What is the time complexity of Remove Duplicate Letters?

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

What data structures are used to solve Remove Duplicate Letters?

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

What are the key insights for solving Remove Duplicate Letters?

Insight 1: The crucial insight is that we can use a stack to maintain a candidate substring and a greedy approach to decide whether to keep or remove characters from the stack to maintain lexicographical order. Insight 2: We need to know the last occurrence of each character in the input string to make informed decisions about removing characters from the stack. This allows us to ensure that we don't remove a character prematurely if it will appear later in the string.

Which companies ask LeetCode 316. Remove Duplicate Letters in interviews?

This problem is frequently asked at ByteDance, DRW, Expedia, FactSet, Goldman Sachs.

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

Remove Duplicate Letters - LeetCode 316 Solution

Q: How to solve Remove Duplicate Letters?

The solution uses a stack and a greedy approach. It iterates through the input string, and for each character, it checks if the character is already in the stack. If not, it compares the character with the character at the top of the stack. If the current character is smaller than the character at the top of the stack, and the character at the top of the stack appears later in the string (based on its last occurrence), then the character at the top of the stack is removed. This process is repeated until the stack is empty, or the conditions for removing the top character are not met. Finally, the current character is added to the stack. A 'seen' set is used to track characters currently in the stack.

Remove Duplicate Letters - Complete Solution Guide

Remove Duplicate Letters is LeetCode problem 316, 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 , remove duplicate letters so that every letter appears once and only once. You must make sure your result is the smallest in lexicographical order among all possible results. Example 1: Input: s = "bcabc" Output: "abc" Example 2: Input: s = "cbacdcbc" Output: "acdb" Constraints: 1 <= s.length <= 10^4 s consists of lowercase English letters. Note: This question is the same as 1081: https://leetcode.com/problems/smallest-subsequence-of-distinct-characters/

Detailed Explanation

The problem asks us to remove duplicate letters from a given string `s` such that each letter appears only once. Additionally, we need to ensure that the resulting string is the smallest in lexicographical order among all possible results. Lexicographical order refers to dictionary order. The input is a string `s` consisting of lowercase English letters. The output should be the string with unique characters in lexicographical order.

Solution Approach

The solution uses a stack and a greedy approach. It iterates through the input string, and for each character, it checks if the character is already in the stack. If not, it compares the character with the character at the top of the stack. If the current character is smaller than the character at the top of the stack, and the character at the top of the stack appears later in the string (based on its last occurrence), then the character at the top of the stack is removed. This process is repeated until the stack is empty, or the conditions for removing the top character are not met. Finally, the current character is added to the stack. A 'seen' set is used to track characters currently in the stack.

Step-by-Step Algorithm

Step 1: Create a `last` array/map to store the last occurrence index of each character in the string `s`.
Step 2: Initialize an empty stack to store the candidate substring.
Step 3: Initialize a `seen` set/array to track the characters currently in the stack.
Step 4: Iterate through the string `s` from left to right.
Step 5: For each character `c`, if `c` is already in `seen`, skip to the next character.
Step 6: While the stack is not empty, `c` is smaller than the top of the stack, and the index `i` of `c` is less than the last occurrence of the top of the stack in `s`, remove the top of the stack from the stack and `seen`.
Step 7: Add `c` to the stack and `seen`.
Step 8: After iterating through the entire string, the stack contains the desired substring. Join the characters in the stack to form the result string.

Key Insights

Insight 1: The crucial insight is that we can use a stack to maintain a candidate substring and a greedy approach to decide whether to keep or remove characters from the stack to maintain lexicographical order.
Insight 2: We need to know the last occurrence of each character in the input string to make informed decisions about removing characters from the stack. This allows us to ensure that we don't remove a character prematurely if it will appear later in the string.
Insight 3: The `seen` set/array is critical to prevent re-adding letters to the stack, ensuring each letter appears only once.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

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

Companies

Asked at: ByteDance, DRW, Expedia, FactSet, Goldman Sachs.

