Leetcode - Decrypt String from Alphabet to Integer Mapping Solution
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
from collections import defaultdict,deque
class Solution:
def freqAlphabets(self, s: str) -> str:
index = -1
arr = deque()
while (len(s)+index)>-1:
if s[index]=="#":
arr.appendleft(s[index-2:index+1 or len(s)])
index-=3
else:
arr.appendleft(s[index])
index-=1
hash_map = defaultdict(str)
for x,y in zip(range(1,10),range(ord("a"), ord("i")+1)):
hash_map[str(x)]= chr(y)
for x,y in zip(range(10,27),range(ord("j"), ord("z")+1)):
hash_map[str(x)+"#"]= chr(y)
return "".join((map(lambda x : hash_map[x], arr)))