## DEV Community

Abhishek Chaudhary

Posted on

# 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.

SOLUTION:

``````class Solution:
def compress(self, chars: List[str]) -> int:
n = len(chars)
i = 0
while i < n:
ctr = 1
while i < n - 1 and chars[i] == chars[i + 1]:
i += 1
ctr += 1
i += 1
if ctr == 1:
chars.extend([chars[i - 1]])
else:
chars.extend([chars[i - 1]] + list(str(ctr)))
for i in range(n):
chars.pop(0)
return len(chars)
``````