwhile, this is the version that fails for 12/23 cases:
def repeatedString(s, n):
return s.count('a') * n//(len(s)) + s[:(n%(len(s)))].count('a')pranavjayachand Do any of you understand the logic behind that?
eurbs The logic for adding parentheses is based on order of operations.
Division and multiplication are executed in left to right order unless parentheses are used to modify the order.
Say we have
s = 'aaa'andn = 10Case 1 (correct, with parens):
s.count('a') *( n//(len(s)) )-->3 * (10 // 3)-->3 * 3-->9Case 2 (without parens):
s.count('a') *( n//(len(s)) )-->3 * 10 // 3-->30 // 3-->10
vimal_mishra_cs1 it worked
b_s_apsara_july // will return the floor(quotient) while performing the division operation but / performs the normal division operation
eg: 10//3 =3 whereas 10/3=3.3333333333333335
I hope this helps.....
akashrai110011 I used the same logic but a simpler code. It's the same thing so you can understand. at first s.count("s") will give the number of a's in one substring. Multiplying that with (n//len(s)), which is the number of complete substrings present in that repeated string, gives the number of a's till that part. Now what's left is a smaller (incomplete) part of the substring which is s[0:n%len(s)] ( length of the remaining = the remainder). We just find the number of a's in that part and add it to the previous value of count
rayedbw s.count("a")
rajatmishra3009 but how do i code this in java
astroid07 Here's the code in Java:
int repeatedString(String s, int n) { int q = n/s.length(); int r = n % s.length(); if(!s.contains("a")) return 0; return s.length()>n? C(s, r) : q*C(s, s.length()) + C(s, r); } int C(String s, int e) { for (int i = 0, int a=0; i < e; i++) { if (s.charAt(i) == 'a') a++; } return a;}
rajatmishra3009 thanks a lott ,but its given that n can be long a;so
gowthamann27 try this
static long repeatedString(String s, long n) { char[] chrIn = s.toCharArray(); int i = 0; long max = 0; long quo = n / chrIn.length; long rem = n % chrIn.length; while (i < chrIn.length) { if (chrIn[i] == 'a') { max++; } i++; } i = 0; max = max * quo; while (i < rem) { if (chrIn[i] == 'a') { max++; } i++;
} return max; }
rahulrajgupta031 this code is not working
gowthamann27 try this
static long repeatedString(String s, long n) { char[] chrIn = s.toCharArray(); int i = 0; long max = 0; long quo = n / chrIn.length; long rem = n % chrIn.length; while (i < chrIn.length) { if (chrIn[i] == 'a') { max++; } i++; } i = 0; max = max * quo; while (i < rem) { if (chrIn[i] == 'a') { max++; } i++;
} return max; }lemanisar23 great solving!!!!!
gowthamann27 Passed all test cases. Can u justify why it is not working?
gowthamann27 try this
static long repeatedString(String s, long n) { char[] chrIn = s.toCharArray(); int i = 0; long max = 0; long quo = n / chrIn.length; long rem = n % chrIn.length; while (i < chrIn.length) { if (chrIn[i] == 'a') { max++; } i++; } i = 0; max = max * quo; while (i < rem) { if (chrIn[i] == 'a') { max++; } i++;
} return max; }
vinod_20_pal this code exceed the time limit
ajay20 thanks rajat for the solution/
xlm13x static long repeatedString(String s, long n) { Matcher matcher = Pattern.compile(String.valueOf('a')).matcher(s); int aInInput = 0; while (matcher.find()) { aInInput++; } long totalA = 0L; if (s.equals("a")) { return n; } else { totalA = totalA + ((n / s.length()) * aInInput); for (int i = 0; i < (n % s.length()); i++) { if (s.charAt(i) == 'a') { totalA++; } } } return totalA; }xlm13x [deleted]rajatmishra3009 what is matcher.find();
xlm13x You can learn it about here: https://docs.oracle.com/javase/7/docs/api/java/util/regex/Matcher.html
preetkewl import java.io.; import java.math.; import java.security.; import java.text.; import java.util.; import java.util.concurrent.; import java.util.regex.*;
public class Solution {
// Complete the repeatedString function below. static long repeatedString(String s, long n) { long check = 0; int i=0,j=0,k=0; long count = 0; for(i=0; i<s.length(); i++){ if(s.charAt(i) == 'a') count++; } long div = n/s.length(); long reminder = n%s.length(); count = div*count; for(i=0; i<reminder; i++){ if(s.charAt(i) == 'a') count++; } return count; } private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) throws IOException { String s = scanner.nextLine(); long n = scanner.nextLong(); long result = repeatedString(s, n); System.out.print(result); scanner.close(); }}
peter_lokey1 I like this solution, just don't know why you declared variables that are never used... check, j, k....
