I am currently practicing for coding interviews and am working on a function that takes in an array and the size of that array and prints out which numbers in it are duplicates. I have gotten this to work using the two for loop method but want an optimized solution using sets. Snippet of the code I have is below,
#include <iostream>
#include <set>
using namespace std;
void FindDuplicate(int integers[], int n){
set<int>setInt;
for(int i = 0; i < n; i++){
//if this num is not in the set then it is not a duplicate
if(setInt.find(integers[i]) != setInt.end()){
setInt.insert({integers[i]});
}
else
cout << integers[i] << " is a duplicate";
}
}
int main() {
int integers [] = {1,2,2,3,3};
int n = sizeof(integers)/sizeof(integers[0]);
FindDuplicate(integers, n);
}
Any helpful advice is appreciated, thanks
I think your comparison is not needed, insert do it for you:
https://en.cppreference.com/w/cpp/container/set/insert
Returns a pair consisting of an iterator to the inserted element (or
to the element that prevented the insertion) and a bool value set to
true if the insertion took place.
Just insert element and check what insert function returns (false on second element of pair in case of duplication) :)
my solution proposal is :
count the frequencies of each element (algo for frequencies are explained here frequency
display elements with frequency more than 1 (it is a duplicate)
In each operation, you do not use imbricated loops.
#include <iostream>
#include <unordered_map>
using namespace std;
void FindDuplicate(int integers[], int n)
{
unordered_map<int, int> mp;
// Traverse through array elements and
// count frequencies
for (int i = 0; i < n; i++)
{
mp[integers[i]]++;
}
cout << "The repeating elements are : " << endl;
for (int i = 0; i < n; i++) {
if (mp[integers[i]] > 1)
{
cout << integers[i] << endl;
mp[integers[i]] = -1;
}
}
}
int main()
{
int integers [] = {1,1,0,0,2,2,3,3,3,6,7,7,8};
int n = sizeof(integers)/sizeof(integers[0]);
FindDuplicate(integers, n);
}
This is my feedback:
#include <iostream>
#include <vector>
#include <set>
// dont' do this, in big projects it's not done (nameclash issues)
// using namespace std;
// pass vector by const reference you're not supposed to change the input
// the reference will prevent data from being copied.
// naming is important, do you want to find one duplicate or more...
// renamed to FindDuplicates because you want them all
void FindDuplicates(const std::vector<int>& input)
{
std::set<int> my_set;
// don't use index based for loops if you don't have to
// range based for loops are more safe
// const auto is more refactorable then const int
for (const auto value : input)
{
//if (!my_set.contains(value)) C++ 20 syntax
if (my_set.find(value) == my_set.end())
{
my_set.insert(value);
}
else
{
std::cout << "Array has a duplicate value : " << value << "\n";
}
}
}
int main()
{
// int integers[] = { 1,2,2,3,3 }; avoid "C" style arrays they're a **** to pass around safely
// int n = sizeof(integers) / sizeof(integers[0]); std::vector (or std::array) have size() methods
std::vector input{ 1,2,2,3,3 };
FindDuplicates(input);
}
You do not need to use a set.
