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".


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

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


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.


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[0]
13    for read in range(1, len(chars)):
14        if chars[read] == curr:
15            counter += 1
16        else:
17            write = compress_char(write, curr, counter)
18            counter = 1
19        curr = chars[read]
20    write = compress_char(write, curr, counter)
21    return write