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Mars Exploration

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  • cys920622 + 26 comments

    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 + 2 comments

      Thank you

      • Phoenix98 + 3 comments

        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 + 1 comment

          No need to introduce counto as distinct from counts

        • nidhi_thakkar66 + 0 comments
          [deleted]

        • souravsaini438 + 0 comments

          include

          using namespace std; int marsExploration(string s){ int i=0,count=0; string str; for(i=0;i 

          int result=marsExploration(s);
cout<<result;
return 0;

          }

      • Naveen_Malla + 5 comments

        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 + 0 comments
          [deleted]

        • phoemur + 1 comment

          In this case its OK, but I just don't like seeing malloc and no free...

        • keshavashourie3 + 0 comments

          good one

    • nayanansh + 0 comments
      [deleted]

    • Ankur_Beginning + 0 comments

      cool

  • voidpro + 2 comments
    #!/bin/python3
s=input().strip()
print(sum(1 for i in range(len(s)) if s[i]!="SOS"[i%3]))

    • mike_buttery + 2 comments

      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 + 1 comment

        can you explain this code

        • mike_buttery + 0 comments

          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 a True (value = 1) or False (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 validation

          Finally we return the sum of all of the comparison results

      • pshastov + 0 comments

        Python, stop! What are you doing to me

    • v110_ + 0 comments

      A faster version using itertools.cycle:

      from itertools import cycle
print(sum(1 for i, j in zip(input(), cycle("SOS")) if i != j))

  • anubrij + 0 comments
    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);
}

  • ajHash + 1 comment

    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 + 1 comment

      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 + 0 comments

        Other solutions dispense of a variable p and use modulus directly on i.

  • aminfara + 2 comments

    in Ruby

    puts gets.strip.each_char.with_index.map{ |c, i| (c == 'SOS'[i % 3]) ? 0 : 1 }.reduce(:+)

    • liahsheep + 0 comments

      some more levels of refactoring... :)

      puts gets.strip.each_char.with_index.count { |c, i| c != (i % 3 % 2 == 0 ? 'S' : 'O') }

    • phderome + 0 comments

      I like this and understand it right away without knowing Ruby. Good for you!

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