Write a c++ program where two or more strings are anagrams if they contain exactly the same letters, ignoring capitalization, punctuation, and spaces. for example, "information superhighway" and "new utopia? horrifying sham" are anagrams, as are "ab123ab" and "%%b b*aa". note that two lines that contain no letters are anagrams.

/*

Problem; to finf s2 strings are anagrams.

(s2 strings are anagrams if they contain exactly the same letters, ignoring capitalization, punctuation,)

and spaces.

Solution: we must take account wich letters are present in each string and how many times. To achieve that we

may create a struct with a char and a char count, some constructors, both geters and an equal operator.

*/

#include <iostream>

#include <vector>

using namespace std;

struct charCount {

private:                                                                    // Internal data structure:

   char letter;                                                            // letter (should be unique)

   unsigned int letterCount;                                               // and is present at least 1 time    

public:

   charCount() {}                                                          // Default Constructor

   charCount(char l) {letter = l; letterCount = 1;}                        // Constructor

   charCount(const charCount & other) {letter = other.letter;

                                       letterCount = other.letterCount;}   // Copy Constructor

   void increments() {letterCount++;}                                      // Counts other occurrence

   char ltr() {return(letter);}                                            // Geter

   unsigned int ltrCount() {return(letterCount);}                          // Geter

   bool operator ==(charCount other) {return(letter == other.letter        // Equal means same char and

                                     && letterCount == other.letterCount);}// same count.

};

/*

We have to analyze each input string returning an ordered vector of charCount elements:

*/

vector<charCount> analysis(string s) {                                // Raw string data

vector<charCount> result;                                             // Result vector

for(char c:s) {                                                       // For each char of string: we take a-z

   if(isalpha(c)) {                                                  // only into account and we normalize it

       char x = tolower(c);                                          // to lowercase, store as x and we are

       bool found = false;                                           // ready to search

       unsigned int pos = 0;                                         // from first position to last

       for(auto & y:result) {                                        // for each vector's element

           if(y.ltr() == x) {                                        // if we find x, we must

               y.increments();                                       // count this occurrence,

               found = true;                                         // turn on the found flag

               break;                                                // and exit from the loop.

               } else if(y.ltr() < x) {                              // Otherwise may be we are not going to

               result.emplace(result.begin()+(pos),(charCount(x)));  // find it in the ordered vector: we insert

               found = true;                                         // it at current vector's position. It is

               break;                                                // present now: we should exit from loop.

               } else pos++;                                         // Otherwise we increment pos and go on.

           }

       if(!found) {                                                  // x doesn't belongs to vector's set: we

           auto it = result.end();                                   // add it at last position, keeping order

           result.insert(it,(charCount(x)));                         // and we may go on for another cycle if any.

           }

       }

   }

return result;

}

template <typename T>                                           // Generic code follows to implement

bool operator ==(vector<T> v1,vector<T> v2) {                   // 2 vectors comparison: they are

bool result = (v1.size() == v2.size());                         // equals if their sizes match,

if(result)                                                      // and each element from first vector

   for(unsigned int i = 0; i < v1.size();i++)                  // matches with each element from the

       result = result && v1[i] == v2[i];                      // second vector at the same position.

return(result);

}

template bool operator ==(vector<charCount>,vector<charCount>); // Orders compiler to generate a charCount version.

bool areAnagrams(string s1,string s2) {                         // Anagrams implies 2 identical analysis vectors.

return(analysis(s1) == analysis(s2));}

int main(int argc, char *argv[]) {                                      // Test code folloes:

string s1,s2;                                                           // 2 strings to store input data.

cout << "Text 1:"; getline(cin,s1);                                     // Reads string 1 till <Enter> key.

cout << "Text 2:"; getline(cin,s2);                                     // Reads string 2 till <Enter> key.

cout << (areAnagrams(s1,s2)?"Are":"Are not") << " anagrams..." << endl; // Are they? We use the ternary operator.

return 0;

}

Explanation: included inside the code as comments.

#include <iostream>

using namespace std;

void areAnagrams(char str1[], char str2[])

{

   int i, flag = 1,  count1[26] = {0}, count2[26] = {0};

   for(i = 0; str1[i] != '\0'; i++) {

       if (isalpha(str1[i])) {

           str1[i] = tolower(str1[i]);

           count1[str1[i] - 'a']++;

       }

   }

   for(i = 0; str2[i] != '\0'; i++) {

       if (isalpha(str2[i])) {

           str2[i] = tolower(str2[i]);

           count2[str2[i] - 'a']++;

       }

   }    

   for(i = 0; i < 26; i++) {

       if (count1[i] != count2[i])

           flag = 0;

   }

   if (flag == 0)

       cout << "The two strings are NOT anagrams";

   else

       cout << "The two strings are anagrams";

}

int main ()

{

   char str1[100], str2[100];

   cout << "Enter the first string: ";

   gets(str1);

   cout << "Enter the second string: ";

   gets(str2);

   areAnagrams(str1, str2);

   return 0;

}

Explanation:

Inside the function:

- In the first loop, each character of the first string is checked if it is a letter. If it is, lower the character and increment its frequency by 1.

- In the second loop, each character of the second string is checked if it is a letter. If it is, lower the character and increment its frequency by 1.

- In the third for loop, compare the frequencies

- If they are equal, that means the strings are anagrams

ιт ιѕ ωнєи ѕσмєσиє ρυтѕ ιи α ¢σ∂є σя ѕσмєтнιиg тнαт вяιиgѕ α νιяυѕ, σя мєѕѕєѕ уσυя gαмє υρ, σя αℓℓσωѕ тнєм тσ ¢нєαт.

αи∂ уєѕ ι нανє вєєи нα¢кє∂ вєfσяє(ιт αιит fυи)

нσρє ι ¢συℓ∂ нєℓρ уσυ.

it is not paas i hope that