Hello:)
Now I'll say what is Sparse Table and I'll show you some problems that I found.
Sparse Table is a data structure, that allows answering range queries. It can answer range minimum queries (or equivalent range maximum queries) in O(1) time, and another queries in O(log n).
You can read how to use it here: GeeksForGeeks, CP-Algorithms
Problems:
Codeforces:
872B - Максимум максимума из минимумов //difficulty 1200 but with sparse table harder
5C - Наибольшая правильная скобочная подстрока //difficulty 1900
475D - CGCDSSQ //difficulty 2000
863E - Выключение телевизора //difficulty 2000
514D - R2-D2 и армия дроидов //difficulty 2100
675E - Электрички и статистика //difficulty 2500
15D - Карта //difficulty 2500
873E - Награждение победителей //difficulty 2500
713D - Зверята и пазл //difficulty 2700
SPOJ:
SPOJ — RMQSQ //easy
Everything you can solve by sparse table.
Hey, Can you tell how can sparse table be used to solve "5C — Longest Regular Bracket Sequence"?Thanks.
Link
This is a video Explained by CodeNcode
Thx for early reply and helpful video (sol with dp) and blog talking about sparse table ''/
really is a thing 3zim.
'_'
первая задача из кф'а не имеет абсолютно никакого отношения к sparse table, Смысл писать спарс если надо найти один раз максимум и минимум на всем массиве.
ASCII4 Can you say how to approach this problem, that you have mentioned: 5C — Longest Regular Bracket Sequence using sparse table ? Thank you.
Treat '(' as +1 and ')' as -1 now build a prefix array.
Now let's say you are at position i so you will try to find some index j having the value same as pre[i] to make substring from i + 1 to j as a regular bracket sequence. Now there is one more condition any value of prefix array between the index i + 1 to j should be greater than or equal to pre[i] so to check that in o(1) we use a sparse table. Let say you have the bracket sequence
(())())( its prefix array will be
( ( ) ) ( ) ) (
0 1 2 1 0 1 0 -1 0
0 1 2 3 4 5 6 7 8
So if I am on position 0 means that I am considering that my substring is starting with position 1 (1 based indexing) now I will find the index of all zeros present in front of 0 which has the value zero so these indexes are 4, 6 and 8 now if I will check all of those starting from 8 that which one is valid and the longest off course here valid means that none of the value between those indices should be smaller than pre[0] means 0 so here 8 is not valid because on position 7 there is -1 present and we can check it easily with space table that whats the minimum value between those indices which is less than 0 but 6 is the valid one and off course, the longest then for every index i have to iterate for about o(n) times which will lead to o(n^2) but if you observe carefully you see it completely monotonic because if 6 is valid then 4 is definitely valid so we can get the valid string in log(n) for each index by binary searching that which one is the valid one so overall complexity will be o(nlog(n))
I tried this approach o(nlogn) but i am getting tle.. can you please guide me how should i optimize? thanks!! submission
Hi, It would help if you wrote:
vll &tobeSearch = f[prefix[i]];notvll tobeSearch = f[prefix[i]];It passes all the test cases you can see in this submission 145175368. And the complexity of your code is n * log(n) * log(n), so it's very slow.
One more recent question you can try is 1547F - Стабилизируй массив (НОД-версия)
This one too , it's from #769 Div 2
900E — easiest sparse table problem