// Source : https://leetcode.com/problems/replace-all-digits-with-characters/
// Author : Hao Chen
// Date   : 2021-05-03

/***************************************************************************************************** 
 *
 * 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 x^th 
 * 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 {
public:
    string replaceDigits(string s) {
        for(int i=0; i<s.size(); i+=2) {
            s[i+1] = s[i] + s[i+1] - '0';
        }
        return  s;
    }
};