🧠 Solving LeetCode Until I Become Top 1% — Day `27`

🧠 Solving LeetCode Until I Become Top 1% — Day 27

🔹 Problem: 2138 Divide a String Into Groups of Size k

Difficulty: #Easy
Tags: #String, #Greedy

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📝 Problem Summary

Given a string s, divide it into groups of size k. If the final group is smaller than k, fill it with the given fill character until it reaches length k.

Return a list of strings, each representing a group of size k.

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🧠 My Thought Process

• Brute Force Idea:

  • Start taking substrings of length k in a loop and append them to a result list.
  • If the final group is smaller than k, pad it manually.

• Optimized Strategy:

  • First, check if s is divisible by k. If not, append the appropriate number of fill characters to make its length divisible by k.
  • Then, iterate over the updated string with a step of size k and slice out chunks.

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⚙️ Code Implementation (Python)

class Solution:
    def divideString(self, s: str, k: int, fill: str) -> List[str]:
        res = []
        n = len(s)

        s = s + (fill*(k-n%k) if n%k else '')

        start = 0

        while start < len(s):
            res.append(s[start:start+k])
            start += k

        return res

⏱️ Time & Space Complexity

• Time: O(N) Traverses the string once to pad (if needed), and once to slice.
• Space: O(N) Output list stores all the k-sized chunks.

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🧩 Key Takeaways

• ✅ Learned how to chunk a string cleanly using modulo logic and list comprehensions.
• 💡 Edge cases like "what if the string isn't divisible by k?" must be handled before chunking.
• 💭 This pattern is helpful for problems involving grouping, formatting, or padding strings to meet constraints.

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🔁 Reflection (Self-Check)

• [x] Could I solve this without help?
• [x] Did I write code from scratch?
• [x] Did I understand why it works?
• [x] Will I be able to recall this in a week?

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📚 Related Problems

• [[68 Text Justification]]

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🚀 Progress Tracker

Metric	Value
Day	27
Total Problems Solved	360
Confidence Today	😃