Tags: "leetcode", "string", "two-pointers", access_time 2-min read

String Compression

Created: March 14, 2020 by [lek-tin]

Last updated: March 14, 2020

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

1. All characters have an ASCII value in [35, 126].
2. 1 <= len(chars) <= 1000.

Solution

Time: O(n)
Space: O(1)

class Solution:
    def compress(self, chars: List[str]) -> int:
        N = len(chars)
        if not chars or N == 0:
            return []

        curr, start, end = 0, 0, 0

        while end < N:
            while end < N and chars[end] == chars[start]:
                end += 1
            if end - start == 1:
                chars[curr] = chars[start]
                curr += 1
            else:
                chars[curr] = chars[start]
                distance = len(str(end-start))
                chars[curr+1:curr+1+distance] = list(str(end-start))
                curr += distance + 1
            start = end

        return len(chars[:curr])