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Replace All Digits with Characters

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' and shift('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 indices i.

Solution Explanation:

This problem involves iterating through a string and performing a character shift operation based on the digits present in the string. The approach used is straightforward: iterate through the string, identify digits (which are always at odd indices), and replace them with the result of the shift operation.

shift(c, x) Operation:

The shift(c, x) operation is essentially adding the integer value of x to the ASCII value of character c. This shifts the character c forward in the alphabet by x positions. For example:

  • shift('a', 1): 'a' (ASCII 97) + 1 = 98, which is 'b'.
  • shift('x', 5): 'x' (ASCII 120) + 5 = 125, which is '}' . Note this will not exceed 'z' as per constraints.

Algorithm:

  1. Iterate: The code iterates through the input string s using a loop. The loop starts at index 1 (the first digit) and increments by 2 in each iteration (to only process odd indices).

  2. Character Shift: Inside the loop, for each digit s[i], the code calculates the shifted character using the ASCII values. It adds the integer value of the digit (s[i] - '0') to the ASCII value of the preceding character (s[i-1]). Then, the result is cast back to a character using chr() (Python) or (char) (Java, C++).

  3. Replacement: The digit s[i] at the odd index is replaced with the shifted character.

  4. Return: Finally, the modified string is returned.

Time Complexity Analysis:

The code iterates through the string once. The operations within the loop (character conversion, addition, and assignment) take constant time. Therefore, the overall time complexity is O(n), where n is the length of the string.

Space Complexity Analysis:

The space complexity depends on the implementation. Some solutions (like the Python solution using list(s)) create a copy of the string, resulting in O(n) space complexity. However, other solutions (like the C++ solution) modify the string in place, leading to O(1) space complexity (excluding the output string). Most of the provided solutions operate in O(1) space if we ignore the output string and consider only the auxiliary space.

Example Walkthrough (Python):

Let's trace the execution for the input string "a1c2e1":

  1. s becomes ['a', '1', 'c', '2', 'e', '1'].
  2. The loop starts at i = 1.
  3. i = 1: s[1] = chr(ord('a') + int('1')) = chr(97 + 1) = 'b'. s becomes ['a', 'b', 'c', '2', 'e', '1'].
  4. i = 3: s[3] = chr(ord('c') + int('2')) = chr(99 + 2) = 'e'. s becomes ['a', 'b', 'c', 'e', 'e', '1'].
  5. i = 5: s[5] = chr(ord('e') + int('1')) = chr(101 + 1) = 'f'. s becomes ['a', 'b', 'c', 'e', 'e', 'f'].
  6. The function returns "abceef".

This demonstrates how the algorithm correctly replaces the digits with the shifted characters.