mishrajianimesh Deserving and understandable 10/10
danishkhanpatha1 pretty good, i liked your approach of solving this problem.
senthil_maran nice
vinuvas321 superb solution
JackDauda Fantastic solution. Makes so much sense and easy to understand. 10/10 mate.
umer_solehri Awsome
burnoutd457 Excellent Solution, easy to understand and works great
aayush_sinha If the String is too long to be iterated with an int variable. How will you count 'a' if the Sting length is larger than max int
chainani_navin97 why you write if s.charAt(i) =='a' two timewe can write one time na instead of second time we can directly write count = div*count +remainder;
mr_dev_b You dont need to run two loops.
static long repeatedString(String s, long n) { Character a = new Character('a'); char ch[] = s.toCharArray(); int len = ch.length; long aInSub=0; long num = n/len; long rem = n%len; long aExtra=0; for( int i =0; i0){ aExtra++; } aInSub++; } rem--; }
return aInSub*num+aExtra;babjoi i spent 5-6hrs to write the below code :( . I have a looong way to go
long total = 0; long l = s.length(); List<Integer> aList = new ArrayList<>(); List<Long> lastList = new ArrayList<>(); long totalFullSplits = n/l; long repeatCounter = totalFullSplits * l; for(int i = 0; i < l; i++) if(s.charAt(i) == 'a') { aList.add(i); long lastIdx = 0; lastIdx = ((i+1) + (l * (totalFullSplits - 1))) - 1; if(repeatCounter + (i+1) <= n) lastIdx = (repeatCounter + (i+1)); lastList.add(lastIdx); } for(int i = 0; i < aList.size(); i++) { total += ((lastList.get(i) - aList.get(i)) / l) + 1; } return total;
brajeshdhakad50 [deleted]
nitinshaily0 hey same approach.. just cleaner code
def repeatedString(s, n):
t=0 t=s.count('a') t*=n//len(s) t+=s[0:n%len(s)].count('a') return thritikrastogi56 Thanks a lot bro
shahirdhananjay What if your code (n // len(s)) returns 0 ?
I think if the quotient is 0 then you should not multiply it with count.
adi_bhutani16 Beautiful!!
suburb4nfilth [deleted]askarovdaulet I would recommend s.count("a",0,n%len(s)) over slicing s. Slicing is not very efficient for very long strings. For 100-character strings will barely matter of course.
snaran Do you think the Python interpreter would do the optimization for you?
eugalt And why count the same thing twice? :)
s = input() n = int(input()) d, r = divmod(n, len(s)) print(s.count('a', 0, r) * (d + 1) + s.count('a', r) * d)
s_ydinesh1177 Can you please explain it? : (
Richard_Khonan what does(n//len(s)) and s[:n%len(s)] mean?
alphasingh I'll explain here with an example, suppose the input is:
aba 10
n = 10, means the string s = "aba" is repeated over the course of 10 letters, i.e. abaabaabaa. Now, we break this string further down. First part, the exactly divisible part, s1 = "abaabaaba" and the second part, the remainder part, s2 = "a". To represent the count of a in s1 and s2, we have, int(n/len(s)) or n//len(s) and s[:n%len(s)], multiplied by count of a in s, respectively.
baymaxlim2204 Thanks. Easy to understand explanation for no matter what language.
[deleted] This is the most clean and articulate explaination/answer to ANY software problem I've never seen online. Thank you.
prash8669 [deleted]
RajShukla why you have used colon ":" can you please explain?
tanagon601 string slicing s[ : : ]
emtodawa The ":" allows Python select (Slice) all the elements before the nth element.
s = (abacde)
s[:3] = (aba)
So if n is 3, this allows Python to select all elements up to, but not including the 3rd item.
brandaohugo More readable in Python3:
def repeatedString(s, n): a_count = s.count('a') reps = n//len(s) rest = n % len(s) count = a_count * reps + s[0:rest].count('a') return count
eugalt If you are calling it more readable you should have managed to format it properly at the least.
