Problem of the Day
A new programming or logic puzzle every Mon-Fri

Comments:

• Anonymous - 10 years, 5 months ago

  Quick and dirty haskell solution:

  revrev :: String -> Char -> String
  revrev [] _ = []
  revrev (x:xs) c 
      | x == c = (x): (revrev' xs [] c)
      | otherwise = (x):(revrev xs c)
  
  
  revrev' :: String -> String -> Char -> String
  revrev' [] ys _ = ys
  revrev' (x:xs) ys c
      | x == c = ys ++ [x] ++ (revrev xs c)
      | otherwise = revrev' xs (x:ys) c
  

  Btw i like that you use "mom" in an example for reversing strings :D

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• Max Burstein - 10 years, 5 months ago

  lol I didn't even realize it was a palindrome. I was talking to my mom at the time and wanted to add a third word. I updated the problem.

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• Nick Krichevsky - 10 years, 5 months ago

  C++ solution

  #include <iostream>
  #include <string>
  
  std::string revrev(std::string s, char c);
  int main(){
      std::string subString = "hello!";
      std::cout<<revrev("hello world taco",' ')<<std::endl;
      return 0;
  }
  
  std::string revrev(std::string s,char c){
      int pos = 0;
      int subCount = 0;
      s+=c; //Add the char to the end of our string, in order to catch the end.
      std::string newString = "";
      for (int i=0; i<s.length(); i++){
          if (s[i] == c){
              subCount++;
              std::string subString = s.substr(pos,i-pos);
              if (subCount%2==0){
                  //Switch the characters around
                  for (int j = subString.length()-1; j>subString.length()/2; j--){
                      char temp = subString[subString.length()-j];
                      subString[subString.length()-j] = subString[j];
                      subString[j]=temp;
                  }
              }
              pos = i+1;
              newString+=(subString+" ");
          }
      }
      return newString;
  }
  

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• bumbleguppy - 10 years, 5 months ago

  Should the last line in the outer loop be

  newString+=(subString+c); ?

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• Nick Krichevsky - 10 years, 5 months ago

  Whoops. Yeah you're right.

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• Eric - 10 years, 5 months ago

  JS

  function revrev(input, delim){ 
  
      var q = input.split(delim); 
  
      for(var i = 1; i < q.length; i+=2){ 
  
          q[i] = q[i].split("").reverse().join(""); 
  
      }
  
      return q.join(delim);
  
  }
  

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