Solving Leetcode Interviews in Seconds with AI: Split Message Based on Limit
Introduction
In this blog post, we will explore how to solve the LeetCode problem "2468" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.
Problem Statement
Explanation
Here's the breakdown of the approach, complexity, and the Python code:
High-level approach:
- Calculate the maximum possible number of parts. If even with this maximum, the suffix length is too great to fit, it's impossible to split the message.
- Iteratively determine the actual number of parts. The key idea is to derive the length of the message portion in each part, considering the limit and suffix length.
- Split the message into parts, constructing the suffix for each part.
Complexity:
- Runtime Complexity: O(n), where n is the length of the message.
- Storage Complexity: O(n), to store the resulting array of strings.
Code
def splitMessage(message: str, limit: int) -> list[str]:
n = len(message)
max_parts = 1
temp = 1
while temp <= n:
max_parts *= 10
temp = len(str(max_parts))
max_parts = max_parts // 10
def suffix_length(parts):
return len(f"<{1}/{parts}>")
def can_split(parts):
message_length_per_part = limit - suffix_length(parts)
if message_length_per_part <= 0:
return False
total_length_possible = message_length_per_part * parts
return total_length_possible >= n
low = 1
high = n
parts = -1
while low <= high:
mid = (low + high) // 2
if can_split(mid):
parts = mid
high = mid - 1
else:
low = mid + 1
if parts == -1:
return []
message_length_per_part = limit - suffix_length(parts)
if message_length_per_part <= 0:
return []
result = []
start = 0
for i in range(1, parts + 1):
end = min(start + message_length_per_part, n)
result.append(message[start:end] + f"<{i}/{parts}>")
start = end
return result