Remove Duplicate Letters
Given a string `s`, remove duplicate letters so that every letter appears once and only once. The result must be the smallest in lexical order among all possible results. Use a Monotonic Stack.
Why Interviewers Ask This
Salesforce evaluates this question to assess a candidate's mastery of greedy algorithms and stack-based data structures. It specifically tests the ability to optimize for lexicographical order while maintaining element uniqueness, requiring candidates to balance immediate character selection with future availability constraints.
How to Answer This Question
1. Clarify requirements: Confirm that the output must contain unique characters in their original relative order but be lexicographically smallest possible.
2. Pre-process: Calculate the last occurrence index of every character to know when a character can safely be skipped or removed.
3. Initialize structures: Use a stack for building the result and a boolean set to track currently included characters.
4. Iterate through string: For each character, if it is already in the stack, skip it.
5. Greedy removal: While the stack is not empty, the top character is greater than the current one, and the top character appears later in the string, pop from the stack and mark as excluded.
6. Push current: Add the current character to the stack and mark as included.
7. Construct result: Join the stack to form the final string. Explain time complexity O(n) and space complexity O(1) since the alphabet size is fixed.
Key Points to Cover
- Demonstrating understanding of the trade-off between lexicographical order and character availability
- Correctly using a hash map to store the last occurrence index of each character
- Implementing the monotonic stack logic to greedily remove larger preceding characters
- Ensuring no character is removed if it does not appear again later in the string
- Achieving optimal O(n) time complexity with a single pass through the string
Sample Answer
To solve the 'Remove Duplicate Letters' problem efficiently, I would employ a monotonic stack approach combined with a greedy strategy. First, I need to understand that we want the lexicographically smallest result, which means smaller characters should appear as early as possible, provided they don't violate the constraint of keeping all unique characters.
My approach starts by pre-calculating the last index where each character appears in the input string. This is crucial because it tells us if a character we are considering removing from our stack will reappear later. If it won't, we cannot remove it.
Next, I initialize an empty stack to build the result and a set to keep track of characters currently in the stack. As I iterate through the string, if a character is already in the stack, I skip it to avoid duplicates. However, if it's new, I check the stack's top. If the top character is lexicographically larger than the current character AND the top character exists later in the string, I pop the top character. This ensures we place the smaller character first without losing the larger one permanently.
After processing the entire string, the stack contains the characters in the correct order. Finally, I join them to return the result. This solution runs in O(n) time because each character is pushed and popped at most once, and uses O(1) space relative to the input size, assuming a fixed alphabet size, which aligns well with Salesforce's focus on scalable, efficient solutions.
Common Mistakes to Avoid
- Removing a character from the stack without checking if it appears later, causing permanent data loss
- Failing to track which characters are currently in the stack, leading to duplicate entries in the result
- Using a sorting approach instead of a stack, which ignores the original relative order constraint
- Not handling the case where the current character is already present in the stack correctly
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