Problem Description

You are given a string s that needs to be divided into groups of a specific size k.

The partitioning works as follows:

• Start from the beginning of the string and take the first k characters to form the first group
• Take the next k characters to form the second group
• Continue this process until you've processed the entire string
• Each character in the string belongs to exactly one group

For the last group, there's a special rule:

• If there aren't enough characters left to form a complete group of size k, you need to pad the group with a given fill character
• The fill character is added repeatedly until the group reaches size k

The important constraint is that if you were to remove any fill characters from the last group and then concatenate all groups back together, you should get the original string s.

Your task is to return an array of strings where each string represents one group of size k.

For example:

• If s = "abcdefghi", k = 3, and fill = "x", the result would be ["abc", "def", "ghi"] (no padding needed)
• If s = "abcdefgh", k = 3, and fill = "x", the result would be ["abc", "def", "ghx"] (last group padded with one 'x')

The solution uses Python's string slicing s[i:i+k] to extract groups of k characters starting at position i, incrementing by k each time. The ljust(k, fill) method ensures each group has exactly k characters by padding with the fill character on the right if needed.

Intuition

The problem is essentially asking us to split a string into equal-sized chunks. This is a common pattern in programming where we need to process data in fixed-size blocks.

The key insight is that we need to iterate through the string in steps of k. Instead of processing character by character, we can jump k positions at a time. This naturally divides the string into groups of the desired size.

For each starting position i (where i is 0, k, 2k, 3k, ...), we need to extract a substring of length k. Python's slicing notation s[i:i+k] perfectly handles this - it gives us up to k characters starting from position i. If there are fewer than k characters remaining, the slice will just return whatever is left.

The challenge is handling the last group when it doesn't have enough characters. We could manually check if the last group is short and then pad it, but there's a more elegant approach: Python's ljust() method. This method left-justifies a string in a field of specified width, padding with a fill character if needed. By calling ljust(k, fill) on each slice, we ensure every group has exactly k characters - if the slice is already k characters long, nothing happens; if it's shorter, it gets padded with the fill character.

Using a list comprehension with range(0, len(s), k) gives us all the starting positions (0, k, 2k, ...) we need to slice from. This approach handles all cases uniformly without special logic for the last group, making the solution both concise and efficient.

Solution Approach

The solution uses a simulation approach to directly implement the partitioning process described in the problem.

The implementation leverages a list comprehension with three key components:

1. Iteration with Step Size: We use range(0, len(s), k) to generate starting indices for each group. This creates the sequence [0, k, 2k, 3k, ...] up to the length of the string. Each value represents the starting position of a new group.

2. String Slicing: For each starting index i, we extract a substring using s[i:i+k]. This slice operation:

   • Takes characters from position i up to (but not including) position i+k
   • Automatically handles the case where there are fewer than k characters remaining (it simply returns whatever is left)

3. Padding with ljust(): The ljust(k, fill) method is applied to each slice to ensure it has exactly k characters:

   • If the slice already has k characters, ljust() returns it unchanged
   • If the slice has fewer than k characters (which only happens for the last group), ljust() pads it on the right with the fill character until it reaches length k

The complete solution [s[i:i+k].ljust(k, fill) for i in range(0, len(s), k)] processes the entire string in one pass:

• Time Complexity: O(n) where n is the length of the string, as we visit each character once
• Space Complexity: O(n) for storing the result array

This approach elegantly handles all edge cases without explicit conditionals - the combination of slicing and ljust() naturally manages both complete groups and the potentially incomplete last group.

Example Walkthrough

Let's walk through the solution with s = "abcdefgh", k = 3, and fill = "x".

