> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# String Compression Python Solution with Tests

> Tested Python solution for LeetCode 443 with 16 pytest cases. Generate a practice environment with lcpy.

LeetCode 443, [Medium](/catalog/medium). Topics: [Two Pointers](/catalog/topics/two-pointers), [String](/catalog/topics/string). [View on LeetCode](https://leetcode.com/problems/string-compression/description/).

Generate this problem as a practice environment: tested reference solution, 16 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 443   # by problem number
lcpy gen -s string_compression   # by problem name
```

## Problem

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.

**Note:** The characters in the array beyond the returned length do not matter and should be ignored.

### Examples

```
Input: chars = ["a","a","b","b","c","c","c"]
Output: 6
Explanation: The groups are "aa", "bb", and "ccc". This compresses to "a2b2c3".
After modifying the input array in-place, the first 6 characters of chars should be ["a","2","b","2","c","3"].
```

```
Input: chars = ["a"]
Output: 1
Explanation: The only group is "a", which remains uncompressed since it is a single character.
After modifying the input array in-place, the first character of chars should be ["a"].
```

```
Input: chars = ["a","b","b","b","b","b","b","b","b","b","b","b","b"]
Output: 4
Explanation: The groups are "a" and "bbbbbbbbbbbb". This compresses to "ab12".
After modifying the input array in-place, the first 4 characters of chars should be ["a","b","1","2"].
```

### Constraints

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

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/string_compression/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/string_compression/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(n)
    # Space: O(1)
    def compress(self, chars: list[str]) -> int:
        write = 0
        read = 0
        n = len(chars)
        while read < n:
            ch = chars[read]
            run_start = read
            while read < n and chars[read] == ch:
                read += 1
            run_length = read - run_start
            chars[write] = ch
            write += 1
            if run_length > 1:
                for digit in str(run_length):
                    chars[write] = digit
                    write += 1
        return write
```

## Complexity

| Time | Space |
| - | - |
| O(n) | O(1) |

## Tags

[NeetCode All](/catalog/neetcode).


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