String compression

Time: O(N); Space: O(1); easy

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: chars = [“a”,“a”,“b”,“b”,“c”,“c”,“c”]

Output: 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: chars = [“a”]

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

Explanation:

  • Nothing is replaced.

Example 3:

Input: chars = [“a”,“b”,“b”,“b”,“b”,“b”,“b”,“b”,“b”,“b”,“b”,“b”,“b”]

Output: 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.

Notes:

  • All characters have an ASCII value in [35, 126].

  • 1 <= len(chars) <= 1000.

[3]:

class Solution1(object):
    def compress(self, chars):
        """
        :type chars: List[str]
        :rtype: int
        """
        anchor, write = 0, 0
        for read, c in enumerate(chars):
            if read+1 == len(chars) or chars[read+1] != c:
                chars[write] = chars[anchor]
                write += 1
                if read > anchor:
                    n, left = read-anchor+1, write
                    while n > 0:
                        chars[write] = chr(n%10+ord('0'))
                        write += 1
                        n //= 10
                    right = write-1
                    while left < right:
                        chars[left], chars[right] = chars[right], chars[left]
                        left += 1
                        right -= 1
                anchor = read+1
        return write

[7]:

s = Solution1()
chars = ["a","a","b","b","c","c","c"]
assert s.compress(chars) == 6
assert chars[:6] == ["a","2","b","2","c","3"]

chars = ["a"]
assert s.compress(chars) == 1
assert chars[:1] == ["a"]

chars = ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
assert s.compress(chars) == 4
assert chars[:4] == ["a","b","1","2"]