What is the time complexity of Partition Labels?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Partition Labels?

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

What are the key insights for solving Partition Labels?

Insight 1: The key is to find the last occurrence of each character in the string. This helps define the boundaries of each partition. Insight 2: Iterate through the string while keeping track of the 'end' of the current partition. The 'end' is the furthest right position of any character encountered so far in the current partition.

Which companies ask LeetCode 763. Partition Labels in interviews?

This problem is frequently asked at InMobi, Sprinklr.

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Partition Labels - LeetCode 763 Solution

Partition Labels - Complete Solution Guide

Partition Labels is LeetCode problem 763, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a string s . We want to partition the string into as many parts as possible so that each letter appears in at most one part. For example, the string "ababcc" can be partitioned into ["abab", "cc"] , but partitions such as ["aba", "bcc"] or ["ab", "ab", "cc"] are invalid. Note that the partition is done so that after concatenating all the parts in order, the resultant string should be s . Return a list of integers representing the size of these parts . Example 1: Input: s = "ababcba

Detailed Explanation

The problem asks us to divide a given string `s` into the maximum number of partitions such that each character appears in at most one partition. The goal is to return a list of integers where each integer represents the size of a partition. This means we want to find the smallest possible partitions while ensuring no character is split across different partitions. For instance, if a character 'a' appears in indices 0 and 5, any partition that includes index 0 must also include index 5, and all indices in between. The input is a string `s` consisting of lowercase English letters, and the output is a list of the sizes of the resulting partitions.

Solution Approach

The solution uses a greedy approach. First, it calculates the last index of each character in the string using a hash map (or an array if we can assume only lowercase letters). Then, it iterates through the string, maintaining a `start` and `end` pointer for the current partition. For each character, the `end` pointer is updated to the maximum of its current value and the last index of the character encountered. If the current index matches the `end`, it means we've reached the end of a valid partition. We record the size of this partition (end - start + 1) and update the `start` to the beginning of the next potential partition.

Step-by-Step Algorithm

Step 1: Create a data structure (hash map/array) `last_indices` to store the last occurrence index of each character in the string `s`.
Step 2: Initialize an empty list `result` to store the sizes of the partitions.
Step 3: Initialize `start` to 0, representing the start index of the current partition.
Step 4: Initialize `end` to 0, representing the end index of the current partition.
Step 5: Iterate through the string `s` with index `i`.
Step 6: For each character `s[i]`, update `end` to `max(end, last_indices[s[i]])`. This extends the current partition to include all occurrences of `s[i]`.
Step 7: If `i == end`, it means we have reached the end of the current partition. Add `end - start + 1` (the size of the partition) to the `result` list.
Step 8: Update `start` to `i + 1` to start a new partition.
Step 9: Return the `result` list.

Key Insights

Insight 1: The key is to find the last occurrence of each character in the string. This helps define the boundaries of each partition.
Insight 2: Iterate through the string while keeping track of the 'end' of the current partition. The 'end' is the furthest right position of any character encountered so far in the current partition.
Insight 3: When the current index `i` matches the `end` of the partition, it means all characters within the current partition only appear within that partition, so it's time to create a new partition.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Hash Table, Two Pointers, String, Greedy.

Companies

Asked at: InMobi, Sprinklr.

Partition Labels | LeetCode 763

Medium

Hash Table Two Pointers String Greedy

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {
    public List<Integer> partitionLabels(String s) {
        Map<Character, Integer> last_indices = new HashMap<>();
        for (int i = 0; i < s.length(); i++) {
            last_indices.put(s.charAt(i), i);
        }

        List<Integer> result = new ArrayList<>();
        int start = 0;
        int end = 0;

        for (int i = 0; i < s.length(); i++) {
            end = Math.max(end, last_indices.get(s.charAt(i)));
            if (i == end) {
                result.add(end - start + 1);
                start = i + 1;
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 763: Partition Labels - Detailed explanation and solution approaches.

Topics

Hash Table Two Pointers String Greedy

Companies

InMobiSprinklr

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:763

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

9 steps

1
Step 1: Create a data structure (hash map/array) `last_indices` to store the last occurrence index of each character in the string `s`.
2
Step 2: Initialize an empty list `result` to store the sizes of the partitions.
3
Step 3: Initialize `start` to 0, representing the start index of the current partition.
4
Step 4: Initialize `end` to 0, representing the end index of the current partition.
5
Step 5: Iterate through the string `s` with index `i`.
6
Step 6: For each character `s[i]`, update `end` to `max(end, last_indices[s[i]])`. This extends the current partition to include all occurrences of `s[i]`.
7
Step 7: If `i == end`, it means we have reached the end of the current partition. Add `end - start + 1` (the size of the partition) to the `result` list.
8
Step 8: Update `start` to `i + 1` to start a new partition.
9
Step 9: Return the `result` list.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(n) because we loop through the string a fixed number of times (twice). The space complexity is O(1) because the maximum number of unique characters is fixed (26 lowercase English letters), thus the size of the `last_indices` map/array is also constant.

📖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

Brute Force

Generate all possible partitions of the string and check if each character appears in at most one part for each partition. Select the partition with the most parts.

Time: ExponentialSpace: O(n) to store the partitions

Trade-offs: This approach is highly inefficient and not practical due to its exponential time complexity. It's useful for understanding the problem but not for implementation.

✨

Key Insights

1Insight 1: The key is to find the last occurrence of each character in the string. This helps define the boundaries of each partition.
2Insight 2: Iterate through the string while keeping track of the 'end' of the current partition. The 'end' is the furthest right position of any character encountered so far in the current partition.
3Insight 3: When the current index `i` matches the `end` of the partition, it means all characters within the current partition only appear within that partition, so it's time to create a new partition.

📚

Key Terms Explained

For Students

Hash Table

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

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. The code correctly handles it by returning an empty list because the loops won't execute.
Edge case 2: String with only one distinct character, e.g., "aaaaa". The algorithm will create a single partition of size 5, which is the correct answer.
Edge case 3: String with all distinct characters, e.g., "abcdefg". The algorithm will create partitions of size 1 for each character.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to update the `end` pointer when a character with a further last index is encountered within a partition.

✗Common mistake 2: Not handling the edge case of a string with only one unique character correctly. If the loop condition or logic isn't carefully constructed, this can lead to incorrect partition sizes.

✗Common mistake 3: Incorrectly calculating partition size (off-by-one error). It should be `end - start + 1`.

✅

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: Task scheduling. Suppose you have tasks that depend on each other, and each task has a deadline. You need to group these tasks into batches so that all dependencies are satisfied within a batch. This problem is similar to partitioning the string based on character dependencies.

🏦Real-world scenario 2: Data compression. In some data compression algorithms, we might want to group repeating sequences together to improve compression ratios. The partition labels problem can be adapted to find the optimal grouping of these sequences.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

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

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: July 14, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

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

@DevExpert92Jun 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.

How do I avoid infinite loops with two pointers?

Always ensure your loop condition guarantees termination (typically left < right or slow != fast). Move at least one pointer in each iteration, and handle edge cases where pointers might not move (like skipping duplicates). Draw out pointer movement on paper first.

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 "Partition Labels"?

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 two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?