What is the optimal time complexity of solving Partition Labels?

The optimized solution achieves a time complexity of O(n) by using sorting, mathematical shortcuts, or hash table lookups.

What is the auxiliary space complexity for Partition Labels?

The space complexity is O(n) since we only use a minimal/constant amount of extra memory for tracking variables (or auxiliary space if dynamic structures are allocated).

What are the typical edge cases we must handle in Partition Labels?

Key edge cases include empty collections, negative or large bounds causing integer overflow, arrays with only one element, or inputs where target criteria are not met.

How can we optimize the brute force approach for Partition Labels?

Brute force methods often inspect all possible combinations. We optimize this by sorting the input list to enable binary search, or tracking processed items in a set to avoid redundant nested loop lookups.

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Partition Labels

Learn how to solve the 'Partition Labels' problem. This detailed resource details brute force and optimized approaches.

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Problem Statement

Easy

You are given a string s. We want to partition the string into as many parts as possible so that each letter appears in at most one part. Return a list of integers representing the size of these parts.

Write a function partitionLabels(s: str) -> List[int].

Constraints

•1 <= len(s) <= 500
•s consists of lowercase English letters

Examples

Example 1

Input

s = "ababcbacadefegdehijhklij"

Output

[9,7,8]

Explanation

The partition is "ababcbaca", "defegde", "hijhklij".

Example 2

Input

s = "eccbbbbdec"

Output

[10]

Explanation

All characters in 'eccbbbbdec' can only be in one group.

Need a Hint?

Analyze the input constraints. Try sorting first (O(n log n)) or using a hash map/set to track seen elements in O(n) time.

Edge Cases to Watch

Empty list or null input variables
Single item lists/arrays
Extremely large input bounds causing integer or stack overflow

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