Revision en8, by humbertoyusta, 2020-12-24 19:31:06

This is a well known technique, but I couldn't find almost any detailed tutorial, so I'll try to help with this one.

Motivational Problem:

You have an array $$$A$$$ and $$$Q$$$ queries, which are, say if the subarray from $$$l$$$ to $$$r$$$, ($$$A[l]$$$, $$$A[l+1]$$$, $$$...$$$, $$$A[r]$$$) forms a permutation.

Some basic observations:

Only the set of elements matters, the order is irrelevant.

You want to know if each element is exactly once.


Assign a random number $$$randval[i]$$$ to each number $$$i$$$, now we can say that if the XOR of the subarray ( $$$randval[A[l]]$$$ xor $$$randval[A[l+1]]$$$ xor $$$...$$$ xor $$$randval[A[r]]$$$ ) is equal to the xor of any permutation of size $$$r - l + 1$$$ (for example $$$1$$$, $$$2$$$, $$$3$$$, ... ,$$$r-l+1$$$, which hash is $$$randval[1]$$$ xor $$$randval[2]$$$ xor $$$...$$$ xor $$$randval[r-l+1]$$$ ) the subarray is a permutation.

Some Insights:

With the XOR we have some issues, we are counting if the number of occurrences of each element is even or odd, not the real number of occurrences, however in this problem if we had repeated elements the XOR-hash will never be the same than the XOR-hash of $$$1$$$, $$$2$$$, $$$3$$$, ... ,$$$r-l+1$$$, because if we have some numbers more than once, since there are only $$$r - l + 1$$$ numbers in the subarray, and also $$$r - l + 1$$$ numbers are needed to form the permutation, some of them cannot appear.

As an alternative to this technique we can use polynomial hash over a binary string that represents the occurrences of each element modulo $$$2$$$ ($$$x$$$-character of this string represents the number of occurrences of $$$x$$$ modulo $$$2$$$), but with XOR hash we can do it faster, with less code and case handling, and with less care about collisions and hacks.

With bitwise-OR we count the number of occurrences of each element of the set modulo $$$2$$$, if we use $$$k$$$-nary xor we will count the number of occurrences modulo $$$k$$$.


Now it's time to talk about collisions, as the numbers are random, the probability of collision in each query if we had only $$$1$$$ bit will be about $$$1/2$$$. And since XOR is independent on each bit, if we use k bits, the probability of collision in each query is $$$2^{-k}$$$. We can use 32-bit random integers or 64-bit random integers depending on the number of queries required. Since the numbers are generated randomly, I think this hashing can't be hacked, but I am not sure, if someone knows more about this, please send some feedback.

More problems using this technique (in no particular order of difficulty):

F. The Number of Subpermutations from Educational Codeforces Round 66 Here editorial doesn't use it, but I have one solution with this technique.

Quick explanation (spoiler)

E. The Untended Antiquity from Codeforces Round #439 Div 2

G. Three Occurrences from Educational Codeforces Round 95

C. Square Subsets from Codeforces Round #448 Div 2 but if we had just 40 numbers up to 10^6.

Quick explanation (spoiler)

Feel free to send any feedback.

Tags #hashing, #xor, #tutorial


  Rev. Lang. By When Δ Comment
en8 English humbertoyusta 2020-12-24 19:31:06 674 I added a brief solution to the fourth problem which uses this technique
en7 English humbertoyusta 2020-12-24 19:20:11 1127 I added a brief solution to the first problem which uses this technique
en6 English humbertoyusta 2020-12-24 07:58:51 0 (published)
en5 English humbertoyusta 2020-12-24 07:55:43 765 Reverted to en3
en4 English humbertoyusta 2020-12-24 07:54:03 765 Tiny change: ' is like\n\nf(l,r)\n\n max ' -> ' is like\nf(l,r)\n max ' (saved to drafts)
en3 English humbertoyusta 2020-12-24 07:16:54 0 (published)
en2 English humbertoyusta 2020-12-24 07:15:58 228 (saved to drafts)
en1 English humbertoyusta 2020-12-24 06:45:32 2734 Initial revision (published)