Introduction

In this blog post, we will explore how to solve the LeetCode problem "316" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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 <= 104 s consists of lowercase English letters. Note: This question is the same as 1081: https://leetcode.com/problems/smallest-subsequence-of-distinct-characters/

Explanation

Here's the breakdown of the approach, complexity, and the Python code:

• Core Idea: Use a stack to maintain a potentially optimal subsequence. Iterate through the string, and for each character, check if it's already in the stack. If not, while the stack is not empty, the current character is lexicographically smaller than the top of the stack, and the top of the stack appears later in the string, pop the stack. Then, push the current character onto the stack.
• Key Data Structures: A stack to store the result and a counter (or last occurrence map) to track character frequencies.

• Lexicographical Optimization: Popping from the stack when encountering a smaller character is the key to ensuring the smallest lexicographical order.

• Complexity: O(N) time complexity, O(1) space complexity (since we only store lowercase English letters which is a fixed size).

Code

    def removeDuplicateLetters(s: str) -> str:
    """
    Removes duplicate letters from a string such that every letter appears once
    and the result is the smallest in lexicographical order.
    """
    last_occurrence = {}
    for i, char in enumerate(s):
        last_occurrence[char] = i

    stack = []
    seen = set()

    for i, char in enumerate(s):
        if char not in seen:
            while stack and char < stack[-1] and i < last_occurrence[stack[-1]]:
                seen.remove(stack.pop())
            stack.append(char)
            seen.add(char)

    return "".join(stack)