radoslav11's blog

RMI 2014 — Minxor
By radoslav11, history, 9 years ago, In English

Hello!

I was solving problem Minxor from second day of RMI 2014. In short the problem is:

You have 3 types of queries.

1 x -> inserts x into the set
2 x -> deletes x from the set
3 -> asks for the smallest xor of any two numbers in the set

I solved the problem offline using a trie + divide and conquer. My solution is similar to link. Again in short: compressing all updates as "number x appears from time I to time J". Here is the code. The complexity is O(M * log(M) * 32);

So my question is:

Is there a solution which runs faster than that or/is online. Thanks in advance :)

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9 years ago, # |
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Yes, there's a simpler, online, O(log M) per query solution. Notice that if a < b < c, the minimum xor pair is never (a, c).

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    9 years ago, # ^ |
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    Can you explain a little bit more the idea? How we get that (log M) per query ?

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      9 years ago, # ^ |
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      Using the observation above, you can prove that members of the minimum xor pair must be adjacent in sorted order. So you only need to keep a simple BST (such as a STL set<>) to keep track of successors and predecessors of inserted/deleted elements and another min-heap-like structure in which you actually store the adjacent xor values in increasing order.

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    9 years ago, # ^ |
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    Thats a nice trick! Thanks

    BTW is there a similar trick but for maximum xor?

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