To find the duplicates:
Sort array with numbers
Iterate over the array (start with second element) and copy elements where previous element equals
current element into a new vector "duplicates"
(Optional) use unique on the "duplicates" if you like to know which number is a duplicate and do not care if it is 2, 3 or 4 times in the numbers array
Example Implementation:
#include <algorithm>
#include <iostream>
#include <vector>
void
printVector (std::vector<int> const &numbers)
{
for (auto const &number : numbers)
{
std::cout << number << ' ';
}
std::cout << std::endl;
}
int
main ()
{
auto numbers = std::vector<int>{ 1, 2, 2, 42, 42, 42, 3, 3, 42, 42, 1, 2, 3, 4, 5, 6, 7, 7 };
std::sort (numbers.begin (), numbers.end ());
auto duplicates = std::vector<int>{};
std::for_each (numbers.begin () + 1, numbers.end (), [prevElement = numbers.begin (), &duplicates] (int currentElement) mutable {
if (currentElement == *prevElement)
{
duplicates.push_back (currentElement);
}
prevElement++;
});
duplicates.erase (std::unique (duplicates.begin (), duplicates.end ()), duplicates.end ());
printVector (duplicates);
}
edit:
If you have no problem with using more memory and more calculations but like it more expressive:
Sort numbers
Create a new array with unique numbers "uniqueNumbers"
Use "set_difference" to calculate (numbers-uniqueNumbers) which leads to an new array with all the duplicates
(Optional) use unique on the "duplicates" if you like to know which number is a duplicate and do not care if it is 2, 3 or 4 times in the numbers array
Example Implementation:
#include <algorithm>
#include <iostream>
#include <vector>
void
printVector (std::vector<int> const &numbers)
{
for (auto const &number : numbers)
{
std::cout << number << ' ';
}
std::cout << std::endl;
}
int
main ()
{
auto numbers = std::vector<int>{ 2, 2, 42, 42, 42, 3, 3, 42, 42, 1, 2, 3, 4, 5, 6, 7, 7 };
std::sort (numbers.begin (), numbers.end ());
auto uniqueNumbers = std::vector<int>{};
std::unique_copy (numbers.begin (), numbers.end (), std::back_inserter (uniqueNumbers));
auto duplicates = std::vector<int>{};
std::set_difference (numbers.begin (), numbers.end (), uniqueNumbers.begin (), uniqueNumbers.end (), std::back_inserter (duplicates));
std::cout << "duplicate elements: ";
printVector (duplicates);
std::cout << "unique duplicate elements: ";
printVector ({ duplicates.begin (), std::unique (duplicates.begin (), duplicates.end ()) });
}
here's a quick solution use an array of size N (try a big number)
and whenever a number is added into the other array on the large array add 1 to the position like:
array_of_repeated[user_input]++;
so if the program asks how many times (for example) number 234 was repeated?
std::cout<<array_of_repeated[requested_number]<<std::endl;
so in this way you wont spend time looking for a number inside the other list
Related
I am new to C++, and I have run into a total lack of understanding on how to sum only even values stored in a vector in C++.
The task itself requests a user to input some amount of random integers, stop when input is 0, and then to return the amount of even values and the sum of those even values.
This is as far as I have managed to get:
#include <algorithm>
#include <functional>
#include <iostream>
#include <vector>
#include <numeric>
using namespace std;
int main()
{
vector<int> vet;
int s = 1;
while (s != 0) {
std::cin >> s;
vet.push_back(s);
}
int n = count_if(vet.begin(), vet.end(),
[](int n) { return (n % 2) == 0; });
cout << n << endl;
//here is the start of my problems and lack of undertanding. Basically bad improv from previous method
int m = accumulate(vet.begin(), vet.end(), 0,
[](int m) { for (auto m : vet) {
return (m % 2) == 0; });
cout << m << endl; //would love to see the sum of even values here
return 0;
}
The function to be passed to std::accumulate takes 2 values: current accumulation value and value of current element.
What you should do is add the value if it is even and make no change when not.
int m = accumulate(vet.begin(), vet.end(), 0,
[](int cur, int m) {
if ((m % 2) == 0) {
return cur + m; // add this element
} else {
return cur; // make no change
}
});
From c++20, you can separate out the logic that checks for even numbers, and the logic for summing up those values:
auto is_even = [](int i) { return i % 2 == 0; };
auto evens = vet | std::views::filter(is_even);
auto sum = std::accumulate(std::begin(evens), std::end(evens), 0);
Here's a demo.
This is my solution(sorry if it's not right I'm writing it on my phone)
You don't need a vector form this, you just need to check right from the input if the number is divisible to 2
My solution:(a littie bit ugly)
#include <iostream>
using namespace std;
int main()
{
int s {1};
int sum{};
int countNum{};
while (s != 0)
{
cin >> s;
if (s % 2 == 0)
{
sum += s;
countNum++;
}
}
cout << countNum << ' ' << sum;
}
i don't realy know what you want to do in the second part of your code but you can sum the even numbers by this way and i want to told you another thing when you using namespace std you don't need to write std::cin you can only write cin directly
#include <iostream>
#include <vector>
using namespace std;
int main()
{
vector<int> vet;
int s = 1;
//Take Input
while (s != 0) {
cin >> s;
vet.push_back(s);
}
//count elements
int elements_count = vet.size(); //vet.size() return the total number of elements of vector
//store the sum here
int sum=0;
//loop on the vector and sum only even numbers
for(int i=0;i<elements_count;i++){
if(vet[i] %2 ==0)
sum += vet[i];//check of the element of index i in the vector is even if it true it will add to sum
}
cout << sum;
return 0;
}
int sumEven=0;
int v[100];
int n;//number of elements you want to enter in the array
do{cout<<"Enter n";
cin>>n;}while(n<=0);
//in a normal 1 dimensional array
for(int i=0;i<n;i++)
if(v[i]%2==0)
sumEven+=v[i];
//in a vector
vector<int> v;
for(vector<int>::iterator it=v.begin();it!=v.end();it++)
if(*it%2==0)
sumEven+=v[i];
Similar to answers above, but if you want to keep the vector of even numbers as well, here are two approaches.
