What is the time complexity of Smallest Subsequence of Distinct Characters?

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

What data structures are used to solve Smallest Subsequence of Distinct Characters?

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

What are the key insights for solving Smallest Subsequence of Distinct Characters?

Insight 1: The problem can be solved using a monotonic stack. The stack helps to maintain the potential smallest subsequence while iterating through the string. Insight 2: The `last_occurrence` map is crucial. It allows us to determine if a character currently in the stack can be removed because it appears later in the string.

Which companies ask LeetCode 1081. Smallest Subsequence of Distinct Characters in interviews?

This problem is frequently asked at ByteDance, FactSet.

JavaScript is required for the full interactive experience. However, all content is accessible without JavaScript.

Smallest Subsequence of Distinct Characters - LeetCode 1081 Solution

Q: How to solve Smallest Subsequence of Distinct Characters?

The solution uses a greedy approach with a stack. We iterate through the input string `s`. For each character, we check if it's already present in our stack (meaning it's already in our potential subsequence). If not, we compare the current character with the top of the stack. If the current character is lexicographically smaller than the top of the stack, AND the character at the top of the stack occurs later in the string, we can pop the top of the stack. We repeat this popping process until either the stack is empty or the conditions are no longer met. Then, we push the current character onto the stack and mark it as 'seen'. Finally, the stack contains the characters of the smallest subsequence in the correct order, which we then convert to a string.

Smallest Subsequence of Distinct Characters - Complete Solution Guide

Smallest Subsequence of Distinct Characters is LeetCode problem 1081, 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

Given a string s , return the lexicographically smallest subsequence of s that contains all the distinct characters of s exactly once . Example 1: Input: s = "bcabc" Output: "abc" Example 2: Input: s = "cbacdcbc" Output: "acdb" Constraints: 1 <= s.length <= 1000 s consists of lowercase English letters. Note: This question is the same as 316: https://leetcode.com/problems/remove-duplicate-letters/

Detailed Explanation

The problem asks us to find the lexicographically smallest subsequence of a given string `s` that contains all distinct characters of `s` exactly once. A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. Lexicographically smallest means that among all possible subsequences satisfying the condition, we want the one that comes first in dictionary order. For example, 'abc' is lexicographically smaller than 'acb'. The input is a string `s` consisting of lowercase English letters, and the output is the smallest subsequence.

Solution Approach

The solution uses a greedy approach with a stack. We iterate through the input string `s`. For each character, we check if it's already present in our stack (meaning it's already in our potential subsequence). If not, we compare the current character with the top of the stack. If the current character is lexicographically smaller than the top of the stack, AND the character at the top of the stack occurs later in the string, we can pop the top of the stack. We repeat this popping process until either the stack is empty or the conditions are no longer met. Then, we push the current character onto the stack and mark it as 'seen'. Finally, the stack contains the characters of the smallest subsequence in the correct order, which we then convert to a string.

Step-by-Step Algorithm

Step 1: Initialize a hash map `last_occurrence` to store the last index of each character in the string `s`. This is done in a single pass through the string.
Step 2: Initialize an empty stack `stack` to store the characters of the potential smallest subsequence.
Step 3: Initialize an empty set `seen` to keep track of which characters are already in the stack.
Step 4: Iterate through the string `s` from left to right.
Step 5: For each character `char` at index `i`, check if `char` is already in the `seen` set. If so, skip to the next character.
Step 6: While the stack is not empty, the current character `char` is less than the character at the top of the stack, and the last occurrence of the character at the top of the stack is greater than the current index `i`, pop the character from the stack and remove it from the `seen` set.
Step 7: Push the current character `char` onto the stack and add it to the `seen` set.
Step 8: After iterating through the entire string, convert the stack to a string and return it.

Key Insights

Insight 1: The problem can be solved using a monotonic stack. The stack helps to maintain the potential smallest subsequence while iterating through the string.
Insight 2: The `last_occurrence` map is crucial. It allows us to determine if a character currently in the stack can be removed because it appears later in the string.
Insight 3: The `seen` set is used to ensure each distinct character appears only once in the result.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Stack, Greedy, Monotonic Stack.

Companies

Asked at: ByteDance, FactSet.

