How do you solve Split Message Based on Limit on LeetCode?

Split Message Based on Limit (LeetCode #2468) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the suffix length is crucial for determining how to split the message.

What is the brute force solution for Split Message Based on Limit?

The brute force approach for Split Message Based on Limit is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying all possible lengths for the total number of parts and checking if we can split the message accordingly. This method is straightforward but inefficient as it may involve multiple checks for each possible part count. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Split Message Based on Limit?

Split Message Based on Limit uses the following concepts: String, Enumeration. The recommended patterns to study are: Enumeration, String Manipulation.

What are the interview tips for Split Message Based on Limit?

Always clarify the constraints and edge cases with the interviewer. Consider the implications of your approach on time and space complexity. Practice breaking down the problem into smaller parts to simplify your thought process.

What are common mistakes when solving Split Message Based on Limit?

Not accounting for the suffix length when calculating maximum part length. Assuming that any split will work without validating the concatenated result.

#2468

Split Message Based on Limit

Hard

StringEnumerationEnumerationString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach involves calculating the total number of digits in the total parts and using that to determine how to split the message efficiently. This reduces unnecessary iterations and checks.

⚙️

Algorithm

4 steps

1Step 1: Calculate the maximum number of parts possible based on the limit and the maximum suffix length.
2Step 2: Iterate through possible total part counts, calculating the suffix length dynamically.
3Step 3: For each total part count, check if the message can be split into valid parts and format them with the appropriate suffix.
4Step 4: Return the first valid split found.

solution.py22 lines

1def splitMessage(message, limit):
2    n = len(message)
3    max_parts = 1
4    while len(f'<{max_parts}/{max_parts}>') <= limit:
5        max_parts += 1
6    max_parts -= 1
7    for total_parts in range(1, max_parts + 1):
8        suffix_length = len(f'<{total_parts}/{total_parts}>')
9        max_part_length = limit - suffix_length
10        if max_part_length <= 0:
11            continue
12        parts = []
13        index = 0
14        for i in range(total_parts):
15            part = message[index:index + max_part_length]
16            if len(part) == 0:
17                break
18            parts.append(part + f'<{i + 1}/{total_parts}>')
19            index += max_part_length
20        if ''.join(p[:-suffix_length] for p in parts) == message:
21            return parts
22    return []

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the message a limited number of times based on the maximum possible parts, which is much more efficient than the brute-force method.

1Understanding the suffix length is crucial for determining how to split the message.
2Iterating through possible part counts efficiently can help avoid unnecessary checks.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.