0443-String-Compression.py

'''
Given an array of characters, compress it in-place.

The length after compression must always be smaller than or equal to the original array.

Every element of the array should be a character (not int) of length 1.

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

Follow up:
Could you solve it using only O(1) extra space?
 

Example 1:

Input:
["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:
"aa" is replaced by "a2". "bb" is replaced by "b2". "ccc" is replaced by "c3".

 

Example 2:

Input:
["a"]

Output:
Return 1, and the first 1 characters of the input array should be: ["a"]

Explanation:
Nothing is replaced.

 

Example 3:

Input:
["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:
Since the character "a" does not repeat, it is not compressed. "bbbbbbbbbbbb" is replaced by "b12".
Notice each digit has it's own entry in the array.

 

Note:

    All characters have an ASCII value in [35, 126].
    1 <= len(chars) <= 1000.
'''
# Time Complexity: O(n)
# Space Complexity: O(1)
class Solution:
    def compress(self, chars: List[str]) -> int:
        index = i = 0
        while i < len(chars):
            cur, length = chars[i], 1
            while (i + 1) < len(chars) and cur == chars[i + 1]:
                length, i = length + 1, i + 1
            chars[index] = cur
            if length  > 1:
                length = str(length)
                for j, v in enumerate(length):
                    chars[index + j + 1] = v
                index += len(length)
            index, i = index + 1, i + 1
        return index