String Compression III

Can you solve this real interview question? String Compression III - Given a string word, compress it using the following algorithm:

• Begin with an empty string comp. While word is not empty, use the following operation:
  • Remove a maximum length prefix of word made of a single character c repeating at most 9 times.
  • Append the length of the prefix followed by c to comp.

Return the string comp.

 

Example 1:

Input: word = "abcde"

Output: "1a1b1c1d1e"

Explanation:

Initially, comp = "". Apply the operation 5 times, choosing "a", "b", "c", "d", and "e" as the prefix in each operation.

For each prefix, append "1" followed by the character to comp.

Example 2:

Input: word = "aaaaaaaaaaaaaabb"

Output: "9a5a2b"

Explanation:

Initially, comp = "". Apply the operation 3 times, choosing "aaaaaaaaa", "aaaaa", and "bb" as the prefix in each operation.

• For prefix "aaaaaaaaa", append "9" followed by "a" to comp.
• For prefix "aaaaa", append "5" followed by "a" to comp.
• For prefix "bb", append "2" followed by "b" to comp.

Solution

class Solution:
    def compressedString(self, word: str) -> str:
        comp = ""
        i = 0
        n = len(word)
        
        while i < n:
            # Get the current character
            char = word[i]
            # Count the number of consecutive chars up to 9
            count = 0
            while i < n and word[i] == char and count < 9:
                count += 1
                i += 1
            # Append the count and character to the result string
            comp += str(count) + char
        
        return comp

      ```