wongj19 [deleted]wongj19 Thank you for the code, and I format it:
def repeatedString(s, n): a_cnt = s.count('a') num = n//len(s) mod = n%len(s) count = a_cnt*num + s[:mod].count('a') return count
DAMwingtsun Yours is the only one that passed all the test cases. Any reason why didn't make it past the 15th test?
if len(s) == 1: if s == 'a': return n else: return 0 elif len(s) in range(1, 101): if n < (10 ** 12) + 1: return ((s * (int(n // len(s)) + 1))[:n]).count('a') else: return 0g_vidhyaa11 Can you please explain me the logic behind this code. I am lil confused here
gowthamann27 // Complete the repeatedString function below. //Consider input is aba 5: -->abaab-->expected op-->3 // first 3 characters is aba. According to 5, we can add only 2 char. So if we can count no.of a in initial array then can multiply with quo and add rem.-->2(length)*1(quo)+2(rem) there are 2 a in aba and rem 1(ab) . In rem only 1 a. totally 2+1=3 // for input aba 10: abaabaabaa, totally 10. 3*3 +1 rem . First aba there are 2 a. then 2*3(quo)=6 then only one a in rem(a) totally 7 static long repeatedString(String s, long n) { char[] chrIn = s.toCharArray(); int i = 0; long max = 0; long quo = n / chrIn.length; // Needed to multiple with count of a long rem = n % chrIn.length;// Needed to add with count of a while (i < chrIn.length) {//will iterate it till inial array then we can multiply it with quo if (chrIn[i] == 'a') { max++; } i++; }//This while loop returns 2 as aba has 2 'a' i = 0; max = max * quo;//2*1=2 while (i < rem) {// here the reminder is 2 so it will iterate it with 0,1 if (chrIn[i] == 'a') { max++; } i++; } return max; }
timmusudheer007 can u please explain this step clearly s[:mod].count.... why u r using that
brennycode Let's take the example of s="aba" and n=10.
The length of "aba" can fit into 10 three times, and "a" occurs twice in the string so 2x3 = 6. However, this only accounts for the first 9 numbers, and we need 10, which means we need to get the remainder. The modulo operation allows us to save that remainder in mod. So s[:mod].count('a') in this case would be s[:1].count('a') which calculates to 1, since there is only one 'a' in a range from s[0] to s[1]
sh_komal1994 Thank you :)
rakeshreddy5566 great by the thought of using remainder
UriyaBA This is brilliant! Gotta love Python
wesleyrichmond this is impossible to understand
umrbek_yakubov what about a case when n < len(s) ?
profnandaa Same concept, in JS:
function repeatedString(s, n) { const counta = (str) => str.split('') .filter(c => c == 'a') .length const d = Math.floor(n / s.length) const r = n - (s.length * d) return d * counta(s) + counta(s.substr(0, r)) }
JShilale This is a great one, thank you.
Forces me to think about the input string as a single unit as opposed to just concatenating the string onto itself until it reached the length of n and checking for the amount of a's in the resulting string like i originally did.
ALI_ASAD Really good solution
jantomm perhaps you don't need to use filter if you split str by 'a'
const repeatedString = (s, n) => { const countA = s => s.split('a').length - 1; let len = s.length; let reps = Math.floor(n / len); let remainder = s.slice(0, n % len); return reps * countA(s) + countA(remainder); };
rAfitiiixxx + 0.5 for better code and + 0.5 for better var naming. Good lord ppl here put awfull names to things. i don't know javascript at all, but seems nice to work with lambdas in the same way i do in C#. I should start to check for how ppl do the algos in js after i'm done with my part in C#.
gparsons Does this work with strings that repeat a?
'abaaaa'
k_klapczynski Yes, it passes all tests when submitted. Tested as well your string and it's all good. It's a nice and clever solution.
stefannhs [deleted]jumavipe I tried this one but i got runtime error in 3 test cases
D1360 nice solution, very similar to mine!
