Leetcode 443. 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.

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.

Code (Java):

class Solution {
    public int compress(char[] chars) {
        if (chars == null || chars.length == 0) {
            return 0;
        }
        
        int newPos = 0;
        int count = 1;
        for (int i = 0; i < chars.length; i++) {
            count = 1;
            while (i < chars.length && i != chars.length - 1 && (chars[i] == chars[i + 1])) {
                count++;
                i++;
            }

            int numChars = getNumberofChars(count);
            newPos = copyArray(chars, newPos, chars[i], count, numChars);
        }
        
        return newPos;
    }

    private int copyArray(char[] chars, int pos, char c, int count, int numChars) {
        chars[pos] = c;
        pos++;

        if (count > 1) {
            int end = pos + numChars - 1;
            while (count > 0) {
                int digit = count % 10;
                chars[end] = (char)(digit + '0');
                end--;

                count /= 10;
            }

            pos += numChars;
        }

        return pos;
    }

    private int getNumberofChars(int num) {
        int ans = 0;
        if (num < 10) {
            ans = 1;
        } else if (num < 100) {
            ans = 2;
        } else if (num < 1000) {
            ans = 3;
        } else {
            ans = 4;
        }

        return ans;
    }
}