 # LeetCode Minimum Swaps to Group All 1's Together Solution

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"`.

Constraints:

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

## Solution

We wish to use two pointers in the same direction so solve this problem. The two pointers are: the `read` pointer (fast), and the `write` pointer (slow). The question is really similar to Move Zeros, where the fast pointer reads the contents and the slow pointer records the writing location.

So we will iterate (read) through `chars` and count the number of consecutive occurrences of the `curr` char. During this process if we reach a new character, we will write the `curr` character and its `count` to the position where the `write` pointer points to. We will use a helper function `compress_char` to help us write the compressed character according to the requirements (i.e. only append the length after the character if the length is greater than 1). At the same time, we update the `write` pointer to the position of the next write, which is exactly the length of the "new" array.

#### Implementation

``````1def compress(self, chars: List[str]) -> int:
2    def compress_char(write, curr, counter):
3        chars[write] = curr
4        write += 1
5        if counter == 1:      # does not append length
6            return write
7        length = str(counter) # convert length to string
8        for c in length:
9            chars[write] = c
10            write += 1
11        return write
12    write, counter, curr = 0, 1, chars
13    for read in range(1, len(chars)):