'''
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.

Follow up:
Could you solve it using only O(1) 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".
Example 4:

Input: chars = ["a","a","a","b","b","a","a"]
Output: Return 6, and the first 6 characters of the input array should be: ["a","3","b","2","a","2"].
Explanation: The groups are "aaa", "bb", and "aa". This compresses to "a3b2a2". Note that each group is independent even if two groups have the same character.
'''

# O(N) TC and O(1) Space

'''
Algorithm

Let's maintain anchor, the start position of the contiguous group of characters we are currently reading.

Now, let's read from left to right. We know that we must be at the end of the block when we are at the last character,
or when the next character is different from the current character.
When we are at the end of a group, we will write the result of that group down using our write head. 
chars[anchor] will be the correct character, and the length (if greater than 1) will be read - anchor + 1. We will write the digits of that number to the array.
'''

class Solution(object):
    def compress(self, chars):
        anchor, write = 0, 0
        for read, char in enumerate(chars):
            if read + 1 == len(chars) or chars[read + 1] != char:
                chars[write] = chars[anchor]
                write += 1
                if read > anchor:
                    for digit in str(read - anchor + 1):
                        chars[write] = digit
                        write += 1
                anchor = read + 1
        return write
