Remove anagrams from a string vector - c++

Question

I am trying to make a program which removes all anagrams from vector. I sorted them alphabetical and i put them in a class, hopefully it would be easier to delete de anagrams, but it didnt actually helped.
INPUT:{"have","ahve","frame","avhe","farme","code","abc","ehav"}
OUTPUT:{"code","abc"}
note- all words are lowercase
#include <iostream>
#include<vector>
#include<algorithm>
using namespace std;
struct Fraza{
string word;
string word_alph;
Fraza(){};
Fraza(string words,string word_alph){
this->word=move(word);
this->word_alph=move(word_alph);
}
void printverify() const{
cout<<word<<" is sorted "<<word_alph<<endl;
}
void print() const{
cout<<word<<endl;
}
};
int main() {
Fraza fraza [100];
vector<Fraza>text;
string tmp;
string tmpa;
// i am taking the input, store the words temporary in tmpa and sort tmpa, then i put them in struct
while(cin>>tmp){
tmpa=tmp;
sort(tmpa.begin(),tmpa.end());
text.emplace_back(Fraza(tmp,tmpa));
}
// trying to delete the words
for(int i=0;i<text.size();i++){
for(int j=i+1;j<text.size();j++) {
if (fraza[i].word_alph==fraza[j].word_alph) {
text.erase(text.begin()+i);
text.erase(text.begin()+j);
}
}
}
for(const auto& it:text)
it.printverify();
for(const auto& it:text)
it.print();
return 0;
}
so far i know that it might work comparing the sorted word. If 2 words are the same, i should try delete the words on the specific position

Answer 1

There are several bugs:
The Fraza constructor has a parameter called "words" that you don't use; this->word = std::move(word) moves from the member to itself. (Get familiar with the initialization list).
You have both an array fraza and a vector text. All the words are in text, but you compare the (empty) strings in fraza.
After you have erased the i:th string, the previously j:th string is the j-1:th string.
Fixing these bugs makes your program almost work, except it leaves behind one copy of anagrams that occur an odd number of times (since you remove an even number of them, there will always be one left).
I would approach this with a table.
First, encapsulate the sorting and give Fraza an ordering that makes anagrams equivalent:
class Fraza
{
public:
Fraza(const std::string& w)
: word(w), sorted(w)
{
std::sort(sorted.begin(), sorted.end());
}
bool operator< (const Fraza& other) const
{
return sorted < other.sorted;
}
const std::string& str() const { return word; }
private:
std::string word;
std::string sorted;
};
Then use std::map as a counting table:
int main()
{
// Count the words.
std::map<Fraza, int> table;
std::string word;
while (std::cin >> word)
{
table[word] += 1;
}
// Print the unique ones.
for (const auto& it: table)
{
if (it.second == 1)
{
std::cout << it.first.str() << std::endl;
}
}
}