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;}
phderome No need to introduce counto as distinct from counts
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.
adityapandey0159 [deleted]
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.
kohsheen_t num=(int)s.charAt(i);
print num and see what you get
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;
kushagrapathak91 nice approach
skchalotra can u explain S[i]-sos[i%3]
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
aiswarya110498 Thank you very much for your simplified code..
amitjoshi438_aj if i=1 or i=2, i%3=0 so how it will compare
gabrielbb0306 Mine was similar. I didn't use %. That was smart. I'm noticing the "%" operator is a good friend of competitive programmers
Xoxopie123 pls explain someone
shah_meit nice solution
hemanthchandu11 nyc solution
kris_tobiasson I had a similar approach with JS here:
let sos = `SOS`; let j = 0; let count = 0; for(let i = 0; i < s.length; i++){ if(sos.charAt(j) !== s.charAt(i)){ count += 1; } (j === 2) ? j = 0 : j++; } return count;
pandeysdr16 The catch is : sos.charAt(i % 3)
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; }
phderome Other solutions dispense of a variable p and use modulus directly on i.
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); }
aminfara in Ruby
puts gets.strip.each_char.with_index.map{ |c, i| (c == 'SOS'[i % 3]) ? 0 : 1 }.reduce(:+)
voidpro #!/bin/python3 s=input().strip() print(sum(1 for i in range(len(s)) if s[i]!="SOS"[i%3]))
mike_buttery Nice work. Can be made even shorter by suming the test (True = 1, False = 0)
def marsExploration(s): return sum(s[i] != "SOS"[i%3] for i in range(len(s)))
rankesh87 can you explain this code
mike_buttery Sure. Here is a breakdown of the comprehension:
for i in range(len(s))is an iterator that provides an index for each letter in the string.s[i] != "SOS"[i%3]is the validation test which results in aTrue(value = 1) orFalse(value = 0) - Not equal is used because we are counting mismatches. (i.e. True we have a mismatch)For each character in the string, using the index we need to test against a single character - so the trick here is to use modulo to get the correct character -
"SOS"[0] = "S", "SOS"[1] = "O", "SOS"[2] = "S"the result of index modulo 3 selects the correct comparator for our validationFinally we return the sum of all of the comparison results
Sort 504 Discussions, By:
Please Login in order to post a comment