CODEWITHSUNDEEP
c++stlfunctor
Marina Golubtsova
Marina Golubtsova

Reputation: 101

how to find duplicates in std::vector<string> and return a list of them?

So if I have a vector of words like: 

Vec1 = "words", "words", "are", "fun", "fun"

resulting list: "fun", "words"

I am trying to determine which words are duplicated, and return an alphabetized vector of 1 copy of them. My problem is that I don't even know where to start, the only thing close to it I found was std::unique_copy which doesn't exactly do what I need. And specifically, I am inputting a std::vector<std::string> but outputting a std::list<std::string>. And if needed, I can use functor.

Could someone at least push me in the right direction please? I already tried reading stl documentation,but I am just "brain" blocked right now.

Upvotes: 9

Views: 17904

Answers (6)

DanielKO
DanielKO

Reputation: 4517

In 3 lines (not counting the vector and list creation nor the superfluous line-breaks in name of readability):

vector<string> vec{"words", "words", "are", "fun", "fun"};
list<string> output;

sort(vec.begin(), vec.end());
set<string> uvec(vec.begin(), vec.end());
set_difference(vec.begin(), vec.end(),
               uvec.begin(), uvec.end(),
               back_inserter(output));

EDIT

Explanation of the solution:

  1. Sorting the vector is needed in order to use set_difference() later.

  2. The uvec set will automatically keep elements sorted, and eliminate duplicates.

  3. The output list will be populated by the elements of vec - uvec.

Upvotes: 7

Jerry Coffin
Jerry Coffin

Reputation: 490148

IMO, Ben Voigt started with a good basic idea, but I would caution against taking his wording too literally.

In particular, I dislike the idea of searching for the string in the set, then adding it to your set if it's not present, and adding it to the output if it was present. This basically means every time we encounter a new word, we search our set of existing words twice, once to check whether a word is present, and again to insert it because it wasn't. Most of that searching will be essentially identical -- unless some other thread mutates the structure in the interim (which could give a race condition).

Instead, I'd start by trying to add it to the set of words you've seen. That returns a pair<iterator, bool>, with the bool set to true if and only if the value was inserted -- i.e., was not previously present. That lets us consolidate the search for an existing string and the insertion of the new string together into a single insert:

while (input >> word)
    if (!(existing.insert(word)).second)
        output.insert(word);

This also cleans up the flow enough that it's pretty easy to turn the test into a functor that we can then use with std::remove_copy_if to produce our results quite directly:

#include <set>
#include <iterator>
#include <algorithm>
#include <string>
#include <vector>
#include <iostream>

class show_copies {
    std::set<std::string> existing;
public:
    bool operator()(std::string const &in) {
        return existing.insert(in).second;
    }
};

int main() {
    std::vector<std::string> words{ "words", "words", "are", "fun", "fun" };
    std::set<std::string> result;

    std::remove_copy_if(words.begin(), words.end(),
        std::inserter(result, result.end()), show_copies());

    for (auto const &s : result)
        std::cout << s << "\n";
}

Depending on whether I cared more about code simplicity or execution speed, I might use an std::vector instead of the set for result, and use std::sort followed by std::unique_copy to produce the final result. In such a case I'd probably also replace the std::set inside of show_copies with an std::unordered_set instead:

#include <unordered_set>
#include <iterator>
#include <algorithm>
#include <string>
#include <vector>
#include <iostream>

class show_copies {
    std::unordered_set<std::string> existing;
public:
    bool operator()(std::string const &in) {
        return existing.insert(in).second;
    }
};

int main() {
    std::vector<std::string> words{ "words", "words", "are", "fun", "fun" };
    std::vector<std::string> intermediate;

    std::remove_copy_if(words.begin(), words.end(),
        std::back_inserter(intermediate), show_copies());

    std::sort(intermediate.begin(), intermediate.end());
    std::unique_copy(intermediate.begin(), intermediate.end(),
        std::ostream_iterator<std::string>(std::cout, "\n"));
}

This is marginally more complex (one whole line longer!) but likely to be substantially faster when/if the number of words gets very large. Also note that I'm using std::unique_copy primarily to produce visible output. If you just want the result in a collection, you can use the standard unique/erase idiom to get unique items in intermediate.

Upvotes: 5

Ben Voigt
Ben Voigt

Reputation: 283684

  1. Make an empty std::unordered_set<std::string>
  2. Iterator your vector, checking whether each item is a member of the set
  3. If it's already in the set, this is a duplicate, so add to your result list
  4. Otherwise, add to the set.

Since you want each duplicate only listed once in the results, you can use a hashset (not list) for the results as well.

Upvotes: 7

Leonid Volnitsky
Leonid Volnitsky

Reputation: 9144

In place (no additional storage). No string copying (except to result list). One sort + one pass:

#include <string>
#include <vector>
#include <list>
#include <iostream>
#include <algorithm>
using namespace std;

int main() {
        vector<string> vec{"words", "words", "are", "fun", "fun"};
        list<string> dup;

        sort(vec.begin(), vec.end());

        const string  empty{""};
        const string* prev_p = &empty;

        for(const string& s: vec) {
                if (*prev_p==s) dup.push_back(s);
                prev_p = &s;
        }

        for(auto& w: dup) cout << w << ' '; 
        cout << '\n';
}

Upvotes: 1

user541686
user541686

Reputation: 210485

Here's a better algorithm than the ones other people have proposed:

#include <algorithm>
#include <vector>

template<class It> It unique2(It const begin, It const end)
{
    It i = begin;
    if (i != end)
    {
        It j = i;
        for (++j; j != end; ++j)
        {
            if (*i != *j)
            { using std::swap; swap(*++i, *j); }
        }
        ++i;
    }
    return i;
}
int main()
{
    std::vector<std::string> v;
    v.push_back("words");
    v.push_back("words");
    v.push_back("are");
    v.push_back("fun");
    v.push_back("words");
    v.push_back("fun");
    v.push_back("fun");
    std::sort(v.begin(), v.end());
    v.erase(v.begin(), unique2(v.begin(), v.end()));
    std::sort(v.begin(), v.end());
    v.erase(unique2(v.begin(), v.end()), v.end());
}

It's better because it only requires swap with no auxiliary vector for storage, which means it will behave optimally for earlier versions of C++, and it doesn't require elements to be copyable.

If you're more clever, I think you can avoid sorting the vector twice as well.

Upvotes: 0

Soron
Soron

Reputation: 448

You can get a pretty clean implementation using a std::map to count the occurrences, and then relying on std::list::sort to sort the resulting list of words. For example:

std::list<std::string> duplicateWordList(const std::vector<std::string>& words) {
    std::map<std::string, int> temp;
    std::list<std::string> ret;
    for (std::vector<std::string>::const_iterator iter = words.begin(); iter != words.end(); ++iter) {
        temp[*iter] += 1;
        // only add the word to our return list on the second copy
        // (first copy doesn't count, third and later copies have already been handled)
        if (temp[*iter] == 2) {
            ret.push_back(*iter);
        }
    }
    ret.sort();
    return ret;
}

Using a std::map there seems a little wasteful, but it gets the job done.

Upvotes: 0

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