Most common element in an array / Finding the relative majority, deterministically in O(n) time and O(1) space?

Question

So for example, the answer for the array:

1, 11, 3, 95, 23, 8, 1

would be 1, since all the other elements only occur once while 1 occurs twice.

A lo

Answer 1

If you want to have fixed space to find the most common element you need to have a maximum number of bits for an element. If you didn't, then large input arrays could have larger input numbers such that the bits to represent the number is bigger than your fixed space to store the result.

Suppose k is the length of the largest number you support. If you try to naively create an array of 2^k buckets to count occurrences of each number (counter sort) you could receive an array consisting of the same number, in which case your algorithm would end up needing log(n) space to store the sum.[*]

If we look at a simpler version of the problem - determine whether or not there are more 1's or 0's in an input, I think you need a stack to do this (you store how much 1 or 0 is leading by), and so constant space just isn't possible, even if we limit the input length to k = 1 bit in size.

Your problem is more general (k > 1, but still fixed), and would also need non-constant space, so it's not possible, as the question is worded.

[*] If you assume counters have O(1) space complexity, then you can take the counter sort approach, although by doing so you've placed an upper-bound on the maximum size of your input array (which may or may not be acceptable): In terms of k, the maximum number of bits for an input element of your array and in terms of c the maximum number of bits in your counter your array can have at most 2^k * 2^c elements (one of the counters would overflow otherwise on the next element). To address this, you could add a O(1) time step to decrement your counters so that the minimum value is always 0 after each element is processed if all counters are non-0, thereby making them relative instead of absolute. This takes O(1) time because if all are non-zero you only need to decrement O(2^k) = O(1) counters by 1 if you perform it on each element. While the algorithm can now process some arbitrarily large inputs, any input array that has a sub-array such that two values a and b are such that count(a) - count(b) > 2^c = max(counter) using a counter strategy will fail for some inputs. In fact a consequence of relying on a O(1) space complexity counter approach is that all arrays that start with 2^c + 1 identical elements cannot be handled by this algorithm.