### Mid0's blog

By Mid0, history, 6 years ago,

i'm trying to solve this problem but i can't find any algorithm better than O(n^2) !

your help will be greatly appreciated.

• 0

 » 6 years ago, # |   0 Auto comment: topic has been updated by Mid0 (previous revision, new revision, compare).
 » 6 years ago, # | ← Rev. 2 →   +1 Consider some substring S[i...j]. Let parity[K] for 0<=K<=9 be the parity of the number of occurrences of digit K in subtring S[i...j]. Notice that the substring will be a palindrome if and only if parity[K] is 0 for all K or there is only one K for which parity[K] is 1 and parity[K] is 0 for all other Ks.Having that in mind, we will loop from 1 to N and at each step we will find the number of those palindromes ending at the current position i. In order to do that, let parity[K] be the described parity for digit K in substring S[1...i]. Consider some j (1<=j<=i). Substring S[j...i] can be made a palindrome if all parities for position j-1 are the same as the parities for position i or there is only one difference.For each position, we can store the parities as a bitmask with 10 bits and that way we can find the answer quickly. Check my code for more details: http://ideone.com/1grO2A :)
•  » » 6 years ago, # ^ |   0 got it , nice usage of bitwise , thank you :)