countAslooks a bit uglier, but I tried to avoid transforming to array.Also, I think it is almost the same, but you could use
%for calculating therestfunction repeatedString(s, n) { countAs = (s, index) => { let count = 0; while (index >= 0) { if (s[index] == 'a') count++; index--; } return count; }; if (n <= s.length) { return countAs(s, n - 1); } else { const parts = Math.floor(n / s.length); const rest = n % s.length; return countAs(s, s.length - 1) * parts + countAs(s, rest - 1); } }
kmario1019 I did the same but with regex:
function repeatedString(s, n) { const count = (s) => s.replace(/[^a]/g, '').length; return count(s) * (parseInt(n / s.length)) + count(s.substr(0, n % s.length)); }
henriquejensen I had the same idea but whithout otimization
function filterAs(arr) { let result = 0; for (let letter of arr) { if (letter === 'a') { result++ } } return result } function repeatedString(s, n) { const lengthString = s.length const numberDivision = Math.floor(n / lengthString) const numberOfAs = filterAs(s) const remainder = n - (numberDivision * lengthString) if (remainder === 0) { return numberDivision * numberOfAs } let moreAs = filterAs(s.substring(0, remainder)) return numberDivision * numberOfAs + moreAs }
manojnataraja Awesome code!!!
chainani_navin97 i not understand what output we get if u substring(0,r) i am confusing in substring plz explain with example
ahmedrabi100 i am using the same idea but the problem i get some cases Wrong here is my Code with C#
static long repeatedString(string s, long n) {
double r = 0; foreach(char c in s) if (c== 'a') r++; double count = Math.Floor(n * r/s.Length); for(int i = 0; i< n%s.Length;i++) if(s[i]== 'a') count++; return Convert.ToInt64(count); }hernandes_silva Based on your code i reached the solution below:
double r = 0; foreach (char c in s) { if (c == 'a') { r++; } } r = ( r * (n/ s.Length)); for (int i = 0; i < n % s.Length; i++) { if (s[i] == 'a') { r++; } } return Convert.ToInt64(r);PS.: Not tested against "n < len(s)".
hernandes_silva Wow, very nice!!!
Now i tested against "n
priyankalokhand2 Thank You
jaishreekulkarni here is my code in C#: I am also getting 10/23 cases wrong.
int len = s.Length; long cnt = 0; string remStr; long repeats = 1, finalCnt,remCnt; if (len > 1) { int rem = (int)n % len; repeats = (n-rem)/len; cnt = s.Split('a').Length - 1; finalCnt = cnt * repeats; remStr = s.Substring(0, rem); remCnt = remStr.Split('a').Length - 1; finalCnt = finalCnt + remCnt; return finalCnt; } else { if ((s == "a")&&(len==1)) { return n; } else return 0; }
akash0744 insted of int take long everywhere.
jaishreekulkarni thanks that solved my problem
nithin9963853245 whats the difference
akshyvrma8 IF you take int then for 10^12 length string It will give negative values ,
prem_cumar plz some one explain this code clearly.i am a beginner
alexa346 Simple yet elegant way of handling it!
chauhanshivam_ Excellent! Thanks;-)
mikelkengni beautiful!!!!cudos
VVIT17BQ1A0513 Please explain briefly about :: s[: n % len(s)].count("a")
kiran2006kundur1 Super !!
juvenn And the same in Ruby:
s.count('a') * n.div(s.size) + s.slice(0,n.remainder(s.size)).count('a')
kingsdeb I used the same logic, but my code sucks, this is beautiful python idiomatic code. <3
rpodlinov l = len(s) print(s.count("a") * (n // l + s.count("a", 0, n%l))
is better in terms of performance due to it do not use slice.
sak7258 the man? el macho?
DAMwingtsun NYJ1qCr, nice! I looked at divmod too but was slower than yours. My is a bit modified for the last part of string but still, I am passing 15 of the 23 cases. Any reason why??? def repeatedString(s, n): if len(s) == 1: if s == 'a': return n if len(s) < 100 and n <= 10 ** 12: return ((s * (int(n // len(s)) + 1))[:n]).count('a') #a, b = divmod(n, len(s))
#return (s * a + s[:b]).count('a') else: return 0avinashgpai98 what does n//len(s) do?
leonardo_daher Very Pythonic solution! Nice!
redmice [deleted]redmice [deleted]
jspades93 [deleted]m_bansal10 Python has count ? wow, poor java, had to loop on char array while comparing them
sabeeh_yunus [deleted]sabeeh_yunus Think I did the same thing in a more explicit way:
def repeatedString(s, n): length = len(s) complete = n // length # total complete strings left = n % length occur = 0 # in single sequence for idx, ch in enumerate(s): if 'a' == ch: occur += 1 occur = occur * complete # in remainder if left > 0: for idx, ch in enumerate(s): if 'a' == ch and idx < left: occur += 1 return occur
michellerodri247 The page is mentioned here for discussing CBD for Nausea Relief the repeated string topic. The repetition of the string is useful in some time, and in some time it is good for our work. So based on the need, we need to choose the option of the string.
