443. 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

Tags: Turing Machines String Two Pointers

Explanation

The solution uses "Two Pointers" method. Imagine that there is a Turing Machine with read_head and write_head to process string chars; the machine is defined as following:

The machine has two states: the character count count and previous character read by the machine prev
The machine has two heads: read_head and write_head
The procedure continues as the following until the read_head reaches the null character of the string:
- If the machine read the same characters as prev, read_head move right and increment the count;
- otherwise (either read different character or read_head reaches the end), keep moving write_head when writing 1 prev and the "count" on the tape. After all, reset the machine state by setting count = 0 and prev = chars[read_head]; moving read_head to right.

Code (Rust)

impl Solution {
    pub fn compress(chars: &mut Vec<char>) -> i32 {
        let mut count = 1;
        let mut prev = chars[0];
        let (mut read_head, mut write_head) = (1,0);
        
        while read_head <= chars.len() {
            if read_head < chars.len() && prev == chars[read_head] {
                count+=1;
            }else {
                // Writing prev character
                chars[write_head] = prev;
                write_head+=1;

                // Writing the "count"
                if count > 1{
                    for c in count.to_string().chars(){
                        chars[write_head] = c;
                        write_head +=1;
                    }
                }

                // Reset the machine state
                if read_head < chars.len() { prev = chars[read_head] };
                count = 1;
            }
            read_head+=1; // read_head move to right
        }

        return write_head as i32;

    }
}

Complexity

\(n\) is length of chars

Time Complexity

\(T(n) = \Theta(n)\)
- read_head and write_head moves at most \(n\) steps.

Auxiliary Space

\(S(n) = O(1)\)

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