#
# Create by Hua on 3/23/22.
#

"""
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' and shift('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 <= 100
    s consists only of lowercase English letters and digits.
    shift(s[i-1], s[i]) <= 'z' for all odd indices i.
"""


class Solution(object):
    def replaceDigits(self, s):
        """
        :type s: str
        :rtype: str

        thought: go through the string by every 2 chars, convert the second
        digit into a char with the formula of chr(ord(ch) + digit).

        03/23/2022 11:29	Accepted	39 ms	13.6 MB	python

        easy, 10 min. char/int conversion took some time.
        """
        n = len(s)
        ret = list()
        for i in range(0,n,2):
            if i + 1 < n:
                chunk = s[i:i+2]
                ch = ord(chunk[0])
                digit = int(chunk[1])
                new_char = chr(ch + digit)
                ret.append(chunk[0])
                ret.append(new_char)

        if len(ret) < n:
            ret.append(s[-1])

        return "".join(ret)