Remove Duplicate Letters | LeetCode 316

Medium

Stack Greedy String Monotonic Stack

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

Solution Code

Available in:

import java.util.*;
class Solution {
    public String removeDuplicateLetters(String s) {
        int[] last = new int[26];
        for (int i = 0; i < s.length(); i++) last[s.charAt(i) - 'a'] = i;
        boolean[] seen = new boolean[26];
        Deque<Character> stack = new ArrayDeque<>();
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (seen[c - 'a']) continue;
            while (!stack.isEmpty() && c < stack.peek() && i < last[stack.peek() - 'a']) {
                seen[stack.pop() - 'a'] = false;
            }
            stack.push(c);
            seen[c - 'a'] = true;
        }
        StringBuilder sb = new StringBuilder();
        for (char c : stack) sb.append(c);
        return sb.reverse().toString();
    }
}

View Full Solution on GitHubGitHub Solution View on LeetCodeLeetCode

Problem Info

Description

Problem 316: Remove Duplicate Letters - Detailed explanation and solution approaches.

Topics

Stack Greedy String Monotonic Stack

Companies

ByteDanceDRWExpediaFactSetGoldman Sachs

Stats

Difficulty:Medium

Languages:4

Companies:5

Problem #:316

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: Create a `last` array/map to store the last occurrence index of each character in the string `s`.
2
Step 2: Initialize an empty stack to store the candidate substring.
3
Step 3: Initialize a `seen` set/array to track the characters currently in the stack.
4
Step 4: Iterate through the string `s` from left to right.
5
Step 5: For each character `c`, if `c` is already in `seen`, skip to the next character.
6
Step 6: While the stack is not empty, `c` is smaller than the top of the stack, and the index `i` of `c` is less than the last occurrence of the top of the stack in `s`, remove the top of the stack from the stack and `seen`.
7
Step 7: Add `c` to the stack and `seen`.
8
Step 8: After iterating through the entire string, the stack contains the desired substring. Join the characters in the stack to form the result string.

💡

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 outer loop iterates through the string once. The inner while loop, although nested, performs at most a constant amount of work per character of the string because a character can be pushed and popped from the stack only once. The space complexity is O(1) because the size of the stack and the other auxiliary data structures (`last` and `seen`) are bounded by the constant number of characters in the English alphabet.

📖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

A recursive approach with backtracking could explore all possible subsequences of the string and choose the lexicographically smallest one that contains all unique characters. However, this approach can be much less efficient.

Time: O(2^n), where n is the length of the string (in the worst case).Space: O(n), due to the recursion depth.

Trade-offs: This approach is less efficient and generally not suitable for larger strings due to its exponential time complexity. It's conceptually simpler to understand for beginners but computationally expensive.

✨

Key Insights

1Insight 1: The crucial insight is that we can use a stack to maintain a candidate substring and a greedy approach to decide whether to keep or remove characters from the stack to maintain lexicographical order.
2Insight 2: We need to know the last occurrence of each character in the input string to make informed decisions about removing characters from the stack. This allows us to ensure that we don't remove a character prematurely if it will appear later in the string.
3Insight 3: The `seen` set/array is critical to prevent re-adding letters to the stack, ensuring each letter appears only once.

📚

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 is empty, the algorithm correctly returns an empty string since the loop never executes.
Edge case 2: String with all duplicate letters: For example, 'aaaa'. The algorithm handles it correctly, returning 'a'.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the last occurrence of a character. Removing a character from the stack without considering if it will appear again later in the string can lead to incorrect results.

✗Common mistake 2: Failing to maintain lexicographical order. Adding characters to the stack without comparing them to existing characters can violate the constraint of finding the smallest lexicographical string.

✅

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: Removing redundant characters in a string while preserving information, relevant in basic forms of compression.

🏦Real-world scenario 2: Optimizing search queries: Removing redundant terms in a search query to improve efficiency, while ensuring the query's meaning is preserved. Though this is a simplified analogy as query optimization is far more complex.

💡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

Stacks are perfect for problems involving matching (parentheses) or "next greater element" patterns. If you need to look back at recent elements, think stack.

🎯 Interview Tip

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

⚠️ Watch Out

Popping from an empty stack. Always check if the stack is non-empty before popping.

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: October 15, 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 14

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

@InterviewProJul 7

❓

Frequently Asked Questions

What problems are best solved with a stack?

Stacks excel at: matching problems (parentheses validation), next greater/smaller element, expression evaluation, backtracking, and undo operations. The key pattern is LIFO (Last In, First Out) - when you need to access the most recently added element.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Remove Duplicate Letters"?

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

🤖 Need help with this problem?