//Written by Mo Kashi

//This program gets all words from an input file as a
//dictionarry and specifies how many of them are Palindromic
//and how many of them are in dictionarry if they are spelled backwards
//The program can also specify how many 1-letter,2-letter,3-letter and
//n-letter-word are in that dictionary.The dictionary is an input file
//"input.txt".The result is sent to an output file"output.txt" in C drive.

#include<iostream>
#include<vector>
#include<string>
#include<fstream>
 vector<string> text;//vector to store all words in dictionarry
 ifstream fin;
 ofstream fout;
 int Freez;
 int number_of_stills;
 int number_of_palindromics;
 string reverse(string);//This method gets a string and returns
 //the reverse of it.

 void Banner(){//This method prompts user to print out
 //the banner
   fout<<endl;
   fout<<"Name:Mo Kashi"<<endl;

}//Method Banner

 void palindromic(string s, vector<string>text){
 //This method gets a word and specifies if it is palindromic
 //is so, it increments the number of the palindromic numbers
   string reversed=reverse(s);
   if(s==reversed){
     number_of_palindromics++;
    }//if
 }//palindromic


bool stillcheck(string s,int position, vector<string>text){
//This methos specifies how many words are still words in
//the dictionarry if they are inversed.It makes use of the
//binary search to specify if the inverse of the word is still
//in the dictionarry .It also includes pladromic
  int lowindex=position;
  int highindex=text.size()-1;
  string reversed=reverse(s);
  int midindex;
  while(highindex>=lowindex){
       midindex=(lowindex+highindex)/2;
       if(text[midindex]==reversed){
         number_of_stills++;
         return true;
       }//if
      else
         if(text[midindex]>reversed)
           highindex=midindex-1;
         else
           lowindex=midindex+1;
 }//while
 return false;
}//still check



string reverse(string s){
//This method gets a string and return its reverse
  if(s.size()==1)
    return s;
  else
    return s.substr(s.size()-1,s.size())
           +reverse(s.substr(0,s.size()-1));
}//revese


int countletters( vector<string>text,int n,int size){
//This method accept "n" as an integer and specifies how
//many n-letter words are there in the dictionary.
  int counter=0;//number of n-letter words
  for(int j=0;j<size;j++){
    if(text[j].size()==n)
      counter++;
  }//for
  return counter;
}//countletters



int main(){
 string word;
 fin.open("c:\\input.txt");
 fout.open("c:\\output.txt");
 Banner();
 fin>>word;//reads one word from dictionarry
 while(fin){
   text.push_back(word);
   fin>>word;
 }//while

 for(int i=1;i<11;i++){
   fout<<"Number of "<<i<<"-letter words is: "<<countletters(text,i,text.size())<<endl;
 }//for
 fout<<"==============================================="<<endl;
 for(int i=0;i<text.size()-1;i++){//checks all words to see if
   //they are still words if they spell backwards
   stillcheck(text[i],i,text);
 }//for

 for(int i=0;i<text.size();i++)
    palindromic(text[i],text);
    number_of_stills=(number_of_stills-number_of_palindromics)*2
                     +number_of_palindromics;

    fout<<"Number Of words that are still words if they spell backwards: "<<number_of_stills<<endl;
    fout<<"====================================================="<<endl;
    fout<<"Number of  palidromics= "<<number_of_palindromics<<endl;
    fout<<"====================================================="<<endl;
    cout<<"Enter an integer to exit"<<endl;
    cin>>Freez;

}