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 <= 100sconsists only of lowercase English letters and digits.shift(s[i-1], s[i]) <= 'z'for all odd indicesi.
Solution: Simulation
Time complexity: O(n)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: string replaceDigits(string s) { for (size_t i = 1; i < s.length(); i += 2) s[i] += s[i - 1] - '0'; return s; } }; |
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