Hi, everyone.
Can someone, please, explain the solution for this problem ?
I really cannot understand what editorial wants to say.# | User | Rating |
---|---|---|
1 | ecnerwala | 3649 |
2 | Benq | 3581 |
3 | orzdevinwang | 3570 |
4 | Geothermal | 3569 |
4 | cnnfls_csy | 3569 |
6 | tourist | 3565 |
7 | maroonrk | 3531 |
8 | Radewoosh | 3521 |
9 | Um_nik | 3482 |
10 | jiangly | 3468 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 162 |
4 | TheScrasse | 159 |
5 | nor | 158 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 151 |
8 | SecondThread | 147 |
9 | orz | 146 |
10 | pajenegod | 145 |
Hi, everyone.
Can someone, please, explain the solution for this problem ?
I really cannot understand what editorial wants to say.Name |
---|
Auto comment: topic has been translated by TigranHakobyan(original revision, translated revision, compare)
Sorry that I can't find editorial. I think you can sort all cars alphabetically and use DP to find maximum length. You have to sort cars to find first-alphabetical one easily.
If you mean just sort a alphabetically, then if we consider {acbd, abcd, bda}. Using your idea we will get answer 2 but the answer is 3, isn't it ?
P.S. editorial
Probelm says, "The front end is the end with the letter that comes first in the alphabet", so you have to flip car "bda" into "adb".
so in case {"acbd", "abcd", "bda"}, the answer is 1.
and if the car A and car B can be linked, A.front <= A.back == B.front <= B.back, so sort like this.
ardiankp's code
sorry about my English — I'm not good at it.
Thank you, so much. I misunderstood the problem.