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 have read all thirty answers and found the simplest one i.e to use a bit array of 100 to be the best. But as the question said we can't use an array of size N, I would use O(1) space complexity and k iterations i.e O(NK) time complexity to solve this.

To make the explanation simpler, consider I have been given numbers from 1 to 15 and two of them are missing i.e 9 and 14 but I don't know. Let the bag look like this:

[8,1,2,12,4,7,5,10,11,13,15,3,6].

We know that each number is represented internally in the form of bits. For numbers till 16 we only need 4 bits. For numbers till 10^9, we will need 32 bits. But let's focus on 4 bits and then later we can generalize it.

Now, assume if we had all the numbers from 1 to 15, then internally, we would have numbers like this (if we had them ordered):

0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

But now we have two numbers missing. So our representation will look something like this (shown ordered for understanding but can be in any order):

(2MSD|2LSD)
00|01
00|10
00|11
-----
01|00
01|01
01|10
01|11
-----
10|00
missing=(10|01) 
10|10
10|11
-----
11|00
11|01
missing=(11|10)
11|11

Now let's make a bit array of size 2 that holds the count of numbers with corresponding 2 most significant digits. i.e

= [__,__,__,__] 
   00,01,10,11

Scan the bag from left and right and fill the above array such that each of bin of bit array contains the count of numbers. The result will be as under:

= [ 3, 4, 3, 3] 
   00,01,10,11

If all the numbers would have been present, it would have looked like this:

= [ 3, 4, 4, 4] 
   00,01,10,11

Thus we know that there are two numbers missing: one whose most 2 significant digits are 10 and one whose most 2 significant bits are 11. Now scan the list again and fill out a bit array of size 2 for the lower 2 significant digits. This time, only consider elements whose most 2 significant digits are 10. We will have the bit array as:

= [ 1, 0, 1, 1] 
   00,01,10,11

If all numbers of MSD=10 were present, we would have 1 in all the bins but now we see that one is missing. Thus we have the number whose MSD=10 and LSD=01 is missing which is 1001 i.e 9.

Similarly, if we scan again but consider only elements whose MSD=11,we get MSD=11 and LSD=10 missing which is 1110 i.e 14.

= [ 1, 0, 1, 1] 
   00,01,10,11

Thus, we can find the missing numbers in a constant amount of space. We can generalize this for 100, 1000 or 10^9 or any set of numbers.

References: Problem 1.6 in http://users.ece.utexas.edu/~adnan/afi-samples-new.pdf