Can I say my solution is in O(1) space if I use a hash array of fixed size say of size 52(for lowercase and uppercase alphabets) to count the number of each alphabet that is present in a string?
# | 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 |
Can I say my solution is in O(1) space if I use a hash array of fixed size say of size 52(for lowercase and uppercase alphabets) to count the number of each alphabet that is present in a string?
Name |
---|
If you assume that the alphabet size is constant, then the space requirement is O(1). However, if a general case for any alphabet is considered, then it is O(m) where m is the number of characters.