Divide a String Into Groups of Size k

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A string s can be partitioned into groups of size k using the following procedure:

The first group consists of the first k characters of the string, the second group consists of the next k characters of the string, and so on. Each element can be a part of exactly one group.
For the last group, if the string does not have k characters remaining, a character fill is used to complete the group.

Note that the partition is done so that after removing the fill character from the last group (if it exists) and concatenating all the groups in order, the resultant string should be s.

Given the string s, the size of each group k and the character fill, return a string array denoting the composition of every group s has been divided into, using the above procedure.

Example 1:

Input: s = "abcdefghi", k = 3, fill = "x"
Output: ["abc","def","ghi"]
Explanation:
The first 3 characters "abc" form the first group.
The next 3 characters "def" form the second group.
The last 3 characters "ghi" form the third group.
Since all groups can be completely filled by characters from the string, we do not need to use fill.
Thus, the groups formed are "abc", "def", and "ghi".

Example 2:

Input: s = "abcdefghij", k = 3, fill = "x"
Output: ["abc","def","ghi","jxx"]
Explanation:
Similar to the previous example, we are forming the first three groups "abc", "def", and "ghi".
For the last group, we can only use the character 'j' from the string. To complete this group, we add 'x' twice.
Thus, the 4 groups formed are "abc", "def", "ghi", and "jxx".

Constraints:

1 <= s.length <= 100
s consists of lowercase English letters only.
1 <= k <= 100
fill is a lowercase English letter.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What should I return if the input string has a length that is not a multiple of k?
Are the input string and the fill character guaranteed to be valid (non-null, correct data type)?
What is the expected range for the length of the input string and the value of k?
If the input string is empty, what should I return?
Are there any specific constraints on the fill character, such as being a single character or ASCII character?

Brute Force Solution

Approach

The brute force method tackles this string division problem by exploring every conceivable way to split the input string into groups of the desired size. It systematically tries each possible division and checks if it adheres to the group size requirement. If a division is valid, it's considered a potential solution.

Here's how the algorithm would work step-by-step:

Begin by taking the first few characters of the string to form the initial group.
Check if the formed group has the required size.
If the group size is correct, then repeat the same process on the remaining portion of the string, attempting to create the next group.
If the group size is incorrect, try a different number of characters for the current group and repeat the size check.
Continue this process of forming groups and verifying their size until the entire string is divided.
Remember to keep track of every possible valid division of the string into groups.
If the string's length is not a multiple of the group size, then pad each group with the specified fill character to ensure each group is complete.

Code Implementation

def divide_string_into_groups(input_string, group_size, fill_character):
    result = []
    string_length = len(input_string)
    current_index = 0

    while current_index < string_length:
        group = ""
        # Form the initial group
        for i in range(min(group_size, string_length - current_index)):
            group += input_string[current_index + i]

        # Pad the group if its length is less than the required size
        while len(group) < group_size:
            group += fill_character

            # Padding is necessary to ensure all groups have the required size.

        result.append(group)
        current_index += group_size

        # Move to the next starting point for the subsequent group.

    # If the string is not a multiple of k, pad the result.
    return result

Big(O) Analysis

Time Complexity

O(k^n) – The brute force approach explores all possible ways to divide the string. In the worst-case scenario, for each position in the string, we may try forming a group of different sizes up to k (the target group size). This branching decision process occurs for each of the n characters in the string. This leads to a branching factor of k at each of the n characters, giving rise to roughly k^n possible divisions to explore. Therefore, the time complexity is O(k^n).

Space Complexity

O(N) – The provided brute force method, as described, explores all possible divisions and keeps track of every possible valid division of the string into groups. This implies storing these divisions in some data structure, like a list of lists. In the worst-case scenario, especially with smaller k values, the number of valid divisions could be substantial, potentially requiring storage proportional to the input string's length, N. Specifically, the list of possible divisions could grow to contain a number of divisions (each of length on the order of N/k), which may approach N. Therefore, the auxiliary space scales linearly with the input size N, resulting in O(N) space complexity.

Optimal Solution

Approach

We need to split a string into groups of a fixed size. The core idea is to walk through the string, collecting characters until we have a group, and then handling any leftover characters at the end. If the string's length is not a multiple of the group size, we pad the last group with a specified character.

Here's how the algorithm would work step-by-step:

Start with an empty list to store our groups.
Go through the original string, character by character.
Build a temporary group by adding characters until the group reaches the desired size.
Once a group is full, add it to our list of groups.
Repeat this process until we reach the end of the original string.
If the last group is not full (because the string's length wasn't a multiple of the group size), add the specified fill character until it is full.
Return the list of groups.

Code Implementation

def divide_string(input_string, group_size, fill_character):
    groups = []
    current_group = ""
    for char in input_string:
        current_group += char
        # Check if the current group is full.
        if len(current_group) == group_size:
            groups.append(current_group)
            current_group = ""

    # Handle the case where the last group is not full.
    if current_group:
        # Pad the last group with fill characters.
        while len(current_group) < group_size:
            current_group += fill_character
        groups.append(current_group)

    return groups

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input string of length n once to form groups. The padding step, if required, adds at most k fill characters to the last group, where k is a constant. Therefore, the time complexity is dominated by the single pass through the string, resulting in O(n) time complexity. The padding operation does not depend on n, so its cost is considered constant relative to n.

Space Complexity

O(N/k) – The algorithm creates a list to store the resulting groups. In the worst-case scenario, where k is a small constant, the number of groups will be proportional to the length of the input string (N) divided by k, i.e., N/k. Each group stores k characters but we are interested in the number of groups and their memory. Therefore, the auxiliary space scales linearly with N/k. The temporary group construction uses a small, fixed amount of memory and can be considered constant and is insignificant relative to the storage needed for the group list.

Edge Cases

Case	How to Handle
Null or empty input string	Return an empty list or throw an IllegalArgumentException as there is nothing to divide.
k is zero or negative	Throw an IllegalArgumentException as a group size cannot be zero or negative.
k is greater than the string length	Return a list containing only the original string padded with the fill character up to length k.
String length is a multiple of k	Ensure the algorithm correctly divides the string without needing padding.
String length is not a multiple of k and fill is null	Throw an IllegalArgumentException as padding is required but no fill character is provided.
String contains only the fill character	The normal logic should still correctly divide and pad the string as needed.
Very large string to test memory usage and performance	Ensure the algorithm does not consume excessive memory or have quadratic time complexity when handling large inputs.
k is equal to the string length	Return a list containing only the original string.

Null or empty input string

How to Handle:

Return an empty list or throw an IllegalArgumentException as there is nothing to divide.

k is zero or negative

How to Handle:

Throw an IllegalArgumentException as a group size cannot be zero or negative.

k is greater than the string length

How to Handle:

Return a list containing only the original string padded with the fill character up to length k.

String length is a multiple of k

How to Handle:

Ensure the algorithm correctly divides the string without needing padding.

String length is not a multiple of k and fill is null

How to Handle:

Throw an IllegalArgumentException as padding is required but no fill character is provided.

String contains only the fill character

How to Handle:

The normal logic should still correctly divide and pad the string as needed.

Very large string to test memory usage and performance

How to Handle:

Ensure the algorithm does not consume excessive memory or have quadratic time complexity when handling large inputs.

k is equal to the string length

How to Handle:

Return a list containing only the original string.