You are given a 0-indexed string s
that has lowercase English letters in its even indices and digits in its odd indices.
There is a function shift(c, x)
, where c
is a character and x
is a digit, that returns the xth
character after c
.
- For example,
shift('a', 5) = 'f'
andshift('x', 0) = 'x'
.
For every odd index i
, you want to replace the digit s[i]
with shift(s[i-1], s[i])
.
Return s
after replacing all digits. It is guaranteed that shift(s[i-1], s[i])
will never exceed 'z'
.
Example 1:
Input: s = "a1c1e1"
Output: "abcdef"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('c',1) = 'd'
- s[5] -> shift('e',1) = 'f'
Example 2:
Input: s = "a1b2c3d4e"
Output: "abbdcfdhe"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('b',2) = 'd'
- s[5] -> shift('c',3) = 'f'
- s[7] -> shift('d',4) = 'h'
Constraints:
-
1 <= s.length <= 100
-
s
consists only of lowercase English letters and digits. -
shift(s[i-1], s[i]) <= 'z'
for all odd indicesi
.
SOLUTION:
class Solution:
def replaceDigits(self, s: str) -> str:
return "".join([chr(ord(s[i - 1]) + int(c)) if i & 1 else c for i, c in enumerate(s)])
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