Problem Description

You are given a string word that you need to compress using a specific algorithm.

The compression algorithm works as follows:

1. Start with an empty result string called comp.
2. While the input string word is not empty, repeatedly perform these operations:
   • Find the longest prefix of word that consists of the same character repeated (e.g., "aaaa" or "bbb")
   • This prefix can be at most 9 characters long
   • Remove this prefix from word
   • Add the length of the prefix followed by the character to comp

For example:

• If word = "aaaaaaaaaaabb", the first prefix would be "aaaaaaaaa" (9 'a's), which gets compressed to "9a"
• The remaining "aaa" would become "3a"
• The "bb" would become "2b"
• Final result: "9a3a2b"

The key constraint is that each compression unit can represent at most 9 consecutive occurrences of a character. If a character appears more than 9 times consecutively, you must split it into multiple compression units.

Your task is to implement this compression algorithm and return the compressed string comp.

Intuition

When looking at this compression problem, we need to identify groups of consecutive identical characters and count them. The natural approach is to traverse the string and group consecutive occurrences of the same character together.

The main challenge is handling the constraint that each group can represent at most 9 characters. If we have more than 9 consecutive identical characters, we need to split them into multiple groups of 9 or less.

Think of it like packing items into boxes where each box can hold at most 9 items of the same type. If you have 15 identical items, you'd fill one box with 9 items and another with 6 items. Similarly, if we encounter "aaaaaaaaaaaaaaa" (15 'a's), we'd compress it as "9a6a".

The solution uses Python's groupby function which automatically groups consecutive identical elements together. For each group:

1. Count the total occurrences (k)
2. Break down k into chunks of at most 9
3. For each chunk, append the count and character to our result

For example, if we have 23 consecutive 'a's:

• First chunk: min(9, 23) = 9, append "9a", remaining = 14
• Second chunk: min(9, 14) = 9, append "9a", remaining = 5
• Third chunk: min(9, 5) = 5, append "5a", remaining = 0

This greedy approach of always taking the maximum possible (up to 9) ensures we get the correct compression while respecting the constraint.

Solution Approach

The solution uses a two-pointer technique combined with Python's groupby function to efficiently compress the string.

Step-by-step implementation:

1. Group consecutive characters: Use groupby(word) to automatically group consecutive identical characters. This function returns an iterator of (key, group) pairs where the key is the character and group contains all consecutive occurrences.

2. Process each group: For each character group (c, v):

   • Convert the group iterator v to a list and get its length k to count consecutive occurrences
   • Example: If we have "aaaaa", k = 5

3. Handle the 9-character limit: Since each compression unit can represent at most 9 characters, we use a while loop to break down k:

   while k:
       x = min(9, k)  # Take at most 9 characters
       ans.append(str(x) + c)  # Add "count + character" to result
       k -= x  # Reduce remaining count

   This greedy approach ensures we always take the maximum allowed (up to 9) in each iteration.

4. Build the result: Each iteration appends a string in the format "[count][character]" to the ans list. For example:

   • 15 'a's → generates "9a" then "6a"
   • 3 'b's → generates "3b"

5. Join and return: Finally, join all compression units with "".join(ans) to create the final compressed string.

Example walkthrough:

• Input: "aaaaaaaaaaaabb"
• After groupby: [('a', 13 a's), ('b', 2 b's)]
• Processing 'a' group (k=13):
  • First iteration: x=9, append "9a", k=4
  • Second iteration: x=4, append "4a", k=0
• Processing 'b' group (k=2):
  • Single iteration: x=2, append "2b", k=0
• Result: "9a4a2b"

The time complexity is O(n) where n is the length of the input string, as we traverse each character once. The space complexity is O(n) for storing the result.

