Given a string s formed by digits (‘0’ – ‘9’) and ‘#’ . We want to map s to English lowercase characters as follows:
- Characters (‘a’ to ‘i’) are represented by (‘1’ to ‘9’) respectively.
- Characters (‘j’ to ‘z’) are represented by (’10#’ to ’26#’) respectively.
Return the string formed after mapping.
It’s guaranteed that a unique mapping will always exist.
Example 1:
Input: s = "10#11#12"
Output: "jkab"
Explanation: "j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".
Example 2:
Input: s = "1326#"
Output: "acz"
Example 3:
Input: s = "25#"
Output: "y"
Example 4:
Input: s = "12345678910#11#12#13#14#15#16#17#18#19#20#21#22#23#24#25#26#"
Output: "abcdefghijklmnopqrstuvwxyz"
Constraints:
- 1 <= s.length <= 1000
- s[i] only contains digits letters (‘0’-‘9’) and ‘#’ letter.
- s will be valid string such that mapping is always possible.
Solution in python:
class Solution:
def freqAlphabets(self, s: str) -> str:
alist = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '10#', '11#',
'1']
i = len(s)-1
result = []
while i >= 0:
if s[i] != '#':
result.append(chr(int(s[i])+96))
i -= 1
else:
result.append(chr(int(s[i-2:i])+96))
i -= 3
return ''.join(result[::-1])
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