LeetCode 443 - String Compression

https://leetcode.com/problems/string-compression/

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.

https://leetcode.com/articles/string-compression/

We will use separate pointers read and write to mark where we are reading and writing from. Both operations will be done left to right alternately: we will read a contiguous group of characters, then write the compressed version to the array. At the end, the position of the write head will be the length of the answer that was written.

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.

    public int compress(char[] chars) {
        int anchor = 0, write = 0;
        for (int read = 0; read < chars.length; read++) {
            if (read + 1 == chars.length || chars[read + 1] != chars[read]) {
                chars[write++] = chars[anchor];
                if (read > anchor) {
                    for (char c: ("" + (read - anchor + 1)).toCharArray()) {
                        chars[write++] = c;
                    }
                }
                anchor = read + 1;
            }
        }
        return write;
    }

https://leetcode.com/problems/string-compression/discuss/92559/Simple-Easy-to-Understand-Java-solution

public int compress(char[] chars) {
        int indexAns = 0, index = 0;
        while(index < chars.length){
            char currentChar = chars[index];
            int count = 0;
            while(index < chars.length && chars[index] == currentChar){
                index++;
                count++;
            }
            chars[indexAns++] = currentChar;
            if(count != 1)
                for(char c : Integer.toString(count).toCharArray()) 
                    chars[indexAns++] = c;
        }
        return indexAns;
    }

https://leetcode.com/problems/string-compression/discuss/92530/Java-O(n)-two-pointers-and-a-counter

public int compress(char[] chars) {        
        int start = 0;
        for(int end = 0, count = 0; end < chars.length; end++) {
            count++;
            if(end == chars.length-1 || chars[end] != chars[end + 1] ) {
                //We have found a difference or we are at the end of array
                chars[start] = chars[end]; // Update the character at start pointer
                start++;
                if(count != 1) {
                    // Copy over the character count to the array
                    char[] arr = String.valueOf(count).toCharArray();
                    for(int i=0;i<arr.length;i++, start++)
                        chars[start] = arr[i];
                }
                // Reset the counter
                count = 0;
            }
        }
        return start;
    }