Remove duplicates from an unsorted array - c++

Question

I'm trying to create a function to remove duplicates from an unsorted int array. I have a solution that works for more examples, but it's failing with the following input:
#include<iostream>
using namespace std;
int removeDuplicates(int arr[], int n)
{
int j = 0;
for (int i=0; i < n; i++){
for(int j=0;j<=i;j++){
if(arr[i]==arr[j]){
n--;
for (int k=i; k<n; k++){
arr[k]=arr[k+1];
}
}
}
}
return n;
}
// Driver code
int main()
{
int arr[] = {0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1};
int n = sizeof(arr) / sizeof(arr[0]);
n = removeDuplicates(arr, n);
for (int i=0; i<n; i++)
cout << arr[i] << " ";
return 0;
}
The output for this arr example is 0 0 1 0 0 and should be 0 1.
Do you see where is the problem? Thank you

Answer 1

Consider using std::set<int> to record numbers you've already seen, and using a STL algorithm to perform the removal:
#include<iostream>
#include<algorithm>
#include<functional>
#include<set>
// Driver code
int main()
{
int arr[] = {0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1};
std::set<int> duplicates;
auto it = std::remove_if(std::begin(arr), std::end(arr), [&duplicates](int i) {
return !duplicates.insert(i).second;
});
size_t n = std::distance(std::begin(arr), it);
for (size_t i = 0; i < n; i++)
std::cout << arr[i] << " ";
return 0;
}
The effect of this code is that all duplicates are moved to the end of the array, and the iterator returned by std::remove_if indicates the end of the new list. So iterating between the beginning and that iterator gives you the array without the duplicates.

Answer 2

look over inner loop you forgot to decrement i
#include<iostream>
using namespace std;
int removeDuplicates(int arr[], int n)
{
int j = 0;
for (int i=0; i < n; i++){
for(int j=0;j<i;j++){
if(arr[i]==arr[j]){
n--;
for (int k=i; k<n; k++){
arr[k]=arr[k+1];
}
i--; // you forgot to decrement i
}
}
}
return n;
}
// Driver code
int main()
{
int arr[] = {0, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1};
int n = sizeof(arr) / sizeof(arr[0]);
n = removeDuplicates(arr, n);
for (int i=0; i<n; i++)
cout << arr[i] << " ";
return 0;
}

Answer 3

In your function
int removeDuplicates(int arr[], int n)
{
int j = 0;
for (int i=0; i < n; i++){
for(j=i+1;j<n;)
{
if(a[i]==a[j])
{
for(int k=j;k<n-1;++k)
arr[k]=arr[k+1];
--n;
}
else
++j;
}
}
return n;
}
Increment j only when the two values do not match. Or else it will skip few values

Answer 4

UPDATE
A possible solution that is O(n log n) in time and requires O(m) extra space, where m is the number of unique elements in the input array:
template <typename RAIter>
size_t remove_duplicates(RAIter first, RAIter last) {
using value_type = typename std::iterator_traits<RAIter>::value_type;
std::map<value_type, size_t> map;
size_t n = 0;
for (auto it = first; it != last; ++it) {
auto & temp = map[*it];
if (temp == 0) temp = ++n;
}
for (auto & e : map)
*(first + e.second - 1) = e.first;
return n;
}
Note also that the contents of the original array is destroyed here, but this is in your attempt as well.
Possible usage:
int main() {
static constexpr size_t n = 26;
std::array<int, n> a = { 0, 0, 1, 0, 3, 2, 1, 1, 0, 1, 0, 0, 2, 2, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1 };
size_t m = remove_duplicates(std::begin(a), std::end(a));
for (size_t i = 0; i < m; i++)
std::cout << a[i] << " ";
std::cout << std::endl;
}
Which prints out 0 1 3 2.
I compared my solution with yours (corrected by #Onk_r). For an input array of 500,000 elements having random values from [0,100). My O(n log n) solution took 19 milliseconds, while your O(n3) solution took 54 seconds! Nice demonstration of how much complexity matters :).

Answer 5

It works, but you have to start with j=1, not 0
wrong:
for (int i=0; i < n; i++){
for(int j=0;j<=i;j++)
solution:
for (int i=0; i < n; i++){
for(int j=1;j<=i;j++){

Answer 6

int arr1[] = {3, 1, 5, 4, 5, 1, 9, 3, 9, 7};
int size = sizeof(arr1) / sizeof(arr1[0]);
int i, j, k = 0;
int arr2[size];
for(i = 0; i < size; i++)
{
for(j = 0; j < k; j++)
{
if(arr1[i] == arr2[j])
{
break;
}
}
if(j == k)
{
arr2[k++] = arr1[i];
}
}

Answer 7

Use a set instead since all elements in a set must be unique. http://www.cplusplus.com/reference/set/set/