viratc Can someone explain me why this fails for round(n/len(s)), but it passes all test cases for n/len(s) ??
yaquake I've solved the task the same way :D perfect pythonic way
b_s_apsara_july Cool Logic :)
anagarzasant My code does exactly the same but here's more readable:
def repeatedString(s, n): length = len(s) div = math.floor(n/length) extra = n%length norm = s.count("a") total = norm*div splice = s[:extra] total += splice.count("a") return total
atharvaphansalk1 great
yago580 Great solution! Solved it exactly the same way in Ruby, but came to the solution with the code organized a little differently.
skd000 Very Nice code broo
c_sudharsan1988 Wonderful.
mthaiseer [deleted]
ujjalbaniya crazy af
singhshagun891 Did the same
def repeatedString(s, n): c=(s.count('a'))*(n//len(s))+s.count('a',0,n%len(s)) return c
coderankitkumar def repeatedString(s, n): l = len(s) if l == 1: return n t = n//l a_cnt = s.count('a') * t + s[:n%l].count('a') return a_cnt
why I am getting two cases not passed?singhshagun891 This is because of the 3rd line that if(l == 1): return n. Let our string s="b" and n=10 then according to your logic its output will also be 10. Try this one(just edited third line in your code):
def repeatedString(s, n): l = len(s) if s == "a": return n a_cnt = s.count('a') t = n//l a_cnt *= t a_cnt += s[:n%l].count('a') return a_cnt
coderankitkumar Thank you! I did it.
vonstuckrad This is funny looking code, but it seems to be such bad practice because its really not readable.
kunalrashinkar99 genius maaaaahN
ritavdas WoW
Sanskritimisra71 excellent bro thnks
shahirdhananjay What if your code (n // len(s)) returns 0 ? if the quotient is 0 then you should not multiply it with count.
suryatejapatkari converting math into python really impressive
suryatejapatkari explanation: example string =aba-- s.count('a') return the number of a's in the original string i.e 2-- since n=10 and len(string)=3 n//3=3(quotient) means the string repeats 3 times completly so no of a's in string is 2 multipy the quotient 3 which is 6-- s[:n % len(s)] where n%len(s) gives reminder which is 1 in this case so s[:1] is ['a'] and it's count is 1 so 6+1=7 ans =7
script_kiddie awesome
Kartik_b190697cs elaborated version: def repeatedString(s, n): k = s.count('a') p = n%len(s) ss=s[:p] return k * int(n/len(s)) + ss.count('a')
karibasava_t_g [deleted]
Raulkg public static void main(String[] args) { Scanner in = new Scanner(System.in); String s = in.next(); long n = in.nextLong(); long count =0; for(char c : s.toCharArray()) if(c == 'a') count++; long factor = (n/s.length()); long rem = (n%s.length()); count = factor*count ; for(int i=0;i<rem;i++) if(s.charAt(i)=='a') count++; System.out.println(count); }parnika182 Nice work
cys920622 for(char c : s.toCharArray()) if(c == 'a') count++;
Instead of allocating space for a charArray you could just do s.charAt(i) like you did below.
rakeshsenapathi [deleted]
vvbhandare Your answer is great but it will more help me if you could explain about deriving tis formula.
AMITARYA int main(){ string s; cin >> s; long n; cin >> n; long count=0; int i,k=s.size(); long p=n/k; for(i=0;i<s.size();i++){ if(s[i]=='a') count++; } count = count * p; for(i=0;i<(n%s.size());i++){ if(s[i]=='a') count++; } cout << count << endl; return 0; }
More simplified cpp solution,i hope this will be heplful
pruvi007 check your logic for ababa 3
it will give output as 3 bt o/p should be 2.
Raulkg are you talking about my solution?
GIWMCoder @Raulkg, nice approach without the complication of running out of heap memory with a StringBuilder.
Please check your solution for the test case of
ababawith 3 as the length. As @pruvi007 pointed out, the output is incorrect.
