Easy interview question got harder: given numbers 1..100, find the missing number(s) given exactly k are missing

Question

I had an interesting job interview experience a while back. The question started really easy:

Q1: We have a bag containing numbers

Answer 1

I think this can be done without any complex mathematical equations and theories. Below is a proposal for an in place and O(2n) time complexity solution:

Input form assumptions :

# of numbers in bag = n

# of missing numbers = k

The numbers in the bag are represented by an array of length n

Length of input array for the algo = n

Missing entries in the array (numbers taken out of the bag) are replaced by the value of the first element in the array.

Eg. Initially bag looks like [2,9,3,7,8,6,4,5,1,10]. If 4 is taken out, value of 4 will become 2 (the first element of the array). Therefore after taking 4 out the bag will look like [2,9,3,7,8,6,2,5,1,10]

The key to this solution is to tag the INDEX of a visited number by negating the value at that INDEX as the array is traversed.

    IEnumerable GetMissingNumbers(int[] arrayOfNumbers)
    {
        List missingNumbers = new List();
        int arrayLength = arrayOfNumbers.Length;

        //First Pass
        for (int i = 0; i < arrayLength; i++)
        {
            int index = Math.Abs(arrayOfNumbers[i]) - 1;
            if (index > -1)
            {
                arrayOfNumbers[index] = Math.Abs(arrayOfNumbers[index]) * -1; //Marking the visited indexes
            }
        }

        //Second Pass to get missing numbers
        for (int i = 0; i < arrayLength; i++)
        {                
            //If this index is unvisited, means this is a missing number
            if (arrayOfNumbers[i] > 0)
            {
                missingNumbers.Add(i + 1);
            }
        }

        return missingNumbers;
    }