cys920622 Straightforward Java solution in constant space and O(n) time.
public static int countChanges(String message) { String sos = "SOS"; int count = 0; for (int i = 0; i < message.length(); i++) { if (message.charAt(i) != sos.charAt(i % 3)) count++; } return count; }
shravankumar1 Thank you
Phoenix98 include
include
include
include
include
include
include
int main(){
int i,len,counto,counts,total; char* S = (char *)malloc(10240 * sizeof(char)); scanf("%s",S); len=strlen(S); len=len/3; counto=0; counts=0; total=0; for(i=0;i<len;i++) { if(S[3*i]!='S') { counts++; } if(S[2*i+2+i]!='S') { counts++; } if(S[3*i+1]!='O') { counto++; } } total=counts+counto; printf("%d",total); return 0;}
Naveen_Malla SIMPLE C CODE
int main(){ char* s = (char *)malloc(10240 * sizeof(char)); int C=0; scanf("%s",s); for(int i=0;i<strlen(s);i=i+3) { if(s[i]!='S') C++; if(s[i+1]!='O') C++; if(s[i+2]!='S') C++; } printf("%d",C); return 0; }
tyagisagar79 [deleted]
keshavashourie3 good one
Prakash221095 its ok, but there is a testcase as input SOSSWEWSSTTS .how can it be handled.
nayanansh [deleted]Ankur_Beginning cool
javedmoh2001 Great soln...
satyakibose98 why the method is static?
jemmydng12 wts wrong in this code public static void main(String[] args) { Scanner in = new Scanner(System.in); String s = in.next(); int l= s.length(); int b=l/3; int count=0; int num; for(int i=0;i<b;i++) { num=(int)s.charAt(i); if(s.charAt(i) != s.charAt(i+b)) { count ++; } } System.out.println(count);
kashyap_rishab91 Why you are converting character to integer?
num=(int)s.charAt(i);
Since radiation can alter any character in string,so u need to scan all characters in string to check the change in message.
chrismin Amazing, thank you. I definitely have to make use of String methods a lot more doing these types of problems.
Srikanth_kalka if (message.charAt(i) != sos.charAt(i % 3)) count++;
Can u explain above condition please....
akhilgautam123 Checking changes between each "SOS". Lets take message = "SOSTRSSOO" and sos = "SOS" for i = 0; i< 9; i++, we will have: message.charAt(i) != sos.charAt(i % 3) is false upto i = 2. Now when i = 3, message[3] != sos[3 % 3 i.e. 0] is true. So count will increase. Again at i =4: message[4] != sos[4 % 3 i.e. 1] is true.
Briefly, the whole message is broken into set of 3 characters and compared to sos
Srikanth_kalka Thank you for your explanation Akhil. Appreciate it.
rishravi this approach is better(atleast better looking and compact) than using 3 if-statements. Thanks!
tyagisagar79 It seems that your code is wrong... For "ROR" the output should be 1.. but your code will give 2...
beckss [deleted]
abhinandkr A slight variant:
string S; cin >> S; string sos = "SOS"; int sum = 0; for (int i = 0; i < S.length(); i++) { if (S[i] - sos[i % 3]) { sum++; } } cout << sum << endl;
MVijayaanand Cute approach
adityabadve123 Did the same thing with c++
int marsExploration(string s,int len) { int error = 0; string sos = "SOS"; for(int i=0;i<len;i++){ if(s[i] != sos[i%3]){ error++; } } return(error); }
hari_4698 [deleted]sanjay228 good approach
ajHash This thing kinda clicked..
int main(){ string s; char te[3]; te[0]='S'; te[1]='O'; te[2]='S'; cin >> s;int i=0,j=0,count=0;; while(s[i]) { if(j==3)j=0; if((s[i])!= (te[j]))count++; i++;j++; } cout<<count; return 0; }
silja_tirronen I like this solution because it can be easily extended to handle an arbitrary pattern, of arbitrary length. In our case the pattern just happens to be "SOS". Here is my implementation in Java, which takes the pattern as a parameter.
