What is the time complexity of Optimal Partition of String?

The optimal solution has a time complexity of O(n), where n is the length of the string 's'. The algorithm iterates through the string once..

What data structures are used to solve Optimal Partition of String?

This problem uses Hash Table, String, Greedy. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Optimal Partition of String?

Insight 1: The greedy approach of forming a new substring whenever a repeating character is encountered is optimal because it minimizes the length of the current substring before starting a new one. Any other strategy would potentially lead to more substrings. Insight 2: A set (hash table) is an efficient data structure for tracking the characters present in the current substring, offering O(1) average time complexity for checking character existence and insertion.

Which companies ask LeetCode 2405. Optimal Partition of String in interviews?

This problem is frequently asked at Docusign.

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Optimal Partition of String - LeetCode 2405 Solution

Q: What is the space complexity of Optimal Partition of String?

The space complexity is O(1). The size of the 'current_set' is at most 26, since there are only 26 lowercase English letters. Therefore, the space used is constant regardless of the input size..

Optimal Partition of String - Complete Solution Guide

Optimal Partition of String is LeetCode problem 2405, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3.

Problem Statement

Given a string s , partition the string into one or more substrings such that the characters in each substring are unique . That is, no letter appears in a single substring more than once . Return the minimum number of substrings in such a partition. Note that each character should belong to exactly one substring in a partition. Example 1: Input: s = "abacaba" Output: 4 Explanation: Two possible partitions are ("a","ba","cab","a") and ("ab","a","ca","ba"). It can be shown that 4 is the minimum n

Detailed Explanation

The problem requires partitioning a given string 's' into the minimum number of substrings, where each substring contains only unique characters. In other words, no character can appear more than once within any single substring. The goal is to find the least number of such substrings needed to cover the entire input string 's'.

Solution Approach

The provided solution employs a greedy approach. It iterates through the string, maintaining a set of characters present in the current substring. For each character, it checks if the character is already in the set. If it is, a new substring is started (the count is incremented), and the current set is reset with only the current character. If the character is not in the set, it's added to the set, extending the current substring. This process continues until the entire string is processed.

Step-by-Step Algorithm

Step 1: Initialize a 'count' variable to 1, representing the initial substring.
Step 2: Initialize an empty set called 'current_set' to store the unique characters of the current substring.
Step 3: Iterate through each character 'char' in the input string 's'.
Step 4: Check if 'char' is already present in 'current_set'.
Step 5: If 'char' is in 'current_set', it means a repeated character is found. Increment 'count' (start a new substring) and reset 'current_set' to contain only the current character 'char'.
Step 6: If 'char' is not in 'current_set', add 'char' to 'current_set'.
Step 7: After iterating through all characters in 's', return the final 'count', representing the minimum number of substrings needed.

Key Insights

Insight 1: The greedy approach of forming a new substring whenever a repeating character is encountered is optimal because it minimizes the length of the current substring before starting a new one. Any other strategy would potentially lead to more substrings.
Insight 2: A set (hash table) is an efficient data structure for tracking the characters present in the current substring, offering O(1) average time complexity for checking character existence and insertion.
Insight 3: There is no need to backtrack or explore different partitioning possibilities; the greedy approach guarantees the optimal solution.

Complexity Analysis

Time Complexity: O(n), where n is the length of the string 's'. The algorithm iterates through the string once.

Space Complexity: O(1). The size of the 'current_set' is at most 26, since there are only 26 lowercase English letters. Therefore, the space used is constant regardless of the input size.

Topics

This problem involves: Hash Table, String, Greedy.

Companies

Asked at: Docusign.

Optimal Partition of String | LeetCode 2405

Medium

Hash Table String Greedy

Time: O(n), where n is the length of the string 's'. The algorithm iterates through the string once.Space: O(1). The size of the 'current_set' is at most 26, since there are only 26 lowercase English letters. Therefore, the space used is constant regardless of the input size.

Solution Code

Available in:

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 2405: Optimal Partition of String - Detailed explanation and solution approaches.

Topics

Hash Table String Greedy

Companies

Docusign

Stats

Difficulty:Medium

Languages:2

Companies:1

Problem #:2405

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize a 'count' variable to 1, representing the initial substring.
2
Step 2: Initialize an empty set called 'current_set' to store the unique characters of the current substring.
3
Step 3: Iterate through each character 'char' in the input string 's'.
4
Step 4: Check if 'char' is already present in 'current_set'.
5
Step 5: If 'char' is in 'current_set', it means a repeated character is found. Increment 'count' (start a new substring) and reset 'current_set' to contain only the current character 'char'.
6
Step 6: If 'char' is not in 'current_set', add 'char' to 'current_set'.
7
Step 7: After iterating through all characters in 's', return the final 'count', representing the minimum number of substrings needed.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n), where n is the length of the string 's'. The algorithm iterates through the string once.

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1). The size of the 'current_set' is at most 26, since there are only 26 lowercase English letters. Therefore, the space used is constant regardless of the input size.

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because each character in the string is processed once. The space complexity is constant because the size of the set is bounded by the number of unique characters (26 lowercase English letters), independent of the string's length.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using an array instead of a set

Instead of using a set, an array of size 26 can be used to track character occurrences in the current substring. The index of the array would correspond to the character (e.g., 'a' -> 0, 'b' -> 1, ..., 'z' -> 25).

Time: O(n), where n is the length of the string s.Space: O(1), as the array size is fixed at 26.

Trade-offs: Using an array might be slightly faster due to potential optimizations in array access compared to set operations, but the difference is usually negligible. The readability might be slightly reduced as character-to-index conversion is necessary.

✨

Key Insights

1Insight 1: The greedy approach of forming a new substring whenever a repeating character is encountered is optimal because it minimizes the length of the current substring before starting a new one. Any other strategy would potentially lead to more substrings.
2Insight 2: A set (hash table) is an efficient data structure for tracking the characters present in the current substring, offering O(1) average time complexity for checking character existence and insertion.
3Insight 3: There is no need to backtrack or explore different partitioning possibilities; the greedy approach guarantees the optimal solution.

📚

Key Terms Explained

For Students

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Greedy Algorithm

Making the best choice at each step without looking ahead.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Empty string. Although the problem statement states that the length of s is at least 1, handling the empty string case would require a check at the beginning and returning 0. The current solution will raise an error as it assumes at least one character.
Edge case 2: String with all same characters (e.g., "ssssss"). The solution handles this case correctly by creating a new substring for each character, resulting in a count equal to the string's length.
Edge case 3: String with all unique characters (e.g., "abcdefg"). The solution creates only one substring, resulting in a count of 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to increment the count when a repeating character is encountered, leading to an incorrect number of substrings.

✗Common mistake 2: Not resetting the 'current_set' after starting a new substring, causing incorrect detection of repeating characters.

✗Common mistake 3: Misunderstanding the greedy approach and attempting to explore different partitioning possibilities, resulting in increased complexity and potential errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data compression. A similar approach can be used to identify repeating patterns in data and compress them by creating smaller, unique chunks.

🏦Real-world scenario 2: Text analysis. Identifying unique word combinations or character sequences in text can be used for sentiment analysis or topic extraction.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

Hash tables trade space for time. Before using one, consider if the problem constraints allow the extra memory. In interviews, always mention this trade-off.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Don't forget to handle key collisions and null/undefined keys. Consider what happens with duplicate values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 5, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Optimal Partition of String"?

For this medium-level Hash Table problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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