#3597

Medium

LeetCode Problem Workspace

Partition String

Partition a string into unique segments using hash table and string manipulation.

Problem Statement

Given a string s, partition it into segments such that each segment contains only unique characters. The partitioning should be done by iterating through the string and grouping consecutive characters that can form unique segments.

Return an array of strings where each string represents one of the segments formed according to the aforementioned rules. The segmentation should be done in such a way that every segment contains distinct characters without any repetition.

Examples

Example 1

Input: s = "abbccccd"

Output: ["a","b","bc","c","cc","d"]

Hence, the final output is ["a", "b", "bc", "c", "cc", "d"] .

Example 2

Input: s = "aaaa"

Output: ["a","aa"]

Hence, the final output is ["a", "aa"] .

Constraints

1 <= s.length <= 105
s contains only lowercase English letters.

Solution Approach

Simulate the Segmentation

To solve this problem, iterate through the string while using a hash table to track characters. For each character, check if it already exists in the current segment. If it does, finalize the current segment and start a new one.

Track Character Occurrences

Utilize a hash table to keep track of character occurrences in the current segment. If a character repeats, it indicates that the segment must be finalized, and a new segment should begin.

Efficiently Form Segments

By checking characters in sequence and updating the hash table efficiently, you can ensure that the partitioning happens in linear time. This avoids the need for additional nested loops, leading to an optimal solution.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the approach is O(n), where n is the length of the string, as we only iterate through the string once. The space complexity is O(k), where k is the number of unique characters, as we store the characters in a hash table.

What Interviewers Usually Probe

Does the candidate understand how to handle string segmentation efficiently?
Is the candidate able to manage a hash table for tracking characters in real-time?
Can the candidate explain the trade-offs involved in choosing hash tables over other data structures?

Common Pitfalls or Variants

Common pitfalls

Not properly managing the segments when characters repeat, leading to incorrect partitions.
Using a brute force approach with nested loops, which results in higher time complexity.
Failing to handle edge cases like strings with all unique characters or all repeated characters.

Follow-up variants

Modify the problem to use uppercase letters only.
Extend the problem to allow a custom rule for segment uniqueness, such as allowing certain characters to repeat a limited number of times.
Solve the problem with a Trie data structure instead of a hash table.

FAQ

What is the core pattern for solving the Partition String problem?

The core pattern for solving the Partition String problem is using a hash table to track the characters in the current segment and simulating the partitioning process.

How do you optimize the solution for time and space complexity in the Partition String problem?

The solution is optimized by using a hash table to track characters in linear time, which reduces the need for nested loops. The space complexity is minimal, proportional to the number of unique characters.

What is the expected output for input 'abbccccd' in the Partition String problem?

The expected output is ["a", "b", "bc", "c", "cc", "d"]. Each segment contains unique characters without repetition.

Can I use a different data structure other than a hash table for solving this problem?

Yes, you can use a Trie, but the hash table approach is more efficient for this problem. A Trie would generally have higher overhead due to its more complex structure.

What should I do if I encounter a string where all characters are unique in the Partition String problem?

In this case, each character will form its own segment. The output will be an array of the individual characters.

terminal

Solution

Solution 1: Hash Table + Simulation

We can use a hash table $\textit{vis}$ to record the segments that have already appeared. Then, we traverse the string $s$, building the current segment $t$ character by character until this segment has not appeared before. Each time we construct a new segment, we add it to the result list and mark it as seen.

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class Solution:
    def partitionString(self, s: str) -> List[str]:
        vis = set()
        ans = []
        t = ""
        for c in s:
            t += c
            if t not in vis:
                vis.add(t)
                ans.append(t)
                t = ""
        return ans

Solution 2: String Hashing + Hash Table + Simulation

We can use string hashing to speed up the lookup of segments. Specifically, we can compute a hash value for each segment and store it in a hash table. In this way, we can determine in constant time whether a segment has already appeared.

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class Solution:
    def partitionString(self, s: str) -> List[str]:
        vis = set()
        ans = []
        t = ""
        for c in s:
            t += c
            if t not in vis:
                vis.add(t)
                ans.append(t)
                t = ""
        return ans

Continue Practicing

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