public static void main(String[] args) { Scanner in = new Scanner(System.in); String s = in.next(); System.out.println(countCorruptSymbols(s, "SOS")); } private static int countCorruptSymbols(String input, String pattern) { int i = 0; int p = 0; int count = 0; while (i < input.length()) { if (input.charAt(i) != pattern.charAt(p)) { count++; } i++; p = (p + 1) % pattern.length(); } return count; }
shrishtee_ruchi java
public static void main(String[] args) { Scanner in = new Scanner(System.in); String S = in.next(); String sos = "SOS"; int count =0; for (int i = 0; i < S.length(); i++) { if (S.charAt(i) != sos.charAt(i % 3)) count++; } System.out.println(count); }
smadden03 why did you do sos.charAt(i % 3)
parviz0492 My complexity is here O(n^2). But eventually in fact are my steps equal to yours ?
public static void main(String[] args) { Scanner scanner = new Scanner(System.in); String message = scanner.nextLine().toUpperCase(); int counter = 0; if(message.length() > 0 && message.length() < 100 && message.length() % 3 == 0){ for(int i = 0; i < message.length() / 3 ; i++){ String str = message.substring(i * 3, (i + 1) * 3); int localCounter = 0; for(int j = 0; j < 3; j++){ if(!(str.charAt(j) == "SOS".charAt(j))) localCounter++; } counter += localCounter; } } System.out.println(counter); scanner.close(); }
muzic_maniac7 You don't need to convert the message into uppercase. It is already stated in the problem that the message will be in uppercase. And why are you checking for constraints! Your 'if' condition will always going to be true as stated in the problem. Also, your complexity isn't O(n^2), it is O(n) or O((n/3)*3). So, eventually you are doing the same thing in a different way. Cheers to different mind! :)
anubrij function main() { var S = readLine(); var n = 0; for(var i = 0; i < S.length ; i += 3){ if(S[i] != "S") n++; if(S[i + 1] != "O") n++; if( S[i + 2] != "S") n++; } console.log(n); }
zee1s Python
import sys altered=0 inp = input().strip() for i in range(len(inp)): x=i%3 if(x==0 and inp[i]!='S'): altered+=1 elif(x==1 and inp[i]!='O'): altered+=1 elif(x==2 and inp[i]!='S'): altered+=1 print(altered)
dipak_majhi import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); String s = in.next(); int count =0; for (int i =0;i<s.length();i+=3){ if(s.charAt(i)!='S') count++; if(s.charAt(i+1)!='O') count++; if(s.charAt(i+2)!='S') count++; } System.out.println(count); } }
mndeep_kmr_shrma for(char sos : "SOS".toCharArray())
could have written this
akhilgautam123 I like this solution more than the one-liners. Comparing whole SOS in one iteration.. Nice logic..
mail_prateek_ag1 Fellow coders, any feedback?
print(sum(1 for x, y in zip(S, 'SOS' * (len(S) // 3)) if x != y))
SebastianNielsen This was my approach:
s = input() print(sum(1 for a, b in zip(s, ['S','O','S'] * int(len(s) / 3)) if a != b))
It is very similiar to your solution except for the 'SOS' part, the string will get converted to a list as in my code anyway. (before getting zipped with 's') So for performance reasons, just write it as in my code from the beginning - even though the performance different in this piece of code is none existent.
But just letting you know, since you asked for feedback.
mail_prateek_ag1 Thanks Sebastian. Appreciate the pointer since I am newly initiated to Python.
prakash_kkpr Is the input 'SRR' is not a valid one ? If it is a valid input above solution will fail, right ?
mail_prateek_ag1 It should not. Did you run it and saw it failing? it shoul output 2 since RR don't match OS
belpoisj123 S does not exist
naaag I suspect there is some issue with the testcases. One of the test case which I downloaded to verify. SOSOSOSOSDSDSKWOSDOSDOASDOASDFAFJDFDOSOSOWNSOSOSNDSKDDOSOSOSJDSDSOOSOSDSDOSASSOASDSAOSOSODSDSOASDWS Answer say: 67
Here, length of string is 99 characters, how could possibly the answer be 67. There can be maximum 99/3=33 SOS signal in this string.
Let me know if I am wrong somewhere here. Thanks.
onehorselee Same problem and same doubt.
gnome_child It's the number of incorrect letters .
dineshkumarkb I have the same doubt too. Should we print the number of letters or the number of times the signal has been corrupted or whatever?
Pretty_Pearl You have placed a pic too? Really(?) admin, you are way too into this :D Thats good though.
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