#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>
int main() {
std::vector<int> vec = {1,2,3,4,5,6,7,8,9,10};
// Hold onto what we know is the right answer.
int known_sum = 2+4+6+8+10;
// Copy only even values into another vector
std::vector<int> even_values;
std::copy_if(vec.begin(), vec.end(),
std::back_inserter(even_values),
[](int val){ return val%2==0; });
// Compute sum from even values vector
int even_value_sum = std::accumulate(even_values.begin(), even_values.end(), 0);
// Compute sum from original vector
int even_value_second = std::accumulate(vec.begin(), vec.end(), 0,
[](int current_sum, int new_value) {
return new_value%2==0 ? current_sum + new_value:current_sum;
}
);
// These should all be the same.
std::cout << "Sum from only even vector: " << even_value_sum << std::endl;
std::cout << "Sum from binary op in std accumulate: " << even_value_second << std::endl;
std::cout << "Known Sum: " << known_sum << std::endl;
}
Range-based for loops
A range-based for loop is arguably always a valid alternative to the STL algos, particularly in cases where the operators for the algos are non-trivial.
In C++14 and C++17
E.g. wrapping a range-based even-only accumulating for loop in an immediately-executed mutable lambda:
#include <iostream>
#include <vector>
int main() {
// C++17: omit <int> and rely on CTAD.
const std::vector<int> v{1, 10, 2, 7, 4, 5, 8, 13, 18, 19};
const auto sum_of_even_values = [sum = 0, &v]() mutable {
for (auto val : v) {
if (val % 2 == 0) { sum += val; }
}
return sum;
}();
std::cout << sum_of_even_values; // 42
}
In C++20
As of C++20, you may use initialization statements in the range-based for loops, as well as the ranges library, allowing you to declare a binary comparator in the initialization statement of the range-based for loop, and subsequently apply it the range-expression of the loop, together with the std::ranges::filter_view adaptor:
#include <iostream>
#include <vector>
#include <ranges>
int main() {
const std::vector v{1, 10, 2, 7, 4, 5, 8, 13, 18, 19};
const auto sum_of_even_values = [sum = 0, &v]() mutable {
for (auto is_even = [](int i) { return i % 2 == 0; };
auto val : v | std::ranges::views::filter(is_even)) {
sum += val;
}
return sum;
}();
std::cout << sum_of_even_values; // 42
}
I want to compare the elements in array whether there is a similar element is present or not?
I am using two loops and it gives TLE can anyone help me in this?
code:-
int main() {
int arr=[1, 2, 3, 3, 4, 5, 6, 5, 6]
int max=0;
int k;
int c=0;
for (int i=0;i<n;i++) {
for (int j= i+1;j<n;j++) {
if (a[i]==a[j]) {
c++;
if (k>max) {
max= k;
}
}
}
}
}
how to optimize this?
The std::sort function time complexity is nlog(n) (According to Wikipedia, it's Intro Sort)
So one approach for your question is to:
Sort the array (It's better to use std::vector instead of C-Style array)
Compare every element with the next one (Which is going to be O(n))
If the two elements are equal, so show it or add it to a std::set to use it later
Here is my implementation for the algorithm that I mentioned before:
#include <vector>
#include <iostream>
#include <set>
#include <algorithm>
int main() {
std::vector<int> data{1, 2, 3, 4, 6, 2, 7, 1, 5, 4};
std::set<int> duplicatedValues;
std::sort(data.begin(), data.end());
for (int i{0}; i < data.size()-1; ++i)
if (data[i] == data[i+1])
duplicatedValues.insert(data[i]);
for (int value : duplicatedValues)
std::cout << value << " ";
std::cout << std::endl;
}
So time complexity for code above is going to be nlog(n)
Note: I should add that if you have n data and all of them are between range of 0 to n-1, We have O(n) for that, take a look at here.
