Problem description

Algorithm walkthrough

Input and output

The divideString method takes three parameters:

• s: The input string to be divided
• k: The size of each group
• fill: The character used to pad the last group if needed And we need to return the grouped strings in groups of size k

1. Initialize variables:

• res = []: An empty list to store the resulting groups
• n = len(s): The length of the input string

2. Iterate through the string in steps of k:

This loop starts at index 0 and jumps by k positions each iteration (0, k, 2k, 3k, etc.)

for i in range(0, n, k):

3. For each iteration, check if a complete group of size k exists:

• if (i+k-1) < n: checks if we can extract a full group of k characters
• If yes: res.append(s[i:i+k]) extracts exactly k characters starting from index i

4. Handle the last group (if incomplete):

• else: branch executes when we can't get a full group of k characters
• s[i:n] gets the remaining characters
• .ljust(k, fill) pads the string on the right with the fill character to make it exactly k characters long

Full solution

class Solution:
    def divideString(self, s: str, k: int, fill: str) -> List[str]:
        # TC: O(n)
        # SC: O(n)
        res = []
        n = len(s)
        for i in range(0,n,k):
            if (i+k-1)<n:
                res.append(s[i:i+k])
            else:
                res.append(s[i:n].ljust(k, fill)) # fill the right with fill character
        return res