**Problem: **Given a massive data stream of $ n$ values in $ \{ 1, 2, \dots, m \}$ and the guarantee that one value occurs more than $ n/2$ times in the stream, determine exactly which value does so.

**Solution: **(in Python)

def majority(stream): held = next(stream) counter = 1 for item in stream: if item == held: counter += 1 elif counter == 0: held = item counter = 1 else: counter -= 1 return held

*Discussion: *Let’s prove correctness. Say that $ s$ is the unknown value that occurs more than $ n/2$ times. The idea of the algorithm is that if you could pair up elements of your stream so that *distinct *values are paired up, and then you “kill” these pairs, then $ s$ will always survive. The way this algorithm pairs up the values is by holding onto the most recent value that has no pair (implicitly, by keeping a count how many copies of that value you saw). Then when you come across a new element, you decrement the counter and implicitly account for one new pair.

Let’s analyze the complexity of the algorithm. Clearly the algorithm only uses a single pass through the data. Next, if the stream has size $ n$, then this algorithm uses $ O(\log(n) + \log(m))$ space. Indeed, if the stream entirely consists of a single value (say, a stream of all 1’s) then the counter will be $ n$ at the end, which takes $ \log(n)$ bits to store. On the other hand, if there are $ m$ possible values then storing the largest requires $ \log(m)$ bits.

Finally, the guarantee that one value occurs more than $ n/2$ times is necessary. If it is not the case the algorithm could output anything (including the most *infrequent* element!). And moreover, if we don’t have this guarantee then *every algorithm *that solves the problem must use at least $ \Omega(n)$ space in the worst case. In particular, say that $ m=n$, and the first $ n/2$ items are all distinct and the last $ n/2$ items are all the same one, the majority value $ s$. If you do not know $ s$ in advance, then you must keep at least one bit of information to know which symbols occurred in the first half of the stream because any of them could be $ s$. So the guarantee allows us to bypass that barrier.

This algorithm can be generalized to detect $ k$ items with frequency above some threshold $ n/(k+1)$ using space $ O(k \log n)$. The idea is to keep $ k$ counters instead of one, adding new elements when any counter is zero. When you see an element not being tracked by your $ k$ counters (which are all positive), you decrement *all* the counters by 1. This is like a $ k$-to-one matching rather than a pairing.