String Compression

• LeetCode: String Compression

• Problem:

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.

• Note:

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

• Follow up:

Could you solve it using only O(1) extra space?

• Examples:

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".

Input:
["a"]

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

Explanation:
Nothing is replaced.

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.

• Analysis:

Use two pointers, one is at start of the specific character and the other is to find the point which fits arr[i] != arr[i - 1].

• Code - Java:

class Solution {
    public int compress(char[] chars) {
        int len = chars.length;
        if (len <= 1) {
            return len;
        }
        int left = 0;
        int pos = 0;
        for (int i = 1; i < len; i++) {
            if (chars[i] != chars[i - 1]) {
                int count = i - left;
                chars[pos++] = chars[left];
                if (count > 1) {
                    String val = String.valueOf(count);
                    int valLen = val.length();
                    for (int j = 0; j < valLen; j++) {
                        chars[pos++] = val.charAt(j);
                    }
                }
                left = i;
            }
        }
        int count = len - left;
        chars[pos++] = chars[left];
        if (count > 1) {
            String val = String.valueOf(count);
            int valLen = val.length();
            for (int j = 0; j < valLen; j++) {
                chars[pos++] = val.charAt(j);
            }
        }
        return pos;
    }
}

• Code - C++:

class Solution {
public:
    int compress(vector<char>& chars) {
        int len = chars.size();
        if (len <= 1) {
            return len;
        }
        int left = 0;
        int pos = 0;
        for (int i = 1; i < len; i++) {
            if (chars[i] != chars[i - 1]) {
                int count = i - left;
                chars[pos++] = chars[left];
                if (count > 1) {
                    string val = to_string(count);
                    int valLen = val.length();
                    for (int j = 0; j < valLen; j++) {
                        chars[pos++] = val[j];
                    }
                }
                left = i;
            }
        }
        int count = len - left;
        chars[pos++] = chars[left];
        if (count > 1) {
            string val = to_string(count);
            int valLen = val.length();
            for (int j = 0; j < valLen; j++) {
                chars[pos++] = val[j];
            }
        }
        return pos;
    }
};

• Code - Python:

class Solution(object):
    def compress(self, chars):
        """
        :type chars: List[str]
        :rtype: int
        """
        charsLen = len(chars)
        if charsLen <= 1:
            return charsLen
        left, pos = 0, 0
        for i in range(1, charsLen):
            if chars[i] != chars[i - 1]:
                count = i - left
                chars[pos] = chars[left]
                pos += 1
                if count > 1:
                    val = str(count)
                    for ch in val:
                        chars[pos] = ch
                        pos += 1
                left = i
        count = charsLen - left
        chars[pos] = chars[left]
        pos += 1
        if count > 1:
            val = str(count)
            for ch in val:
                chars[pos] = ch
                pos += 1
        return pos

• Time Complexity: O(N), N is the length of input array

• Space Complexity: O(1)