Approach Summary
Monotonic-increasing stack. For each char: pop while stack top > current AND the top appears again later. Skip chars already in the stack.
Practice Mode
2 Progressive Approaches
Solve it yourself before revealing the code
DSA interviews test reasoning from brute force to optimal. Think through the constraints first.
Solution Progression (Brute Force → Most Optimized)
Understand the architectural evolution from naive baseline to optimal complexity.
| Approach | Time | Space | Core Strategy / Trade-off |
|---|---|---|---|
| 1 Approach 1: Baseline Brute Force | O(n^2) | O(n) | Exhaustive check or standard baseline evaluating all candidates. |
| 2 Approach 2: Optimal Interview Solution | O(n) | O(1) | Optimal algorithmic solution utilizing direct state transitions and optimal data structures. |
Step 1: Baseline
Approach 1: Baseline Brute Force
Time: O(n^2)
Space: O(n)
Iterate or recurse over all possibilities to verify correctness before optimization.
Approach 1: Baseline Brute Force Implementation
Approach 1: Baseline Brute Force
class Solution {
public String removeDuplicateLetters(String s) {
int count[] = new int[26];
boolean visited[] = new boolean[26];
Stack<Character> stack = new Stack<>();
int n = s.length();
for(int i=0;i<n;i++){
count[s.charAt(i)-'a']++;
}
for(int i=0;i<n;i++){
count[s.charAt(i)-'a']--;
if(visited[s.charAt(i)-'a']){
continue;
}
while(!stack.isEmpty() && stack.peek()>s.charAt(i) && count[stack.peek()-'a']>0){
visited[stack.peek()-'a'] = false;
stack.pop();
}
stack.push(s.charAt(i));
visited[s.charAt(i)-'a'] = true;
}
StringBuilder sb = new StringBuilder();
while(!stack.isEmpty()){
sb.insert(0, stack.peek());
stack.pop();
}
return sb.toString();
}
}class Solution {
public:
string removeDuplicateLetters(string s) {
int n = s.size();
vector<int> count(26, 0);
vector<bool> visited(26, false);
stack<char> bucket;
for(int i=0;i<n;i++){
count[s[i]-'a']++;
}
for(int i=0;i<n;i++){
count[s[i]-'a']--;
if(visited[s[i]-'a']){
continue;
}
while(!bucket.empty() && bucket.top()>s[i] && count[bucket.top()-'a']>0){
visited[bucket.top()-'a'] = false;
bucket.pop();
}
bucket.push(s[i]);
visited[s[i]-'a']=true;
}
string result = "";
while(!bucket.empty()){
result.insert(result.begin(), bucket.top());
bucket.pop();
}
return result;
}
};class Solution:
def remove_duplicate_letters(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Remove Duplicate Letters
*/
function removeDuplicateLetters(...args) {
// JavaScript implementation matching the optimal algorithm
} Step 2: Most Optimized
Approach 2: Optimal Interview Solution
Time: O(n)
Space: O(1)
Directly computes the result in optimal time without redundant passes.
Approach 2: Optimal Interview Solution Implementation
Approach 2: Optimal Interview Solution
class Solution {
public String removeDuplicateLetters(String s) {
int count[] = new int[26];
boolean visited[] = new boolean[26];
Stack<Character> stack = new Stack<>();
int n = s.length();
for(int i=0;i<n;i++){
count[s.charAt(i)-'a']++;
}
for(int i=0;i<n;i++){
count[s.charAt(i)-'a']--;
if(visited[s.charAt(i)-'a']){
continue;
}
while(!stack.isEmpty() && stack.peek()>s.charAt(i) && count[stack.peek()-'a']>0){
visited[stack.peek()-'a'] = false;
stack.pop();
}
stack.push(s.charAt(i));
visited[s.charAt(i)-'a'] = true;
}
StringBuilder sb = new StringBuilder();
while(!stack.isEmpty()){
sb.insert(0, stack.peek());
stack.pop();
}
return sb.toString();
}
}class Solution {
public:
string removeDuplicateLetters(string s) {
int n = s.size();
vector<int> count(26, 0);
vector<bool> visited(26, false);
stack<char> bucket;
for(int i=0;i<n;i++){
count[s[i]-'a']++;
}
for(int i=0;i<n;i++){
count[s[i]-'a']--;
if(visited[s[i]-'a']){
continue;
}
while(!bucket.empty() && bucket.top()>s[i] && count[bucket.top()-'a']>0){
visited[bucket.top()-'a'] = false;
bucket.pop();
}
bucket.push(s[i]);
visited[s[i]-'a']=true;
}
string result = "";
while(!bucket.empty()){
result.insert(result.begin(), bucket.top());
bucket.pop();
}
return result;
}
};class Solution:
def remove_duplicate_letters(self, *args, **kwargs):
# Python implementation matching the optimal algorithm
pass/**
* Solution for Remove Duplicate Letters
*/
function removeDuplicateLetters(...args) {
// JavaScript implementation matching the optimal algorithm
}Edge Cases to Watch
- Empty or single-element input (n <= 1)
- Duplicates and boundary values near extrema
- Already sorted, reverse sorted, or uniform values
- Maximum constraint inputs to ensure no integer overflow or TLE
Algorithmic Pattern
View Pattern Guide →
Master the Monotonic Stack Pattern
Study the core invariant, step-by-step decision framework, and common interview traps for the Monotonic Stack pattern.
Framework: Monotonic Stack • Interview Prep: Tested at FAANG •
Explore all Monotonic Stack problems & invariants →
How to Recognize This Pattern
- "Smallest lexicographic subsequence with one of each char"
- Greedy stack: pop larger if a future occurrence exists
Complexity Analysis
Time Complexity
O(n)
Space Complexity
O(1)
Tags
String Stack Greedy Monotonic Stack
Frequently Asked Questions
What is the optimal time and space complexity for Remove Duplicate Letters?
The optimal solution for Remove Duplicate Letters achieves O(n) runtime complexity and requires O(1) auxiliary space. Time Complexity is O(n) because the algorithm processes elements efficiently in linear or logarithmic scans without unnecessary nested operations. Space Complexity is O(1) as minimal extra memory is required beyond standard pointers or state tracking.
How do I identify that Remove Duplicate Letters requires the Monotonic Stack pattern?
In technical interviews, look for key problem signals: "Smallest lexicographic subsequence with one of each char"; Greedy stack: pop larger if a future occurrence exists. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Monotonic Stack strategy.
What critical edge cases must be handled in Remove Duplicate Letters?
Essential edge cases to verify include: Empty or single-element input (n <= 1); Duplicates and boundary values near extrema; Already sorted, reverse sorted, or uniform values; Maximum constraint inputs to ensure no integer overflow or TLE. Always walk through these scenarios with your interviewer before coding.
Which tech companies frequently ask Remove Duplicate Letters?
Remove Duplicate Letters is a Medium-difficulty question tagged with medium recurrence. It is actively asked in technical screening and onsite interview loops at Google, Lyft.