Smallest Subsequence of Distinct Characters | LeetCode 1081

Medium

String Stack Greedy Monotonic Stack

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

Solution Code

Available in:

import java.util.HashMap;
import java.util.Stack;
import java.util.HashSet;

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

        Stack<Character> stack = new Stack<>();
        HashSet<Character> seen = new HashSet<>();

        for (int i = 0; i < s.length(); i++) {
            char charAtI = s.charAt(i);
            if (seen.contains(charAtI)) {
                continue;
            }

            while (!stack.isEmpty() && charAtI < stack.peek() && last_occurrence.get(stack.peek()) > i) {
                seen.remove(stack.pop());
            }

            stack.push(charAtI);
            seen.add(charAtI);
        }

        StringBuilder sb = new StringBuilder();
        for (char c : stack) {
            sb.append(c);
        }

        return sb.toString();
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 1081: Smallest Subsequence of Distinct Characters - Detailed explanation and solution approaches.

Topics

String Stack Greedy Monotonic Stack

Companies

ByteDanceFactSet

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1081

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

8 steps

1
Step 1: Initialize a hash map `last_occurrence` to store the last index of each character in the string `s`. This is done in a single pass through the string.
2
Step 2: Initialize an empty stack `stack` to store the characters of the potential smallest subsequence.
3
Step 3: Initialize an empty set `seen` to keep track of which characters are already in the stack.
4
Step 4: Iterate through the string `s` from left to right.
5
Step 5: For each character `char` at index `i`, check if `char` is already in the `seen` set. If so, skip to the next character.
6
Step 6: While the stack is not empty, the current character `char` is less than the character at the top of the stack, and the last occurrence of the character at the top of the stack is greater than the current index `i`, pop the character from the stack and remove it from the `seen` set.
7
Step 7: Push the current character `char` onto the stack and add it to the `seen` set.
8
Step 8: After iterating through the entire string, convert the stack to a string and return it.

💡

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 the main loop iterates through the string once, and the while loop pops characters from the stack. Although the while loop is nested, each character can only be pushed onto the stack and popped off the stack once. Thus, the while loop operations take at most O(n) time in total. The space complexity is O(1) because the size of `last_occurrence`, `stack`, and `seen` is bounded by the number of lowercase letters (26), which is a 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

Recursion with Backtracking

We can potentially use recursion to explore different subsequences. However, we need to ensure that we only consider distinct characters and find the lexicographically smallest subsequence.

Time: O(2^n), where n is the length of the string `s`. This is because, in the worst case, we might need to explore all possible subsequences.Space: O(n), due to the recursion depth.

Trade-offs: Recursion with backtracking is generally less efficient than the stack-based solution, especially for larger input strings. It is more suitable when the input string size is relatively small. The main advantage of recursion might be in its readability for some, but in this case, the iterative approach is pretty easy to grasp.

✨

Key Insights

1Insight 1: The problem can be solved using a monotonic stack. The stack helps to maintain the potential smallest subsequence while iterating through the string.
2Insight 2: The `last_occurrence` map is crucial. It allows us to determine if a character currently in the stack can be removed because it appears later in the string.
3Insight 3: The `seen` set is used to ensure each distinct character appears only once in the result.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

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: If the input string `s` is empty, the code will not enter the loop and return an empty string, which is the correct output.
Edge case 2: String with only one unique character: If the input string contains only one unique character, the stack will contain only that character, and the code will return the string with that character, which is also correct.
Edge case 3: String with duplicate characters: The algorithm correctly handles strings with duplicate characters by maintaining the lexicographically smallest subsequence containing all distinct characters exactly once.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using the `last_occurrence` map correctly. It is essential to check if a character on the stack appears later in the string before popping it.

✗Common mistake 2: Forgetting to check if a character is already in the stack (`seen` set) before adding it.

✗Common mistake 3: Not handling the lexicographical order correctly, which can lead to a non-optimal subsequence.

✅

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: Optimizing data compression algorithms. This algorithm can be adapted to reduce the size of data by finding the smallest representation that retains essential information.

🏦Real-world scenario 2: Designing efficient search algorithms. The concept of finding the smallest subsequence can be applied to prioritize the most relevant search results based on certain criteria.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: June 2, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

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

@CodeNinjaJun 29

❓

Frequently Asked Questions

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.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

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 "Smallest Subsequence of Distinct Characters"?

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

🤖 Need help with this problem?