1844. Replace All Digits with Characters
Description
You are given a 0-indexed string s
that has lowercase English letters in its even indices and digits in its odd indices.
You must perform an operation 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 the result of the shift(s[i-1], s[i])
operation.
Return s
after replacing all digits. It is guaranteed that shift(s[i-1], s[i])
will never exceed 'z'
.
Note that shift(c, x)
is not a preloaded function, but an operation to be implemented as part of the solution.
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
.
Solutions
Solution 1: Simulation
Traverse the string, for characters at odd indices, replace them with the character that is a certain number of positions after the previous character.
Finally, return the replaced string.
The time complexity is $O(n)$, where $n$ is the length of the string $s$. Ignoring the space consumption of the answer, the space complexity is $O(1)$.
1 2 3 4 5 6 |
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1 2 3 4 5 6 7 8 9 |
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1 2 3 4 5 6 7 8 9 10 |
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1 2 3 4 5 6 7 |
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1 2 3 4 5 6 7 8 |
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1 2 3 4 5 6 7 8 9 10 11 12 |
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1 2 3 4 5 6 7 |
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