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.
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
# | User | Contrib. |
---|---|---|
1 | cry | 167 |
2 | maomao90 | 163 |
2 | Um_nik | 163 |
4 | atcoder_official | 161 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
9 | nor | 153 |
9 | Dominater069 | 153 |
Name |
---|
Auto comment: topic has been updated by Mid0 (previous revision, new revision, compare).
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 :)
got it , nice usage of bitwise , thank you :)