Here's my modified code:
public static void main(String[] args) { Scanner in = new Scanner(System.in); String s = in.next(); long n = in.nextLong(); long count = 0; if(n > s.length()) { long repeat = n/s.length(); long diff = n % repeat; for(char c: s.toCharArray()) { if(c == 'a') count++; } count = count * repeat; for(int i=0;i<diff;i++) { if(s.charAt(i) == 'a') count++; } } else { int len = (int) n; s = s.substring(0,len+1); for(char c: s.toCharArray()) { if(c == 'a') count++; } } System.out.println(count); }
Oxobo [deleted]
abderrahmane_be1 static long repeatedString(String s, long n) { int i = 0; long lCountA = 0, rCountA = 0; long sTimes = n / s.length(); long remainChars = n % s.length(); if(remainChars > 0){ for(; i < remainChars; i++){ if(s.charAt(i) == 'a') lCountA++; } } if(sTimes > 0){ for(; i < s.length(); i++){ if(s.charAt(i) == 'a') rCountA++; } } return sTimes * (rCountA + lCountA) + lCountA; }Monish_Narendra Nice... Could you please Explain How your code works...
ricardimgyn static long repeatedString(String s, long n) { int strLength = s.length(); long aOccurrence = 0; long factor = n / strLength; long remainder = n % strLength; for (int i=0; i < strLength; i++) { if (s.charAt(i) == 'a') { aOccurrence += (i < remainder) ? factor + 1 : factor; } } return aOccurrence; }
rckskelton Very clean and smart.
abishekdevaraj Your solution is smart and nice... Could you please explain how your code works?
pbilla1 Without understanding you commented it is smart and nice..WOW
amitkumar80a without this one test case fails
int len = (int) n; s = s.substring(0,len+1);
and including this all test case passed ..what's the difference
mandalsambeet It's giving me the correct output. My logic is similar to @Raulkg.
rvrishav7 try this out happy coding #include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { string s; long long int i=0,cpl=0,n; cin>>s>>n; long long int r,q; while(s[i]!='\0') { if(s[i]=='a') { cpl++; } i++; } long long int l=s.length(); q=n/l; cpl=cpl*q; r=n%l; for(i=0;i<r;i++) { if(s[i]=='a') { cpl++; } } cout<<cpl; return 0; }
rvrishav7 [deleted]shivaisno1 now it works fa ababa 3
if(n
shobhitvj1311 how are you able to make such logics??
aadarsh9600 Please Explain Your Logic. @AMITARYA
maheshvangala191 Tried it by not considering modulous loop but I didn't get it thank you
maheshvangala191 Tried it by not considering modulous loop but I didn't get it thank you
selva1996 ur code is fine , but it is good to store s.size() value to a variable and then use that variable for calculations , because s.size() everytime runs 'n' times the length of the string when for loop checks the condition.
indsonusharma i want to know where or how did u think ths logic
CSociety You won't need to run for loop twice in else case using this code. Time Complexity will decrease
int i = 0, count = 0, count1 = 0; if(n%s.size()==0){ for(int i = 0; i<s.size(); i++){ if(s.at(i)=='a'){ count++; } } return n/s.size()*count; } else{ for(int i = 0; i<s.size(); i++){ if(s.at(i)=='a'){ count++; if(i<n%s.size()){ count1++; } } } return (n/s.size()*count)+count1; } }
indsonusharma thank
virginielgb I don't think you need the first "if", do you ? This is my (accepted) JS solution :
function repeatedString(s, n) { var inString = 0; var added = 0; for(var i = 0 ; i < s.length ; i++) { if(s[i] == 'a') { inString++; if(i < (n%s.length)) added++; } } var repeats = (n- n%s.length)/s.length; return repeats*inString+added; }
elvis_dukaj A better way to write...
#include <iostream> #include <string> #include <algorithm> using namespace std; int main() { string s; cin >> s; long n; cin >> n; auto repetition = n / s.size(); auto partial = n % s.size(); auto aCountInString = count(begin(s), end(s), 'a'); auto aCountInPartial = count(begin(s), begin(s) + partial, 'a'); cout << aCountInString * repetition + aCountInPartial; return 0; }
nshukl23 [deleted]
shubham24976 you should use long long !!!!!!#amitarya
khannaa153 I had the same code, but made a small mistake and your code really helped me rectified it.
raptor_inc2018 [deleted]
lifekills Nice, but it can be done with a single loop:
public static void main(String[] args) { Scanner in = new Scanner(System.in); String s = in.next(); long n = in.nextLong(); if (!s.contains("a")) { System.out.println(0); return; } long count = 0; long mod = s.length() > 0 ? n % s.length() : 0; long remainder = 0; boolean strLonger = false; for (int i = 0; i < s.length(); ++i) { if (i >= n) { strLonger = true; break; } if (s.charAt(i) == 'a') { if (mod > i) remainder++; count++; } } if (!strLonger) count = count * Math.max(n / s.length(), 1) + remainder; System.out.println(count); }
Dheeraj571 Great work .Easy for rookies to understand.
