443. String Compression

Problem Statement

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".```

<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 &lt;= chars.length &lt;= 2000</code></li>
<li><code>chars[i]</code> is a lowercase English letter, uppercase English letter, digit, or symbol.</li>
</ul>


::: details _Click to open Hints_

- How do you know if you are at the end of a consecutive group of characters?

:::

## Solution:

::: code-group

```rs [Rust]
impl Solution {
    pub fn compress(chars: &mut Vec<char>) -> i32 {
        let (mut i, n) = (0, chars.len());
        let mut new_len = 0;

        while i < n {
            let mut j = i;

            // increment j until we find a different character or reach the end
            while j < n && chars[j] == chars[i] {
                j += 1;
            }

            // place the character at the new position
            // e.g. if we have aabbccc, we place a at the start of the array
            chars[new_len] = chars[i];
            new_len += 1;

            // if the length of the group of characters is greater than 1
            // i.e. suppose if new_len is 12, we need to place 12 as characters [..., '1','2', ...]
            if j - i > 1 {
                for c in (j - i).to_string().chars() {
                    chars[new_len] = c;
                    new_len += 1;
                }
            }

            // place i at same position as j,
            // i.e. the start of the next group of characters
            i = j;
        }

        new_len as i32
    }
}

:::

443. String Compression #

Problem Statement #

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443. String Compression

Problem Statement

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