Leetcode #2468: Split Message Based on Limit
In this guide, we solve Leetcode #2468 Split Message Based on Limit in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a string, message, and a positive integer, limit. You must split message into one or more parts based on limit.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String, Binary Search, Enumeration
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: message = "this is really a very awesome message", limit = 9
Output: ["thi<1/14>","s i<2/14>","s r<3/14>","eal<4/14>","ly <5/14>","a v<6/14>","ery<7/14>"," aw<8/14>","eso<9/14>","me<10/14>"," m<11/14>","es<12/14>","sa<13/14>","ge<14/14>"]
Explanation:
The first 9 parts take 3 characters each from the beginning of message.
The next 5 parts take 2 characters each to finish splitting message.
In this example, each part, including the last, has length 9.
It can be shown it is not possible to split message into less than 14 parts.
Python Solution
class Solution:
def splitMessage(self, message: str, limit: int) -> List[str]:
n = len(message)
sa = 0
for k in range(1, n + 1):
sa += len(str(k))
sb = len(str(k)) * k
sc = 3 * k
if limit * k - (sa + sb + sc) >= n:
ans = []
i = 0
for j in range(1, k + 1):
tail = f'<{j}/{k}>'
t = message[i : i + limit - len(tail)] + tail
ans.append(t)
i += limit - len(tail)
return ans
return []
Complexity
The time complexity is , where is the length of the string message. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.