Divide a String Into Groups of Size k — LeetCode 2138 Python Solution
- Problem
- #2138
- Pattern
- Hash Map
- Reading time
- 2 min
- Source
- leetcode.com
The problem
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.
Example
- Input
- s = "abcdefghi", k = 3, fill = "x"
- Output
- ["abc","def","ghi"]
- Explanation
- The first 3 characters "abc" form the first group.
Python solution
class Solution:
def divideString(self, s: str, k: int, fill: str) -> List[str]:
return [s[i : i + k].ljust(k, fill) for i in range(0, len(s), k)]Complexity
| Measure | Complexity |
|---|---|
| Time | O(n) |
| Space | O(n), where n is the length of the string s auxiliary |
Pattern: Hash Map
Trade memory for time: remember what you have seen so the second pass never happens. LeetCode 2138. Divide a String Into Groups of Size k is filed here because the reference solution below belongs to the algorithm family this hub collects, even though its LeetCode tags point elsewhere.
The hash map guide has the Python template for the pattern and the 709 LeetCode problems that use it.
Related problems
Frequently asked questions
- How hard is LeetCode 2138. Divide a String Into Groups of Size k?
- LeetCode 2138. Divide a String Into Groups of Size k is rated Easy on LeetCode.
- What is the time complexity of LeetCode 2138. Divide a String Into Groups of Size k?
- The Python solution on this page runs in O(n).
- What is the space complexity of LeetCode 2138. Divide a String Into Groups of Size k?
- The Python solution on this page uses O(n), where n is the length of the string s auxiliary space.
- What topics does LeetCode 2138. Divide a String Into Groups of Size k cover?
- LeetCode 2138. Divide a String Into Groups of Size k is tagged String and Simulation on LeetCode.