Second Approach: (By using std::unordered_map)
(It takes a little shorter time than the previous approach, but roughly 2 times more memory!)
We can add every element in a dictionary and if we had that same data again in our array, increase its number in dictionary; and finally, show elements that have number more than 1.
#include <iostream>
#include <unordered_map>
#include <vector>
#include <stdlib.h>
int main() {
std::vector<int> data{1, 2, 8, 2, 5, 4, 1, 9, 2, 3, 8};
std::unordered_map<int, int> duplicatedData;
for (int i{0}; i < data.size(); ++i) {
if (duplicatedData.find(data[i]) == duplicatedData.end()) {
duplicatedData[data[i]] = 1;
}
else {
duplicatedData[data[i]] += 1;
}
}
for (auto x : duplicatedData) {
if (x.second > 1) {
std::cout << x.first << " ";
}
}
std::cout << std::endl;
}
EDIT: I recorded a video for this, with performance measurement in here
i will do it like
#include <algorithm>
#include <vector>
#include <execution>
#include <atomic>
#include <iostream>
int main()
{
int arr[] = {1,2,3,5,6};
std::atomic<bool> com_tr = false;
std::for_each(std::execution::par, &arr[0], (&arr[0] + sizeof(arr)/sizeof(int)),
[&](auto a)
{ com_tr = std::count(&arr[0], (&arr[0] + sizeof(arr)/sizeof(int)), a) == 1 ? (bool)com_tr : !false; });
if (com_tr) std::cout << "have one " << std::endl;
}
compiled with c++ -std=c++17 -O3 '/home/alex/etr/b.cpp' -o t -ltbb
but if it's really all you need to accomplish better would be std::any_of()
As the title says, I'm looking for a way to sort a vector without modifying the original one.
My first idea is of course to create a copy of the vector before the sort, e.g.:
std::vector<int> not_in_place_sort(const std::vector<int>& original)
{
auto copy = original;
std::sort(copy.begin(), copy.end());
return copy;
}
However, maybe there is a more efficient way to perform the sort using C++ standard algorithm (maybe a combination of sort and transform?)
Use partial_sort_copy. Here is an example:
vector<int> v{9,8,6,7,4,5,2,0,3,1};
vector<int> v_sorted(v.size());
partial_sort_copy(begin(v), end(v), begin(v_sorted), end(v_sorted));
Now, v remains untouched but v_sorted contains {0,1,2,3,4,5,6,7,8,9}.
Here is my favorite. Sort an index and not the original array/vector itself.
#include <algorithm>
int main() {
int intarray[4] = { 2, 7, 3, 4 };//Array of values
//or you can have vector of values as below
//std::vector<int> intvec = { 2, 7, 3, 4 };//Vector of values
int indexofarray[4] = { 0, 1, 2, 3 };//Array indices
std::sort(indexofarray, indexofarray + 4, [intarray](int index_left, int index_right) { return intarray[index_left] < intarray[index_right]; });//Ascending order.
//have intvec in place of intarray for vector.
}
After this, indexofarray[] elements would be 0, 2, 3, 1, while intarray[] is unchanged.
As suggested in the comments pass the function argument by value std::vector<int> original:
#include <iostream>
#include <vector>
#include <algorithm>
std::vector<int> not_in_place_sort(std::vector<int> original) {
std::sort(original.begin(), original.end());
return original;
}
int main() {
std::vector<int> v = { 8, 6, 7, 2, 3, 4, 1, 5, 9 };
std::vector<int> v2 = not_in_place_sort(v); // pass the vector by value
std::cout << "v1: " << '\n';
for (auto el : v) {
std::cout << el << ' ';
}
std::cout << "\nv2: " << '\n';
for (auto el : v2) {
std::cout << el << ' ';
}
}
That will sort a copy of your original vector leaving the original intact.
As pointed out below this might restrict some optimizations such as RVO but will call vector's move constructor in the return statement instead.
