Using a hash to find one duplicated and one missing number in an array

I had this question during an interview and am curious to see how it would be implemented.

Given an unsorted array of integers from 0 to x. One number is missing and one is duplicated. Find those numbers.

Here is what I came up with:

int counts[x+1];

for(int i =0;i<=x; i++){
    counts[a[i]]++;
    if(counts[a[i]] == 2)
        cout<<”Duplicate element: “<<a[i];  //I realized I could find this here
}

for(int j=0; j<=x; j++){
    if(counts[j] == 0)
        cout<<”Missing element: “<<j; 
    //if(counts[j] == 2)
    //  cout<<”Duplicate element: “<<j;   //No longer needed here.
}

My initial solution was to create another array of size x+1, loop through the given array and index into my array at the values of the given array and increment. If after the increment any value in my array is two, that is the duplicate. However, I then had to loop through my array again to find any value that was 0 for the missing number.

I pointed out that this might not be the most time efficient solution, but wasn't sure how to speed it up when I was asked. I realized I could move finding the duplicate into the first loop, but that didn't help with the missing number. After waffling for a bit, the interviewer finally gave me the idea that a hash would be a better/faster solution. I have not worked with hashes much, so I wasn't sure how to implement that. Can someone enlighten me? Also, feel free to point out any other glaring errors in my code... Thanks in advance!

If x were small enough (such that the sum of 0..x can be represented), you could compute the sum of the unique values in a, and subtract that from the sum of 0..x, to get the missing value, without needing the second loop.