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### YouKn0wWho's blog

By YouKn0wWho, history, 20 months ago, ,

I am stuck on this following problem:

#### Statement:

Given a string of size n and q queries of type (l,r) you need to find how many palindromes are there in the substring of range [l...r].

#### Constraints:

1<=n,q<=100000

1<=l<=r<=n

Thanks for reading the problem. It will be really helpful if you can give me a solution.

• +15

 » 20 months ago, # | ← Rev. 2 →   0 Mistake in approach.
•  » » 20 months ago, # ^ |   +6 Thanks for your reply.But I am not clear about the merging procedure. Can you please tell me more about this part "number of palindromes between this parts (left end in first half and right in second)."
•  » » 20 months ago, # ^ | ← Rev. 2 →   0 why even O((r - l) * logn)? You can preprocess string using Manachar's algorithm, and then for each i in [l, r], you can find the number of palindromes in between the parts, centered at i in O(1), so it's just O(r - l).but I don't understand, how it will be O(nlog 2 n) in total, because you will have to merge even for queries. Can you please elaborate.
•  » » » 20 months ago, # ^ | ← Rev. 2 →   +2 [deleted]
•  » » » » 20 months ago, # ^ |   +3 We know that there can be at most n palindromes in a string of size nNo there can be n 2 palindromes, for example consider aaaaaa.so that won't work :|
•  » » » » » 20 months ago, # ^ |   0 Sorry, you are right.
•  » » » 20 months ago, # ^ | ← Rev. 3 →   0 Mistake in approach.
•  » » » » 20 months ago, # ^ |   +8 "each interval will be smaller in at most logn merges"No, you may even have to merge already merged intervals, with a simple interval. So that won't be O(nlogn)
•  » » » » » 20 months ago, # ^ |   0 You are absolutely right! I will try to invent something better and correct :)
 » 20 months ago, # | ← Rev. 2 →   0 Sqrt-decomposition solution is costly but may pass, depending on TL.Whenever you add/remove some index, you have to add/remove palindromic substrings ending at that index, which you can find using binary search (hash the string and it's reverse).Total complexity will be
•  » » 20 months ago, # ^ |   0 How will you find palindromic substrings ending at some index by binary search?
•  » » » 20 months ago, # ^ |   0 HashingHash the given string and it's reverse, Now you can binary search on the maximum length that can be a palindrome ending at that index (in the direction you want). To check if the substring of given length is palindrome, you just compare hash of string with it's reverse, for that substring.
•  » » » » 20 months ago, # ^ | ← Rev. 2 →   0 That is not correct, you can do it for center of string, otherwise function is not monotonic and you can not use binary search: Just look at string aba, there is two palindromes and binary search will check first ba and say, ok it is not palindrome answer is smaller than 2.
•  » » » » » 20 months ago, # ^ |   0 oh ok.. missed it :(
•  » » » » » » 20 months ago, # ^ |   +3 1:1 in hacks, we are playing until 2 ;)
 » 20 months ago, # |   +20 It is easier to explain by considering only palindromes centered at indicies (so, odd length), the idea is the same anyway. For each index i, r i will be the longest radius of a palindrome centered there (in other words, the amount of palindromes centered at index i). Directly from manacher, this takes to calculate.For a query [l, r], we first compute . Now we want to calculateAgain, thanks to symmetry, I'll only explain how to compute the left term..The sum over i can be found in constant time. As for the other term, if we create some array r i' = r i - i during the preprocessing, then the queries are asking for some over range of min(C, r i') where C is constant. You can solve this in per query using wavelet tree, and I guess it's also solvable with a persistent segtree.
•  » » 20 months ago, # ^ | ← Rev. 2 →   +5 Thanks for your reply. This is an interesting approach.Well, I am interested to know about this idea- min(i-l+1,r[i])=i+min(1-l,r[i]-i).I have brute forced for different i,l and r[i] and found that this is true. But curious mind wants to know the proof of this. If you can come up with a proof feel free to share this with me.Thanks again.
•  » » » 20 months ago, # ^ |   +3 min(i - l + 1, r i) = min(i + (1 - l), i + (r i - i)) = i + min(1 - l, r i - i)
•  » » » 20 months ago, # ^ |   +3 min(x + a, y + a) = a + min(x, y). I think this is fairly obvious, but you can do case analysis anyway: x < y: min(x + a, y + a) = x + a = a + min(x, y). x ≥ y: min(x + a, y + a) = y + a = a + min(x, y). Now just plug a = i, x = 1 - l, y = r i - i
•  » » » » 20 months ago, # ^ |   0 thanks. got it
 » 20 months ago, # | ← Rev. 2 →   0 Palindrome tree can tall you, how many palindrome add s[i]. You can create partial sum array and say on query O(1).Sorry for my English.
 » 5 weeks ago, # |   +14 Please provide problem link.I want to solve it.
•  » » 5 weeks ago, # ^ |   +4
 » 5 weeks ago, # |   0 My approach takes O(n ^ 2) precalculation to get O(1) queries.Can I improve furthur more with dynamic programming ? My codevoid solve() { string s; cin >> s; vector > precal(2, vector(n, +INF)); /// precal0[x] - min position (p) | (0 + p)..(x)_(x + 1)..(x + p) is palindrome /// precal1[x] - min position (p) | (0 + p)......(x)......(x + p) is palindrome const int lower = -1, upper = n; for (int i = 0; i < n; ++i) /// Precalculation O(n ^ 2) { for (int l = i, r = i + 0; lower < l && r < upper; --l, ++r) if (s[l] == s[r]) precal[1][i] = l; else break; for (int l = i, r = i + 1; lower < l && r < upper; --l, ++r) if (s[l] == s[r]) precal[0][i] = l; else break; } for (int q = readInt(); q--; ) /// O(q) | O(1) per query { int l = readInt() - 1; int r = readInt() - 1; bool type = (r - l + 1) & 1; /// Odd or Even Size palindrome int m = (l + r) / 2; /// Middle of palindrome cout << (precal[type][m] <= l ? 1 : -1) << '\n'; } }