Check std::vector has duplicates - c++

Question

I want to check if a vector of integers has any duplicates or not, and have to return true if it does. So I try to do something like this:
vector<int> uGuess = {1,2,3,3,4,5}
vector<int> a = uGuess;
sort(a.begin(), a.end());
bool d = unique(a.begin(), a.end());
And this will not work since unqiue cannot be assigned as a bool value.
How should I proceed towards this?
If I were to write a for loop to perform the same action, how should I do that?

Answer 1

The algorithm you're looking for is std::adjacent_find.
// The container must be sorted!
const std::vector<int> sortedVector = {1,2,3,3,4,5};
const bool hasDuplicates = std::adjacent_find(sortedVector.begin(), sortedVector.end()) != sortedVector.end();
Unlike std::unique, std::adjacent_find doesn't modify the container.
As a bonus, std::adjacent_find returns an iterator to the first element in the duplicate "pair":
const auto duplicate = std::adjacent_find(sortedVector.begin(), sortedVector.end());
if (duplicate != sortedVector.end())
std::cout << "Duplicate element = " << *duplicate << "\n";

Answer 2

Looking at google for std::unique I found this page std::unique. I looked at what it did:
Eliminates all except the first element from every consecutive group of equivalent elements from the range [first, last)
So it looks like it does what you want - removes the duplicates.
I then looked at what it returns...
... returns a past-the-end iterator for the new logical end of the range
So the result from std::unique is a sequence which is not necessary the same as the whole vector.
If nothing was removed, the return value would be the end of the vector.
So you want:
vector<int>::iterator it = std::unique(a.begin(), a.end());
bool wasUnique = (it == a.end());
Or for C++11:
auto it = std::unique(a.begin(), a.end());
bool wasUnique = (it == a.end());
Finally for the unique function to work, the vector needs to be sorted, so the complete code would be:
sort(a.begin(), a.end());
auto it = std::unique(a.begin(), a.end());
bool wasUnique = (it == a.end());

Answer 3

You should using set
set<int> s(a.begin(), a.end());
return s.size() != a.size();

Answer 4

If someone is forced to write own algorithm:
bool hasDuplicates(const std::vector<int>& arr) {
for (std::size_t i = 0; i < arr.size(); ++i) {
for (std::size_t j = i + 1; j < arr.size(); ++j) {
if (arr[i] == arr[j])
return true;
}
}
return false;
}
But in real code you should use things that already exist, and in the standard library.

Answer 5

Sort the vector if it's not already sorted, and then use std::unique(), like this:
#include <iostream>
#include <vector>
#include <algorithm>
int main() {
std::vector<int> v = {3, 1, 3, 4, 5};
sort(v.begin(), v.end());
auto it = std::unique(v.begin(), v.end());
std::cout << ((it == v.end()) ? "Unique\n" : "Duplicate(s)\n");
return 0;
}
Output:
Duplicate(s)

Answer 6

So far all these solutions either modify the container or have O(n²) complexity. You can put a std::map to much better use:
#include <algorithm>
#include <iterator>
#include <map>
template <typename Iterator>
bool has_duplicates( Iterator first, Iterator last )
{
std::map <typename std::iterator_traits <Iterator> ::value_type, std::size_t> histogram;
while (first != last)
if (++histogram[ *first++ ] > 1)
return true;
return false;
}
#include <iostream>
#include <vector>
int main()
{
using std::begin;
using std::end;
int a[] = { 2, 3, 5, 7, 11 };
int b[] = { 2, 3, 5, 5, 7 };
std::vector <int> c( begin(a), end(a) );
std::vector <int> d( begin(b), end(b) );
std::cout << std::boolalpha;
std::cout << "a has duplicates false : " << has_duplicates( begin(a), end(a) ) << "\n";
std::cout << "b has duplicates true : " << has_duplicates( begin(b), end(b) ) << "\n";
std::cout << "c has duplicates false : " << has_duplicates( begin(c), end(c) ) << "\n";
std::cout << "d has duplicates true : " << has_duplicates( begin(d), end(d) ) << "\n";
}

Answer 7

If your vector is small, say < 32 objects, or if copying and sorting the objects is expensive or impossible due to lack of move or copy constructor/assignment then a straight O(n^2) compare everything against everything else algorithm is the way to go.
Here is my solution:
template <typename Iterator>
bool has_duplicates( Iterator first, Iterator end ) {
for (auto i = first; i != end; ++i) {
for (auto j = first; i != j; ++j) {
if (*i == *j) return true;
}
}
return false;
}
template <typename Container>
bool has_duplicates(const Container &v) {
for (const auto & i : v) {
for (const auto & j : v) {
if (&i == &j) break;
if (i == j) return true;
}
}
return false;
}