s0033 I think we could avoid one loop
static long repeatedString(String s, long n) { String sh = s.replaceAll ("[^a]", ""); long ost = n%s.length(); long qty = ((n - ost)/ s.length())*sh.length(); for (int i = 0; i < ost; i++) { if (s.charAt(i) == 'a') { qty++; } } return qty; }
chanakauomfit Great..!
hegde_vishal1 It was not mentioned in the question but your solution doesn't work if n is less than s.length()
CSociety Thanks for the suggestion
Highlander_Monk nice solution :)
ankitshah_tech Excellent bro
Abdullah104 Very smart and clear solution, Very nice to have got the chance to see your solutionn. Thank you
jaletechs Beautiful!
sonikakumari_s1 wow
Kprashanth047 thanks a lot its very clear
umesh_technotes What happens if n < s.length?
alexfilth186 Leave loops for JDK:
static long repeatedString(String s, long n) { long canPlace = n / s.length(); long missingPart = n % s.length(); long qtyPerString = s.chars().filter(ch -> ch == 'a').count(); long missingChars = (missingPart > 0) ? s.substring(0, (int)missingPart).chars().filter(ch -> ch == 'a').count() : 0; return (canPlace * qtyPerString) + missingChars; }
andrepd I did it similar:
static long repeatedStringB(String s, long n) { return s.chars().filter(ch -> ch == 'a').count() * (n / s.length()) + s.chars().limit(n % s.length()).filter(ch -> ch == 'a').count(); }
lukezawada 2 for loops is kind of unneeded here. Here's a refactor (in C#): int As = 0; int modAs = 0;
for(int i = 0; i < s.Length; i++) { if(s[i] == 'a') { As++; if(i < (n % s.Length)) modAs++; } } return As * (n/s.Length) + modAs;
awesomeanimesh Can u please explain what is the use of factor and rem
EDIT-Got it.Great logic.awesome
federicocapucci pretty genious! i spent a day and couldn't figure out some of the cases without iterating throughout a string the size of 'n' .
Thank you!
ezeq_1 Nice solution.
jhpar25 Java Solution
static long count(String s, long n) { long numOfS = n/s.length(); long rest = n % s.length(); if(!s.contains("a")) return 0; return s.length()>n? aCounter(s, rest) : numOfS*aCounter(s, s.length()) + aCounter(s, rest); } private static long aCounter(String s, long end) { int a=0; for (int i = 0; i < end; i++) { if (s.charAt(i) == 'a') a++; } return a; }
aneri_masi Elegant!
pablohcl93 Terrific!!
3r1ck_442 Hi, I dont get why you are getting numOfS and rest, coud you please explain?
jhpar25 numOfS calculates how many times string s has appeared as a complete string, up to nth character in the infinite string . rest represents the number of characters left after finding out number of s.
For example, let's say we have a string s "abcd" and n is equal to 10. The infinite string will be something like "abcdabcdabcd...". Then, we can break this infinite string into substrings like this up to 10th character, abcd/abcd/ab. In this case, we have 2 complete sets of s and a partial s which contains only 2 characters.
numOfS gives us 2.5, but long is a type of integer so the result becomes 2 and 0.5 is gone. The value 2.5 means we have 2 complete sets of s and 0.5, half of s. Since ab is not a complete s, we need to figure out how to pick up that 0.5 we dropped above.
Now, rest counts what's left, remainders. 10 % 4 is 2. Finally, these results will be passed onto aCounter, which literally counts the number of As in the string, depending on the algorithm.
The rest is pretty easy to understand :)
Sharangpure Excellent solution !
JosiahKing Learnt a new thing here apart from logic. Ternary operator usage. Thank you!!
yamany good job bro
raheezrahman Excellent Work!
davebuchholz13 Adding that I learned a lot of tricks from this one solution. Bravo!
davebuchholz13 String manipulation can be deceptive! I was going through with the mindset during my planning phase of splitting the string with a substring() at long index but quickly realized that is not allowed.
Tried a few different tricks and came upon this solution. I was 1/2 of the way there. Thank you for this.
