#128 Medium Time O(n) Space O(1) Animated

Partition Labels

Partition string into as many parts as possible with unique characters.

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1 Predict
2 Trace
3 Explain
4 Implement

Pattern Greedy choice with a local proof

Commit to the locally best safe choice only after identifying why it cannot block an optimal answer.

Approach: Greedy / Two Pointers

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Core Insight

Commit to the locally best safe choice only after identifying why it cannot block an optimal answer.

Common Pitfall

Sort or define the selection order before applying the greedy rule.

Description

Partition string into as many parts as possible with unique characters.

Constraints

Time Complexity: O(n)
Space Complexity: O(1)

Solution — Greedy / Two Pointers

TimeO(n) SpaceO(1)

Implementation Plan

Translate the invariant into these small, testable moves.

Order candidates by the greedy criterion.
Inspect the next candidate.
Accept it only when it preserves feasibility.
Return the accumulated result.

def partitionLabels(s):
    lastIndex = {} # char -> last index in s
    for i, c in enumerate(s):
        lastIndex[c] = i
        
    res = []
    size, end = 0, 0
    for i, c in enumerate(s):
        size += 1
        end = max(end, lastIndex[c])
        
        if i == end:
            res.append(size)
            size = 0
            
    return res

Merge Triplets to Form Target Triplet

Valid Parenthesis String

Partition Labels

Steps

Description

Constraints

Tags

Solution — Greedy / Two Pointers