# 443 String Compression

Given an array of characters `chars`, compress it using the following algorithm:

Begin with an empty string `s`. For each group of consecutive repeating characters in `chars`:

• If the group's length is `1`, append the character to `s`.
• Otherwise, append the character followed by the group's length.

The compressed string `s` should not be returned separately, but instead, be stored in the input character array `chars`. Note that group lengths that are 10 or longer will be split into multiple characters in `chars`.

After you are done modifying the input array, return the new length of the array.

You must write an algorithm that uses only constant extra space.

Example 1:

``````Input: chars = ["a","a","b","b","c","c","c"]
Output: Return 6, and the first 6 characters of the input array should be:
["a","2","b","2","c","3"]
Explanation: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".
``````

Example 2:

``````Input: chars = ["a"]
Output: Return 1, and the first character of the input array should be:
["a"]
Explanation: The only group is "a", which remains uncompressed since it's a single character.
``````

Example 3:

``````Input: chars = ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
Output: Return 4, and the first 4 characters of the input array should be:
["a","b","1","2"].
Explanation: The groups are "a" and "bbbbbbbbbbbb". This compresses to "ab12".
``````

Constraints:

• `1 <= chars.length <= 2000`
• `chars[i]` is a lowercase English letter, uppercase English letter, digit, or symbol.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 `````` ``````class Solution: def compress(self, chars: List[str]) -> int: n = len(chars) i, j = 0, 0 while j < n: # set char chars[i] = chars[j] i += 1 # count count = 1 while j + 1 < n and chars[j] == chars[j+1]: count += 1 j += 1 # set count if count > 1: for c in str(count): chars[i] = c i += 1 j += 1 return i``````