Example Walkthrough

Let's walk through the compression of word = "aaabbccccc" step by step:

Step 1: Group consecutive characters Using groupby, we get:

• Group 1: ('a', ['a', 'a', 'a']) → 3 'a's
• Group 2: ('b', ['b', 'b']) → 2 'b's
• Group 3: ('c', ['c', 'c', 'c', 'c', 'c']) → 5 'c's

Step 2: Process each group

Processing Group 1 - 'a' with k=3:

• Since k=3 < 9, we take all 3
• x = min(9, 3) = 3
• Append "3a" to result
• k = 3 - 3 = 0 (done with this group)

Processing Group 2 - 'b' with k=2:

• Since k=2 < 9, we take all 2
• x = min(9, 2) = 2
• Append "2b" to result
• k = 2 - 2 = 0 (done with this group)

Processing Group 3 - 'c' with k=5:

• Since k=5 < 9, we take all 5
• x = min(9, 5) = 5
• Append "5c" to result
• k = 5 - 5 = 0 (done with this group)

Step 3: Join results

• ans = ["3a", "2b", "5c"]
• Final result: "3a2b5c"

Example with splitting (exceeding 9 limit): Let's also see word = "aaaaaaaaaaaa" (12 'a's):

Processing 'a' with k=12:

• First iteration: x = min(9, 12) = 9
  • Append "9a" to result
  • k = 12 - 9 = 3
• Second iteration: x = min(9, 3) = 3
  • Append "3a" to result
  • k = 3 - 3 = 0 (done)

Final result: "9a3a"

This demonstrates how the algorithm handles both simple cases and cases where we need to split groups due to the 9-character limit.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the input string word. This is because:

• The groupby function iterates through the string once to create groups of consecutive identical characters, which takes O(n) time
• Converting the grouped values to a list with len(list(v)) consumes each group exactly once across all iterations
• The while loop inside processes each character in the original string exactly once (distributing them into chunks of at most 9)
• String concatenation operations in the loop are O(1) for each append operation
• The final "".join(ans) operation is O(n) since it needs to concatenate all characters in the result

The space complexity is O(n), where n is the length of the input string word. This is because:

• The ans list stores the compressed representation, which in the worst case (when no consecutive characters exist or when characters repeat more than 9 times) can be proportional to the input size
• The groupby iterator and the intermediate list created by list(v) use space proportional to the size of each group, which sums up to O(n) across all groups
• The final joined string also requires O(n) space

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

Common Pitfalls

1. Forgetting the 9-Character Limit

One of the most common mistakes is overlooking the constraint that each compression unit can represent at most 9 consecutive characters. A naive implementation might simply count all consecutive characters and write them directly:

Incorrect Implementation:

def compressedString(self, word: str) -> str:
    result = []
    i = 0
    while i < len(word):
        char = word[i]
        count = 1
        while i + 1 < len(word) and word[i + 1] == char:
            count += 1
            i += 1
        result.append(str(count) + char)  # Wrong! count could be > 9
        i += 1
    return "".join(result)

Why it fails: For input like "aaaaaaaaaaaaa" (13 a's), this would produce "13a" instead of the correct "9a4a".

Solution: Always split counts greater than 9 into multiple chunks:

while count > 0:
    chunk_size = min(9, count)
    result.append(str(chunk_size) + char)
    count -= chunk_size

2. Inefficient Manual Character Counting

Another pitfall is implementing character counting manually with nested loops, which can lead to off-by-one errors and complexity:

Error-Prone Implementation:

def compressedString(self, word: str) -> str:
    result = []
    i = 0
    while i < len(word):
        j = i
        while j < len(word) and word[j] == word[i]:
            j += 1
        count = j - i  # Easy to miscalculate
        # ... rest of logic

Common bugs in manual counting:

• Forgetting to handle the last character group
• Index out of bounds when checking word[j]
• Incorrectly updating the loop counter

Solution: Use Python's itertools.groupby() which handles the grouping logic cleanly and avoids index manipulation errors. Alternatively, if implementing manually, be careful with boundary conditions and test with edge cases like single characters and strings ending with different patterns.

3. String Concatenation Performance Issue

Building the result string through repeated concatenation can be inefficient:

Inefficient approach:

result = ""
for character, group in groupby(word):
    count = len(list(group))
    while count > 0:
        chunk = min(9, count)
        result += str(chunk) + character  # String concatenation in loop
        count -= chunk

Why it's problematic: Strings are immutable in Python, so each concatenation creates a new string object, leading to O(n²) time complexity in worst case.

Solution: Use a list to collect parts and join them at the end:

result = []
# ... append to result list
return "".join(result)  # Single join operation at the end