More elegant way to check for duplicates in C++ array?

Question

I wrote this code in C++ as part of a uni task where I need to ensure that there are no duplicates within an array:

// Check for duplicate numbers in user in         


        

Answer 1

The following solution is based on sorting the numbers and then removing the duplicates:

#include <algorithm>

int main()
{
    int userNumbers[6];

    // ...

    int* end = userNumbers + 6;
    std::sort(userNumbers, end);
    bool containsDuplicates = (std::unique(userNumbers, end) != end);
}

Answer 2

#include<iostream>
#include<algorithm>

int main(){

    int arr[] = {3, 2, 3, 4, 1, 5, 5, 5};
    int len = sizeof(arr) / sizeof(*arr); // Finding length of array

    std::sort(arr, arr+len);

    int unique_elements = std::unique(arr, arr+len) - arr;

    if(unique_elements == len) std::cout << "Duplicate number is not present here\n";
    else std::cout << "Duplicate number present in this array\n";

    return 0;
}

Answer 3

//std::unique(_copy) requires a sorted container.
std::sort(cont.begin(), cont.end());

//testing if cont has duplicates
std::unique(cont.begin(), cont.end()) != cont.end();

//getting a new container with no duplicates
std::unique_copy(cont.begin(), cont.end(), std::back_inserter(cont2));

Answer 4

I think @Michael Jaison G's solution is really brilliant, I modify his code a little to avoid sorting. (By using unordered_set, the algorithm may faster a little.)

template <class Iterator>
bool isDuplicated(Iterator begin, Iterator end) {
    using T = typename std::iterator_traits<Iterator>::value_type;
    std::unordered_set<T> values(begin, end);
    std::size_t size = std::distance(begin,end);
    return size != values.size();
}

Answer 5

You could sort the array in O(nlog(n)), then simply look until the next number. That is substantially faster than your O(n^2) existing algorithm. The code is also a lot cleaner. Your code also doesn't ensure no duplicates were inserted when they were re-entered. You need to prevent duplicates from existing in the first place.

std::sort(userNumbers.begin(), userNumbers.end());
for(int i = 0; i < userNumbers.size() - 1; i++) {
    if (userNumbers[i] == userNumbers[i + 1]) {
        userNumbers.erase(userNumbers.begin() + i);
        i--;
    }
}

I also second the reccomendation to use a std::set - no duplicates there.

Answer 6

It's ok, specially for small array lengths. I'd use more efficient aproaches (less than n^2/2 comparisons) if the array is mugh bigger - see DeadMG's answer.

Some small corrections for your code:

• Instead of int j = i writeint j = i +1 and you can omit your if(j != i) test
• You should't need to declare i variable outside the for statement.