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.

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'

Approach

C++

#include <bits/stdc++.h>
using namespace std;

string replaceDigits(string s)
{

    int n = s.size();
    string str = "";
    int i = 0;
    while (i < n)
    {
        str += s[i];
        i++;
        if (i < n)
            str += s[i - 1] + s[i] - '0';
        i++;
    }
    return str;
}

int main()
{
    string s = "a1c1e1";

    cout << replaceDigits(s) << "\n";

    return 0;
}