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

Yet another way is using residual graph filtering.

Suppose we have numbers 1 to 4 and 3 is missing. The binary representation is the following,

1 = 001b, 2 = 010b, 3 = 011b, 4 = 100b

And I can create a flow-graph like the following.

                   1
             1 -------------> 1
             |                | 
      2      |     1          |
0 ---------> 1 ----------> 0  |
|                          |  |
|     1            1       |  |
0 ---------> 0 ----------> 0  |
             |                |
      1      |      1         |
1 ---------> 0 -------------> 1

Note that the flow graph contains x nodes, while x being the number of bits. And the maximum number of edges are (2*x)-2 .

So for 32 bit integer it will take O(32) space or O(1) space.

Now if I remove capacity for each number starting from 1,2,4 then I am left with a residual graph.

0 ----------> 1 ---------> 1

Finally I shall run a loop like the following,

 result = []
 for x in range(1,n):
     exists_path_in_residual_graph(x)
     result.append(x)

Now the result is in result contains numbers that are not missing as well(false positive). But the k <= (size of the result) <= n when there are k missing elements.

I shall go through the given list one last time to mark the result missing or not.

So the time complexity will be O(n) .

Finally, it is possible to reduce the number of false positive(and the space required) by taking nodes 00,01,11,10 instead of just 0 and 1.