c++ - Repeated elements in a std::vector

Question

I have a std::vector and I want to check all the elements in it. If a certain element appears more than once, I signal an error.

This is how I did it:

std::vector<std::string> test;
test.push_back("YES");
test.push_back("YES");

for(int i = 0; i < test.size(); i++)
{
    if(test[i] > 1)
    {
        DCS_LOG_DEBUG("ERROR WITH COUNT")
    }
}

This did not work though I know how to count using the std::vector::count() method. But I want to get the count for each element, as opposed to counting everything... any ideas?

Answer 1

The simplest way is to std::sort the vector and then use std::adjacent_find.

---

However, if you don't want to sort the vector, you can do something like this in C++11:

#include <unordered_map>
#include <functional> // For std::hash<std::string>.
#include <string>
#include <iostream>

int main() {

    // Test data.
    std::vector<std::string> v;
    v.push_back("a");
    v.push_back("b");
    v.push_back("c");
    v.push_back("a");
    v.push_back("c");
    v.push_back("d");
    v.push_back("a");

    // Hash function for the hashtable.
    auto h = [](const std::string* s) {
        return std::hash<std::string>()(*s);
    };

    // Equality comparer for the hashtable.
    auto eq = [](const std::string* s1, const std::string* s2) {
        return s1->compare(*s2) == 0;
    };

    // The hashtable:
    //      Key: Pointer to element of 'v'.
    //      Value: Occurrence count.
    std::unordered_map<const std::string*, size_t, decltype(h), decltype(eq)> m(v.size(), h, eq);

    // Count occurances.
    for (auto v_i = v.cbegin(); v_i != v.cend(); ++v_i)
        ++m[&(*v_i)];

    // Print strings that occur more than once:
    for (auto m_i = m.begin(); m_i != m.end(); ++m_i)
        if (m_i->second > 1)
            std::cout << *m_i->first << ": " << m_i->second << std::endl;

    return 0;

}

This prints:

a: 3
c: 2

I didn't actually benchmark it, but this has a chance for being rather performant, for following reasons:

• Assuming the actual vector elements do not produce pathologically lopsided hashes, this is actually an O(n) algorithm, as opposed to O(n*log(n)) for sorting.
• We are using the hashtable of pointers to strings, not strings themselves, so there is no unnecessary copying taking place.
• We can "pre-allocate" hashtable buckets (we pass v.size() when constructing m), so hashtable resizes are minimized.