[deleted] Simple and elegant !!! Awesome job !!!
linmeijiang I don't really understand the syntax here: return s.length()>n? aCounter(s, rest) : numOfS*aCounter(s, s.length()) + aCounter(s, rest)
Cound you explain? thank you.
dennisstancu static long repeatedString(String s, long n) { long count = s.length()-s.replace("a", "").length(); long reps = n/s.length(); long rest = n%s.length(); count = count*reps; for(int i = 0; i<rest; i++){ if(s.charAt(i) == 'a') count++; } return count; }
juaxix Simple code for c++14
#include <iostream> #include <algorithm> using namespace std; int main() { string s; unsigned long int n; cin >> s; cin >> n; cout << ( //string length * a's count + rest of the string a's cut ((n/s.size()) * count(s.begin(),s.end(),'a')) + count(s.begin(),s.begin() + n%s.size(),'a') ); return 0; }
jose_hernro So Great!!! How did you get the equation?
inwerp [deleted]
paulius_gasp Can you explain it?
ashoka_tanvi Good, but not all test cases pass.
gabrielgiordano Efficient JavaScript solution
The algorithm doesn't count the string a second time, both best and worst case is O(n) where n = length of the string
From my repo https://github.com/gabrielgiordan/HackerRank
function repeatedString(s, n) { let c = 0, ca = 0, r = n % s.length for (let i = s.length; i-- > 0;) { if (s.charAt(i) == 'a') { ++c if (i < r) ++ca } } return ((n - r) / s.length * c) + ca }
hmndz Could you explain the variable ca?
gabrielgiordano Of course, 'ca' is the remaining 'a' characters, while 'c' is the total amount of characters in the given string without the remaining. 'r' is the remaining that doesn't fit into the 'n', so ((n - r) / s.length * c) is the amount of characters whitout the remainings of 'a' characters. The remaining must be counted only if is not greater than the index (i < r). So the total amount without the remaining + the remaining 'ca' is the result.
247galindo Nicely done!
warliang The return value got me confused, but your explanation was very helpful! Thanks!
function repeatedString(s, n) {
let count = (s.match(/a/g) || []).length * Math.floor(n/s.length) let remainder = n % s.length let remainderCount = (s.substring(0, remainder).match(/a/g) || []).length return count + remainderCount}
This was what I used!
fmuiin14 This is work for javascript ...
but can you explain aboout this code?
evgeniy_aksyonov Hi, thank you for sharing this.
But can you please explain how we can get answer 7 for given s = 'aba' and n = 10?
return statement will hold: ((10 - 1) / 3 * 2) + 1 which is: (9 / 6) + 1 = 2.5
Can't get my head around it =\
blacr7 ((10 - 1) / 3 * 2) + 1is(9)/3 * 2 + 1not(9/6) +1evgeniy_aksyonov omg, I forgot about operations order =\ so obvious...
thanks for reaching out, blacr7
blacr7 no problem its a easy mistake to make
DesertFaux Curious about the for loop here:
for (let i = s.length; i-- > 0;)
The use of postfix incrementing on the i variable caught my eye. Is this structure considered good practice? Usually we see:
for (let i = s.length; i > 0; i--)
Which does the same thing but is a bit easier to read as it follows the standard 3 statement structure.
geekrittr first one is not good from the readability perspective
the_goodone This two are not the same...it would not even passed the code for let i = s.length; i > 0; i--
unni_navi for (let i = s.length; i >= 0;i--)
sindhuanurag2 it will be good, if you can put useful names instead of c, ca, r.... Because that will help me to relate more with your algorithm.
jon_dow Here is my solution with comments. There are additional variables declared for ease of understanding.
// number of times string can be repeated within limit n const numOfRepeats = Math.floor( n / s.length ); // additional number of strings to get to the limit n const remainderString = n % s.length; // find number of matches in a string let matches = ( s.match(/a/g) || [] ).length; // multiply number of matches in a string with number of repeatations matches = matches * numOfRepeats; // find number of matches in remainder string const remainderMatches = (s.substring(0,remainderString).match(/a/g) || [] ).length; // add it up return matches + remainderMatches;
ericchi_311 Thanks for your solution and great comments!
meetbrij Thanks for your solution and the comments. Makes it easier to understand.
sooz1 Great comments. Helps so much. Thanks!
anirbandr This is realy helpful. The more I go through the suggested solutions, the more I realize how much better it is to provide such solutions with easily understable comments. It helps create a good mental model. Appreciate your effort.
meetbrij Thank you so much for this solution. I did learn a lot.
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