c++ Remove the duplicates in sorted array - print the new array

Question

I am testing out this piece of code in c++

#include <iostream>

using namespace std;

int removeDuplicates(int A[], int n) {
    if(n == 0) return 0;

    int index = 0;

    for(int i=0; i<n; i++)
    {
        if(A[index] == A[i])
        {

            continue;
        }
        index++;
        A[index] = A[i];
    }
    cout<<"New array is "<<A[index]<<endl;
    return index+1;
}


int main(){

    int x[10] = {1,1,6};
    int n = 3;
    int z = removeDuplicates(x, n);
    cout<<"Length is "<<z;
}

The result of length is right. However, when cout the new array. The result is only 6. 1 is missing.

Where should I put cout<<"New array is "<<A[index]<<endl;

To give me the result as New array is 16

More code I need to add? Please help, anyone?

cout<<"New array is "<<A[index]<<endl; outputs the element at position index of A. — Jonny Henly
– Jonny Henly, Commented Feb 27, 2015 at 22:16
Use a for loop to print contents of array. You can't directly print a whole array. — hyde
– hyde, Commented Feb 27, 2015 at 22:21

Cheers and hth. - Alf · Accepted Answer · 2015-02-27 22:14:41Z

4

std::unique and std::erase are your friends.

std::vector is also a good friend.

By the way, your program does remove duplicates correctly; it's just the output statement that doesn't show more of the array than the single value 6.

answered Feb 27, 2015 at 22:14

Cheers and hth. - Alf

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Comments

miachm · Accepted Answer · 2015-02-27 22:31:17Z

for(int i=0; i<n; i++)
{
    if(A[index] == A[i])
    {

        continue;
    }
    index++;
    A[index] = A[i];
}

Let's go trace: A = {1,1,6}; n = 3;

1º i = 0; index = 0 -> a[0] == a[0]? -> Continue

2º i = 1; index = 0 -> a[0] == a[1]? -> Continue

3º i = 2; index = 0 -> a[0] == a[2]? -> Nope, so index++ -> a[1] = a[2];

The new array is: {1,6,6}. index has the value of = 1, so you print:

cout<<"New array is "< 6

You return index = 2 and A = {1,6,6}. This is a correct result, your algorithm works, but the print doesn't have sense. You should print with for in the return:

#include <iostream>

using namespace std;

int removeDuplicates(int A[], int n)
{
    if(n == 0) return 0;

    int index = 0;

    for(int i=0; i<n; i++)
    {
        if(A[index] == A[i])
        {

            continue;
        }
        index++;
        A[index] = A[i];
    }
    return index+1;
}


int main()
{

    int x[10] = {1,1,6};
    int n = 3;
    int size = removeDuplicates(x, n);
    cout<<"Length is "<<size<<endl;

    for (int i = 0; i<size; i++)
    {
        cout<<x[i]<<' ';
    }
}

This works great :). But, you have one function of standar library for this:

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{

    int x[10] = {1,1,6};
    int n = 3;
    int* end = unique(x,x+n); // returns reference to the end

    for (int i = 0;(x+i) != end; i++)
    {
        cout<<x[i]<<' ';
    }
}

You can check reference of unique here: http://www.cplusplus.com/reference/algorithm/unique/

0x6773 · Accepted Answer · 2015-02-28 00:45:49Z

0

Instead of using std::unique and std::erase, do it in one line using std::unique_copy and std::ostream_iterator<>. Let me do that:

unique_copy(begin(ar),end(ar),ostream_iterator<T>(cout,"\n"));

where ar is any STL sequence container, and T is its value_type

answered Feb 28, 2015 at 0:45

0x6773

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