# 0443. String Compression

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

## Description

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".
```

**Example 4:**

```
**Input:** chars = ["a","a","a","b","b","a","a"]
**Output:** Return 6, and the first 6 characters of the input array should be: ["a","3","b","2","a","2"].
**Explanation:** The groups are "aaa", "bb", and "aa". This compresses to "a3b2a2". Note that each group is independent even if two groups have the same character.
```

**Constraints:**

* `1 <= chars.length <= 2000`
* `chars[i]` is a lowercase English letter, uppercase English letter, digit, or symbol.

## ac

```java
class Solution {
    public int compress(char[] chars) {
        // edge cases
        if (chars.length <= 1) return chars.length;

        // count chars
        int cnt = 1, l = 0;
        for (int i = 0; i < chars.length; i++) {
            if (i == chars.length - 1 || chars[i] != chars[i+1]) {
                l = write(chars, l, chars[i], cnt);
                cnt = 1; // reset
            } else {
                cnt++;
            }
        }

        return l;
    }

    private int write(char[] chars, int start, char c, int cnt) {
        chars[start++] = c;
        if (cnt > 1) {
            String s = String.valueOf(cnt);
            for (char ch : s.toCharArray()) {
                chars[start++] = ch;
            }
        }

        return start;
    }
}
```
