What is the brute force approach and complexity for Partition Labels?

For each cut, check whether any letter appears on both sides with time complexity O(N²) and space complexity O(1).

What is the optimal approach and complexity for Partition Labels?

Record each letter's last index; cut when the index reaches the furthest last index seen in this part with optimal time complexity O(N) and space complexity O(1).

How do you identify the pattern for Partition Labels?

Look for key signals in the prompt: "each letter appears in at most one part" indicates A letter can't be cut in two. Its span, from first to last occurrence, has to stay inside one part.; "as many parts as possible" indicates Maximise the number of cuts: cut as often as is safe.; "size of these parts, in order" indicates Output the lengths of consecutive blocks, left to right..

Partition Labels

Medium·LeetCode #763 Pro OnlySlight Variation

LeetCode

Problem Statement

You get a string s of lowercase letters. Cut it into consecutive parts so that each letter appears in only one part.

Make as many parts as you can. Return the length of each part, from left to right.

Input

A string s of lowercase English letters.

Output

A list of part lengths, in order. The lengths add up to s.length.

Example 1

Input: s = "ababcbacadefegdehijhklij"

Output: [9, 7, 8]

The parts are ababcbaca, defegde and hijhklij. No letter is in two parts. Merging the last two parts would give fewer parts, so that is not the best answer.

Example 2

Input: s = "eccbbbbdec"

Output: [10]

The letters e and c show up on both sides of every possible cut, so the only option is one part.

Constraints

• 1 ≤ s.length ≤ 500
• s consists of lowercase English letters

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“each letter appears in at most one part”

→ A letter can't be cut in two. Its span, from first to last occurrence, has to stay inside one part.

“as many parts as possible”

→ Maximise the number of cuts: cut as often as is safe.

“size of these parts, in order”

→ Output the lengths of consecutive blocks, left to right.

Approach Complexity Ladder

Brute Force — test every cut positionO(N²)O(1)

For each cut, check whether any letter appears on both sides

Optimal — cut as soon as the reach closesO(N)O(1)

Record each letter's last index; cut when the index reaches the furthest last index seen in this part

Brute Force Intuition

A cut after index i is allowed only if no letter appears both at or before i and after i. Test every position, cut at each allowed one, and return the part lengths.

Recognition rule for this sub-pattern:

To fit the most activities, always take the one that finishes earliest.

Remember this · Greedy

Trigger

Maximise count / minimise cost with a natural "best next step", sorting makes the choice obvious, reachability.

Think

Is there a choice I can always make safely without looking ahead?

Key question

Why can't the greedy choice be worse than the optimal one? (Try to break it with a small example.)

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