Problem Description

You are given a string compressed representing a compressed version of a string. This format consists of a character followed by its frequency. For example, the string "a3b1a1c2" is a compressed form and represents the original string "aaabacc". The task is to create a better compression of this string with the following conditions:

1. Each character should appear only once in the compressed version.
2. The characters should be in alphabetical order.

Intuition

To solve this problem, we need to transform the given compact string representation into a more optimized form without losing any frequency information. Here's how we can think through the solution:

1. Count Frequencies: First, we need an efficient way to count the frequencies of each character in the compressed format. A hash table (or dictionary) is well-suited for this task, where each unique character serves as a key and its total frequency as the value.

2. Parse the String: We can traverse the input string using two pointers. One points to the current character and the other captures the number following it, which denotes its frequency. By reading characters and their respective frequencies, we can update our hash table accordingly.

3. Combine and Sort: Once we have computed the cumulative frequency for each character, the next step is to build a new string. The characters need to be arranged in alphabetical order. We achieve this by sorting the hash table's keys and concatenating them along with their corresponding frequencies to form the final string.

By following these steps, we turn the input string into a more compact and alphabetically organized version while preserving the original frequency tallies of each character.

Learn more about Sorting patterns.

Solution Approach

To implement the solution for the better compression of the given input string, we'll employ a hash table alongside a two-pointer technique. Here's how the solution works step-by-step:

1. Initialize a Counter: We use a Counter from Python's collections module to keep track of the cumulative frequencies of each character.

   cnt = Counter()

2. Traverse the String: Set up two variables, i and n, where i points to the current position in the string and n is the total length of the string.

   i, n = 0, len(compressed)

3. Process Characters and Frequencies: As we traverse the string, we use another pointer j that starts just after i to read and compute the frequency of the current character.

   while i < n:
       j = i + 1
       x = 0
       while j < n and compressed[j].isdigit():
           x = x * 10 + int(compressed[j])
           j += 1
       cnt[compressed[i]] += x
       i = j

   • Start with j at i + 1 to check subsequent digits and collect the complete frequency number.
   • Convert each extracted digit into an integer and update the total frequency count for the character in the Counter.

4. Generate the Result String: After populating the Counter, we create the better compression string by sorting the keys (characters) alphabetically and joining each character with its cumulative frequency.

   return "".join(sorted(f"{k}{v}" for k, v in cnt.items()))

   • Use a generator to iterate over the sorted items of the Counter.
   • For each character k with frequency v, construct the string f"{k}{v}", then concatenate all parts to form the final result.

This approach efficiently compresses the string by using a combination of a hash table for counting and sorting to achieve the desired output format.

Example Walkthrough

Let's walk through the solution approach using the example compressed string "a3b1a1c2".

Step 1: Initialize a Counter

We start by setting up a Counter from Python's collections module to keep track of character frequencies.

cnt = Counter()

Step 2: Traverse the String

Next, we set up pointers i and n to help traverse the string.

i, n = 0, len("a3b1a1c2")

Step 3: Process Characters and Frequencies

We iterate over the string with a while loop. For each character, i points to the character, and j reads the frequency.

• At the start, i = 0:

  • compressed[i] = 'a'
  • Move j to read 3.
  • Update cnt['a'] += 3.

• Move i to the next character, i = 2:

  • compressed[i] = 'b'
  • j moves to read 1.
  • Update cnt['b'] += 1.

• Move i to the next character, i = 4:

  • compressed[i] = 'a'
  • j moves to read 1.
  • Update cnt['a'] += 1.

• Move i to the next character, i = 6:

  • compressed[i] = 'c'
  • j moves to read 2.
  • Update cnt['c'] += 2.

After processing, our Counter is:

cnt = {'a': 4, 'b': 1, 'c': 2}

Step 4: Generate the Result String

Sort the characters in alphabetical order and concatenate each character with its frequency.

result = "".join(sorted(f"{k}{v}" for k, v in cnt.items()))

The final sorted and compressed string is:

• a4b1c2

Thus, the original compressed string "a3b1a1c2" is transformed into a more compact and ordered format, "a4b1c2".

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n + |\Sigma| \log |\Sigma|), where n is the length of the string compressed, and |\Sigma| is the size of the character set, which is 26 for lowercase letters. This is due to the need to iterate through the string compressed (O(n)) and sort the characters in cnt (O(|\Sigma| \log |\Sigma|)).

The space complexity of the code is O(|\Sigma|). This is because the Counter dictionary cnt stores counts for each unique character from the character set, leading to a space requirement proportional to the size of the character set.

Learn more about how to find time and space complexity quickly.