1844. Replace All Digits with Characters
Easy82.6% acceptance113,205 / 136,970 submissions
Asked by 1 company
Topics
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 <= 100sconsists only of lowercase English letters and digits.shift(s[i-1], s[i]) <= 'z'for all odd indicesi.
Hints
Hint 1
We just need to replace every even positioned character with the character s[i] positions ahead of the character preceding it
Hint 2
Get the position of the preceeding character in alphabet then advance it s[i] positions and get the character at that position