For the case where you are interested in proxy sorting (sorting an index list), you may want to implement a more flexible algorithm that allows you to deal with containers which do not support random access (such as std::list). For example:
#include <algorithm>
#include <iostream>
#include <list>
#include <numeric>
#include <vector>
template <typename Container>
auto sorted_indices(const Container& c) {
std::vector<typename Container::size_type> indices(c.size());
std::iota(indices.begin(), indices.end(), 0);
std::sort(indices.begin(), indices.end(), [&c](auto lhs, auto rhs) {
return (*(std::next(c.begin(), lhs)) < *(std::next(c.begin(), rhs)));
});
return indices;
}
template <typename Container, typename Indices>
auto display_sorted(const Container& c, const Indices& indices) {
std::cout << "sorted: ";
for (auto&& index : indices) {
std::cout << *(std::next(c.begin(), index)) << " ";
}
std::cout << std::endl;
}
template <typename Container>
auto display_sorted(const Container& c) {
return display_sorted(c, sorted_indices(c));
}
template <typename Container>
auto display(const Container& c) {
std::cout << "as provided: ";
for (auto&& ci : c) std::cout << ci << " ";
std::cout << std::endl;
}
int main() {
// random access
const std::vector<int> a{9, 5, 2, 3, 1, 6, 4};
display(a);
display_sorted(a);
display(a);
std::cout << "---\n";
// no random access
const std::list<int> b{9, 5, 2, 3, 1, 6, 4};
display(b);
display_sorted(b);
display(b);
}
Sample run:
$ clang++ example.cpp -std=c++17 -Wall -Wextra
$ ./a.out
as provided: 9 5 2 3 1 6 4
sorted: 1 2 3 4 5 6 9
as provided: 9 5 2 3 1 6 4
---
as provided: 9 5 2 3 1 6 4
sorted: 1 2 3 4 5 6 9
as provided: 9 5 2 3 1 6 4
As you would expect, relying on proxy sorting could have important performance implications. For example: every time you want to traverse in order, you will possibly incur cache misses. In addition, the traversal will have the same complexity as the underlying container for random access: In the case of std::vector, std::next(v.begin(), n) is O(1), but in the case of std::list, std::next(l.begin(), n) is O(n).
For int's it doesn't make much difference if you're sorting an index or making a copy & sorting the copy; the data still needs to be initialized, and in the case of the indexes, this will involve a loop assigning values rather than faster memcpy routines; so may end up slower; in addition you're going to be jumping around memory lots more; so now the cache can't do its job nicely.
For larger objects I'd not sort the index, but use a vector of pointers. The copy of the pointers is cheap compared to copying the objects themselves; the containers are still obvious because they're containing pointers of your object; and the sort isn't attempting to reference another vector.
You can create another vector to store the indices. Here is the code:
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main()
{
vector<int> numbers = {50,30,20,10,40};
vector<int> indexOfNumbers;
for(int i = 0; i < numbers.size(); i++)
{
indexOfNumbers.push_back(i);
}
// Now, indexOfNumbers = [0,1,2,3,4]
std::sort(
indexOfNumbers.begin(), indexOfNumbers.end(),
[numbers](int leftIndex, int rightIndex)
{
return numbers[leftIndex] < numbers[rightIndex]; // sort in ascending order
}
);
// After sorting, indexOfNumbers = [3, 2, 1, 4, 0]
// Access the sorted elements
cout << "Accessing the sorted elements : ";
for(int i = 0; i < numbers.size(); i++)
{
cout << numbers[indexOfNumbers[i]] << " ";
}
// prints numbers in sorted order i.e. [10,20,30,40,50]
return 0;
}
Source: Made slight modification according to Tyrer's answer (https://stackoverflow.com/a/47537314)
I've recently began to work with C++ std::sets, so there is a question I did not find answer to in Google.
I have a std::set of some int values (e.g., let it be 1, 2, 3, 4, 5). The task is to calculate difference between two first elements. Is it possible to do with C++?
I'm using data structure std::set; And I know that first element can be got like that:
int diff = *arSeq.begin();
Where arSeq is mentioned set.
Is there any way to get the second element?
Yes it is possible. Access the first element via the std::set::begin iterator and the next one using the std::next function.
#include <iostream>
#include <set>
int main() {
std::set<int> s = { 1, 2, 3, 4, 5 };
auto result = *s.begin() - *std::next(s.begin());
std::cout << result;
}
Please note that you will again get the same result of -1 even if you defined your set as:
std::set<int> s = { 5, 3, 1, 4, 2 };
because the std::set is a container that:
contains a sorted set of unique objects...
