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
- For example,
shift('a', 5) = 'f'and
shift('x', 0) = 'x'.
For every odd index
i, you want to replace the digit
s after replacing all digits. It is guaranteed that
shift(s[i-1], s[i]) will never exceed
Input: s = "a1c1e1" Output: "abcdef" Explanation: The digits are replaced as follows: - s -> shift('a',1) = 'b' - s -> shift('c',1) = 'd' - s -> shift('e',1) = 'f'
Input: s = "a1b2c3d4e" Output: "abbdcfdhe" Explanation: The digits are replaced as follows: - s -> shift('a',1) = 'b' - s -> shift('b',2) = 'd' - s -> shift('c',3) = 'f' - s -> shift('d',4) = 'h'
1 <= s.length <= 100
sconsists only of lowercase English letters and digits.
shift(s[i-1], s[i]) <= 'z'for all odd indices
Iterate the string, if at odd position, keep the original character, otherwise at even position, base on the value to replace with a new character.
class Solution: def replaceDigits(self, s: str) -> str: result = "" for i, c in enumerate(s): if i % 2 == 0: result += c else: result += chr(ord(s[i - 1]) + int(c)) return result
- Time Complexity: O(N).
- Space Complexity: O(N).