How can I find repeated words in a vector of strings in C++?

user_12670795 0
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I have a std::vector<string> where each element is a word. I want to print the vector without repeated words!

I searched a lot on the web and I found lots of material, but I can't and I don't want to use hash maps, iterators and "advanced" (to me) stuff. I can only use plain string comparison == as I am still a beginner.

So, let my_vec a std::vector<std::string> initialized from std input. My idea was to read all the vector and erase any repeated word once I found it:

  for(int i=0;i<my_vec.size();++i){
    for (int j=i+1;j<my_vec.size();++j){
      if(my_vec[i]==my_vec[j]){
        my_vec.erase(my_vec.begin()+j); //remove the component from the vector
      }
    }
  }

I tried to test for std::vector<std::string> my_vec{"hey","how","are","you","fine","and","you","fine"}

and indeed I found

hey how are you fine and

so it seems to be right, but for instance if I write the simple vector std::vector<std::string> my_vec{"hello","hello","hello","hello","hello"}

I obtain

hello hello

The problem is that at every call to erase the dimension gets smaller and so I lose information. How can I do that?

vector c++ string
5 Answers

Minimalist approach to your existing code. The auto-increment of j is what is ultimately breaking your algorithm. Don't do that. Instead, only increment it when you do NOT remove an element.

I.e.

for (int i = 0; i < my_vec.size(); ++i) {
    for (int j = i + 1; j < my_vec.size(); ) {  // NOTE: no ++j
        if (my_vec[i] == my_vec[j]) {
            my_vec.erase(my_vec.begin() + j);
        }
        else ++j; // NOTE: moved to else-clause
    }
}

That is literally it.

user_13229720

Why don't you use std::unique?

You can use it as easy as:

std::vector<std::string> v{ "hello", "hello", "hello", "hello", "hello" };
std::sort(v.begin(), v.end());
v.erase(std::unique(v.begin(), v.end()), v.end());

N.B. Elements need to be sorted because std::unique works only for consecutive duplicates.

In case you don't want to change the content of the std::vector, but only have stable output, I recommend other answers.

user_55173780

You can store the element element index to erase and then eliminate it at the end. Or repeat the cycle until no erase are performed.

First code Example:

std::vector<int> index_to_erase();

for(int i=0;i<my_vec.size();++i){
    for (int j=i+1;j<my_vec.size();++j){
      if(my_vec[i]==my_vec[j]){
        index_to_erase.push_back(j);
        
      }
    }
  }
//starting the cycle from the last element to the vector of index, in this 
//way the vector of element remains equal for the first n elements
for (int i = index_to_erase.size()-1; i >= 0; i--){
   my_vec.erase(my_vec.begin()+index_to_erase[i]); //remove the component from the vector
}

Second code Example:

bool Erase = true;
while(Erase){
  Erase = false;
  for(int i=0;i<my_vec.size();++i){
    for (int j=i+1;j<my_vec.size();++j){
      if(my_vec[i]==my_vec[j]){
        my_vec.erase(my_vec.begin()+j); //remove the component from the vector
        Erase = true;
      }
    }
  }
}
user_124911750

Erasing elements from a container inside a loop is a little tricky, because after erasing element at index i the next element (in the next iteration) is not at index i+1 but at index i.

Read about the erase-remove-idiom for the idomatic way to erase elements. However, if you just want to print on the screen there is a much simpler way to fix your code:

for(int i=0; i<my_vec.size(); ++i){
   bool unique = true;
   for (int j=0; j<i; ++j){
       if(my_vec[i]==my_vec[j]) {
           unique = false;
           break; 
       }
       if (unique) std::cout << my_vec[i];
   }
}

Instead of checking for elements after the current one you should compare to elements before. Otherwise "bar x bar y bar" will result in "x x bar" when I suppose it should be "bar x y".

Last but not least, consider that using the traditional loops with indices is the complicated way, while using iterators or a range-based loop is much simpler. Don't be afraid of new stuff, on the long run it will be easier to use.

user_41177280

You can simply use the combination of sort and unique as follows.

#include <iostream>
#include <algorithm>
#include <vector>

int main() {
    std::vector<std::string> vec{"hey","how","are","you","fine","and","you","fine"};
    sort(vec.begin(), vec.end());
    vec.erase(unique(vec.begin(), vec.end() ), vec.end());
    
    for (int i = 0; i < vec.size(); i ++) {
        std::cout << vec[i] << " ";
    }
    std::cout << "\n";

    return 0;
}
user_82028500
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