Step 1: Generate starting indices Using range(0, len(s), k) where len(s) = 8 and k = 3:

• Starting indices: [0, 3, 6]

Step 2: Process each group

Group 1 (i = 0):

• Extract slice: s[0:3] → "abc" (characters at positions 0, 1, 2)
• Apply padding: "abc".ljust(3, "x") → "abc" (already 3 characters, no padding needed)

Group 2 (i = 3):

• Extract slice: s[3:6] → "def" (characters at positions 3, 4, 5)
• Apply padding: "def".ljust(3, "x") → "def" (already 3 characters, no padding needed)

Group 3 (i = 6):

• Extract slice: s[6:9] → "gh" (only 2 characters left at positions 6, 7)
• Apply padding: "gh".ljust(3, "x") → "ghx" (padded with one 'x' to reach length 3)

Step 3: Collect results The list comprehension produces: ["abc", "def", "ghx"]

Verification: If we remove the fill character 'x' from the last group and concatenate: "abc" + "def" + "gh" = "abcdefgh" ✓

This matches our original string, confirming the solution works correctly.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm iterates through the string s once using a list comprehension with range(0, len(s), k), which creates approximately n/k iterations where n is the length of string s. For each iteration, it performs:

• String slicing s[i : i + k] which takes O(k) time
• The ljust(k, fill) operation which takes O(k) time in the worst case (when padding is needed)

Since we have n/k iterations and each iteration takes O(k) time, the total time complexity is (n/k) * O(k) = O(n).

Space Complexity: O(n)

The algorithm creates a new list to store the result. This list contains approximately ceil(n/k) strings, where each string has exactly k characters. The total number of characters stored is at most n plus some padding characters (at most k-1 additional characters for the last group). Therefore, the space complexity is O(n + k) = O(n) since k is typically much smaller than n and can be considered a constant in most practical cases.

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

Common Pitfalls

1. Misunderstanding the Fill Parameter Type

One common mistake is assuming fill is always a single character when implementing custom padding logic. While the problem typically provides a single character, the code should handle the parameter correctly.

Incorrect approach:

# Manually padding - assumes fill is single char
if len(group) < k:
    group += fill * (k - len(group))  # Works only if fill is single char

Correct approach:

# Using ljust() handles fill parameter properly
group.ljust(k, fill)  # Works regardless of fill's actual type

2. Off-by-One Errors in Manual Implementation

When implementing the solution without using ljust(), developers often make indexing errors:

Incorrect approach:

result = []
for i in range(0, len(s), k):
    if i + k <= len(s):
        result.append(s[i:i+k])
    else:
        # Wrong: might access beyond string bounds
        remaining = s[i:len(s)]
        padding_needed = k - (len(s) - i)
        result.append(remaining + fill * padding_needed)

Correct approach:

# Simpler and less error-prone
result = []
for i in range(0, len(s), k):
    group = s[i:i+k]  # Slicing handles bounds automatically
    result.append(group.ljust(k, fill))

3. Attempting to Modify Strings In-Place

Since strings are immutable in Python, trying to modify them directly will cause errors:

Incorrect approach:

for i in range(0, len(s), k):
    group = s[i:i+k]
    if len(group) < k:
        group[len(group):] = fill  # TypeError: strings don't support item assignment

Correct approach:

# Create new string with padding
for i in range(0, len(s), k):
    group = s[i:i+k]
    if len(group) < k:
        group = group + fill * (k - len(group))

4. Forgetting Edge Cases

Not handling edge cases like empty strings or when k equals 1:

Incomplete approach:

# Doesn't explicitly handle edge cases
return [s[i:i+k].ljust(k, fill) for i in range(0, len(s), k)]

Better approach with validation:

def divideString(self, s: str, k: int, fill: str) -> List[str]:
    if not s:  # Handle empty string
        return []
    if k <= 0:  # Handle invalid k
        return []
    return [s[i:i+k].ljust(k, fill) for i in range(0, len(s), k)]

5. Inefficient String Concatenation

Building the padded string character by character is inefficient:

Inefficient approach:

for i in range(0, len(s), k):
    group = s[i:i+k]
    while len(group) < k:
        group += fill  # Multiple string concatenations
    result.append(group)

Efficient approach:

for i in range(0, len(s), k):
    group = s[i:i+k]
    if len(group) < k:
        group += fill * (k - len(group))  # Single concatenation
    result.append(group)