Decrypt String from Alphabet to Integer Mapping

This can be solved by iterating the input string from right to left. If we encounter a ‘#’, it means the last two digits before ‘#’ represents a character from ‘j’ to ‘z’. If there’s no ‘#’, then the digit represents a character from ‘a’ to ‘i’.

Python solution:

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class Solution:
    def freqAlphabets(self, s: str) -> str:
        i = len(s) - 1
        result = []

        while i >= 0:
            if s[i] == '#':
                # Extract the two digits before '#'
                num = int(s[i-2:i])
                # Map the number to its corresponding character and append to the result
                result.append(chr(ord('j') + num - 10))  # 'j' corresponds to 10
                i -= 3
            else:
                # Map the single digit number to its corresponding character
                result.append(chr(ord('a') + int(s[i]) - 1))  # 'a' corresponds to 1
                i -= 1
        # Since we processed the string from right to left, we need to reverse the result to get the correct order
        return ''.join(result[::-1])

In this code:

We initialize an index i to the end of the string and an empty list result to store the characters.
We iterate over the string from the end to the start. If the current character is ‘#’, we get the previous two digits and map it to a character from ‘j’ to ‘z’. We then decrease the index by 3 to skip the processed digits and the ‘#’.
If the current character is not ‘#’, it’s a single digit, which we map to a character from ‘a’ to ‘i’. We then decrease the index by 1.
After processing the entire string, we join the characters in result into a string. We also reverse the string as we initially processed the characters from right to left.
Finally, we return the resulting string.

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class Solution:
    def freqAlphabets(self, s: str) -> str:
        i = len(s) - 1
        result = []

        while i >= 0:
            if s[i] == '#':
                num = int(s[i-2:i])
                char = chr(ord('j') + num - 10)
                result.append(char)
                i -= 3
            else:
                num = int(s[i])
                char = chr(ord('a') + num - 1)
                result.append(char)
                i -= 1

        result_reversed = result[::-1]
        result_str = ''.join(result_reversed)

        return result_str