How do you solve Partition String on LeetCode?

Partition String (LeetCode #3597) 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: Segments must be unique, so tracking seen segments is crucial.

What is the time complexity of Partition String ?

The optimal solution for Partition String runs in O(n) time and uses O(n) space. Each character is processed once, leading to O(n). The space complexity accounts for the segments and seen set.

What is the brute force solution for Partition String ?

The brute force approach for Partition String is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach generates all possible segments and checks if they are unique. It can be inefficient as it may repeatedly check the same segments. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Partition String ?

Partition String uses the following concepts: Hash Table, String, Trie, Simulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Partition String ?

Explain your thought process clearly. Use examples to illustrate your approach. Practice coding without an IDE to simulate interview conditions.

What are common mistakes when solving Partition String ?

Not resetting the segment correctly after a duplicate. Forgetting to add the last segment after the loop.

#3597

Partition String

Medium

Hash TableStringTrieSimulationHash MapArray

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)

This approach uses a single pass through the string while maintaining a set of seen segments, ensuring each segment is built only once and checked for uniqueness efficiently.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty list for segments and a set for seen segments.
2Step 2: Use a loop to build segments character by character, checking against the seen set.
3Step 3: If a segment is unique, add it to the list and update the seen set.

solution.py11 lines

1def partition_string(s):
2    segments, seen = [], set()
3    segment = ''
4    for char in s:
5        segment += char
6        if segment in seen:
7            segments.append(segment[:-1])
8            seen.add(segment[:-1])
9            segment = char
10    segments.append(segment)
11    return segments

ℹ

Complexity note: Each character is processed once, leading to O(n). The space complexity accounts for the segments and seen set.

1Segments must be unique, so tracking seen segments is crucial.
2Building segments dynamically helps avoid unnecessary checks.

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