You can do it the following way
#include <iostream>
#include <set>
#include <iterator>
int main()
{
std::set<int> s({ 1, 2, 3, 4, 5 });
long long int diff = 0;
if (not (s.size() < 2))
{
auto first = s.begin();
auto second = std::next(first);
diff = *first < *second ? ( long long int )*second - *first : ( long long int )*first - *second;
}
std::cout << "difference = " << diff << std::endl;
return 0;
}
What did you mean by first two elements. First two element inserted? then its not possible.
Set elements are sorted with given comparator. By default primitive elements are sorted in ascending order. See the output of following program.
// set::begin/end
#include <iostream>
#include <set>
#include<functional>
int main ()
{
int myints[] = {75,23,65,42,13};
std::set<int, std::greater<int>> myset (myints,myints+5);
auto first = myset.begin();
auto second = std::next(first);
std::cout <<*first<< ", " <<*second<<"\n";
return 0;
}`enter code here`
This question already has answers here:
Checking for duplicates in a vector [duplicate]
(5 answers)
Closed 9 years ago.
I have a vector of int which can include maximum 4 elements and minimum 2, for example :
std::vector<int> vectorDATA(X); // x means unknown here
What I want to do is to erase the elements that are repeated for example :
vectorDATA{1,2,2} to vectorDATA{1,2}
vectorDATA{1,2,3} to nothing changes
vectorDATA{2,2,2} to vectorDATA{2}
vectorDATA{3,2,1,3} to vectorDATA{3,2,1}
vectorDATA{1,2,1,2} to vector{1,2}
and so on
here a code simple :
cv::HoughLines(canny,lineQ,1,CV_PI/180,200);
std::cout << " line Size "<<lineQ.size()<< std::endl;
std::vector<int> linesData(lineQ.size());
std::vector<int> ::iterator it;
if(lineQ.size() <=4 && lineQ.size() !=0 ){
if(lineQ.size()==1){
break;
}else {
for ( int i = 0; i<lineQ.size();i++){
linesData[i] = lineQ[i][1]; // my comparison parameter is the lineQ[i][1]
}
// based on the answer I got I'm trying this but I really don't how to continue ?
std::sort(lineQ.begin(),lineQ.end(),[](const cv::Vec2f &a,const cv::Vec2f &b)
{
return ????
}
I tried use a for and do while loop, but I didn't get it, and the function std::adjacent_find this has a condition that the elements should be consecutive.
Maybe it's easy but I just don't get it !
thanks for any help !
The easy way is sort then unique-erase, but this changes order.
The c++11 order preserving way is to create an unordered_set<int> s; and do:
unordered_set<int> s;
vec.erase(
std::remove_if( vec.begin(),vec.end(), // remove from vector
[&](int x)->bool{
return !std::get<1>(s.insert(x)); // true iff the item was already in the set
}
),
vec.end() // erase from the end of kept elements to the end of the `vec`
);
which is the remove-erase idiom using the unordered_set to detect duplicates.
I didn't see a sort-less source code in the already mentioned answers, so here it goes. Hash table for checking duplicates, shifting unique elements towards the front of the vector, note that src is always >= dst and dst is the number of copied, i.e. unique elements at the end.
#include <unordered_set>
#include <vector>
#include <iostream>
void
uniq (std::vector<int> &a) {
std::unordered_set<int> s;
size_t dst = 0;
for (size_t src = 0; src < a.size(); ++src) {
if (s.count (a[src]) == 0) {
s.insert (a[src]);
a[dst++] = a[src];
}
}
a.resize (dst);
}
int
main () {
std::vector<int> a = { 3, 2, 1, 3, 2, 1, 2, 3, 4, 5 ,2, 3, 1, 1 };
uniq (a);
for (auto v : a)
std::cout<< v << " ";
std::cout << std::endl;
}
If you want to realy remove repeated elements, you may try something like this:
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int main () {
int data[] = {1,2,3,2,1};
vector<int> vectorDATA = (&data[0], &data[0] + 5);
sort(vectorDATA.begin(),vectorDATA.end());
for(int i = 0; i < vectorDATA.size()-1; ++i)
{
if(vectorDATA[i] == vectorDATA[i+1])
vectorDATA.erase(vectorDATA.begin()+i+1);
}
for(int i = 0; i < vectorDATA.size();++i)
{
cout << vectorDATA[i] << " ";
}
cout << endl;
return 0;
}
Lack of of this method is then elements lost his order.