### Sammmmmmm's blog

By Sammmmmmm, 2 weeks ago,

 Given an array of N integers. Count the number of subarrays so that every value that appears in the subarray appears an odd number of times. I think this can be solved using XOR hashing but do not know how. Can someone help? Thanks!

 N, Ai <= 1e5

• +38

 » 2 weeks ago, # | ← Rev. 2 →   0 l
 » 2 weeks ago, # |   +18 I think there's a O(n * sqrt(n) * log(n)) solution. Let p[i] be xor hash on prefix ending at i. Then the condition for [l, r] can be rewritten as p[r] ^ p[l — 1] = (xor of hashes of numbers that appear at least once in [l, r]) or in other words p[r] = p[l — 1] ^ (xor of ... in [l, r]). Create another array t of length n. Initially t[i] = p[i — 1]. We will move r and support this array. Let last[i] be the right most position of i (it will change as we move r). Then i will appear in [l, r] iff l <= last[i]. In transition (r — 1) -> r : last[a[r]] will change to r, so we have to xor t[i] with hash(a[r]) for 1 <= i <= last[a[r]], change last[a[r]] to r and xor t[i] with hash(a[r]) for 1 <= i <= r. After this we need to know the number of 1 <= i <= r such that t[i] = p[r]. So we need a DS that supports 2 operations: 1) Xor of Prefix. 2) Count on Prefix. This can be done with sqrt-decomposition by supporting how much we've xored each block with (lazily) and how many times each number appears in a block.
•  » » 2 weeks ago, # ^ |   +8 Why are people downvoting this? Its a valid (if not the best) solution given a hashmap with O(1) update/query
 » 2 weeks ago, # |   0 In fact, there is a problem the same as this one: link.
 » 2 weeks ago, # |   +2 In fact, there is a problem the same as this one: link
 » 2 weeks ago, # |   +9 In fact, I don't know a problem the same as this one: No Link
 » 2 weeks ago, # |   0 I really feel like Persistent segtree would be able to count suitable L for fixed R. But can't put my finger on how.