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
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3831 |
3 | Radewoosh | 3646 |
4 | jqdai0815 | 3620 |
4 | Benq | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | gamegame | 3386 |
10 | ksun48 | 3373 |
# | User | Contrib. |
---|---|---|
1 | cry | 164 |
1 | maomao90 | 164 |
3 | Um_nik | 163 |
4 | atcoder_official | 160 |
5 | -is-this-fft- | 158 |
6 | awoo | 157 |
7 | adamant | 156 |
8 | TheScrasse | 154 |
8 | nor | 154 |
10 | Dominater069 | 153 |
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
Name |
---|
l
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.
Why are people downvoting this? Its a valid (if not the best) solution given a hashmap with O(1) update/query
In fact, there is a problem the same as this one: link.
In fact, there is a problem the same as this one: link
In fact, I don't know a problem the same as this one: No Link
I really feel like Persistent segtree would be able to count suitable L for fixed R. But can't put my finger on how.