#2405

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Optimal Partition of String

Given a string s, partition it into substrings with unique characters and return the minimum number of substrings.

Problem Statement

You are given a string s consisting of lowercase English letters. Partition the string into one or more substrings such that no character appears more than once in any substring.

Return the minimum number of substrings that can partition the string while ensuring each character is unique within a substring. Each character should belong to exactly one substring.

Examples

Example 1

Input: s = "abacaba"

Output: 4

Two possible partitions are ("a","ba","cab","a") and ("ab","a","ca","ba"). It can be shown that 4 is the minimum number of substrings needed.

Example 2

Input: s = "ssssss"

Output: 6

The only valid partition is ("s","s","s","s","s","s").

Constraints

1 <= s.length <= 105
s consists of only English lowercase letters.

Solution Approach

Greedy Approach with Hash Set

The most efficient way to solve this problem is through a greedy approach where you keep track of characters in a set. As you iterate through the string, you add characters to the set until you encounter a repeated character, at which point a new substring is started. This ensures each substring has unique characters.

Iterating and Updating Substring Count

You start with an empty set and iterate through the string, adding characters to the set. When a character is encountered that is already in the set, it signals the end of the current substring and the start of a new one. You increment the substring count accordingly.

Edge Case Handling

You must also handle edge cases such as strings with all unique characters, strings with only one character, or strings where all characters are the same. In these cases, the algorithm should efficiently handle the partitions, ensuring the output is accurate.

Complexity Analysis

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

The time complexity is O(n), where n is the length of the string. This is because we iterate through the string once, with constant time operations for set additions and lookups. The space complexity is O(min(n, 26)), as we store characters in a set, which can hold up to 26 unique characters at most (for lowercase English letters).

What Interviewers Usually Probe

Can the candidate identify and implement a greedy solution to handle substrings with unique characters?
Does the candidate consider edge cases in the problem, such as all characters being the same or all characters being unique?
Is the candidate able to explain how set operations play a role in partitioning the string?

Common Pitfalls or Variants

Common pitfalls

Failing to handle strings with all unique or all same characters properly.
Misunderstanding the need to track only unique characters within each substring.
Not using an efficient data structure like a set to track characters within the substring.

Follow-up variants

What if the string contains digits instead of lowercase letters?
Can the algorithm handle uppercase letters or a mixture of uppercase and lowercase?
How would you adapt this solution for strings with a larger character set, like Unicode characters?

FAQ

What is the optimal approach for partitioning a string into substrings with unique characters?

The optimal approach is to use a greedy strategy, where you iterate through the string and track characters using a set. When a repeat character is found, a new substring starts.

How can I improve the efficiency of my solution for partitioning strings?

By using a set to track unique characters, you can check for duplicates in constant time, ensuring the solution runs in O(n) time.

What is the time complexity of the optimal partition of string problem?

The time complexity is O(n), where n is the length of the string, as each character is processed once.

How do I handle edge cases for the optimal partition of string problem?

Edge cases include strings with all unique characters or strings with repeated characters. Handle them by properly managing the set to track unique characters in substrings.

What does 'Greedy choice plus invariant validation' mean for this problem?

It refers to making a greedy choice at each step (choosing unique characters for each substring) and validating the partition's correctness by ensuring no character repeats within a substring.

terminal

Solution

Solution 1: Greedy

According to the problem description, each substring should be as long as possible and contain unique characters. Therefore, we can greedily partition the string.

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class Solution:
    def partitionString(self, s: str) -> int:
        ans, mask = 1, 0
        for x in map(lambda c: ord(c) - ord("a"), s):
            if mask >> x & 1:
                ans += 1
                mask = 0
            mask |= 1 << x
        return ans

Continue Practicing

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