[LeetCode 1081] Smallest Subsequence of Distinct Characters

Return the lexicographically smallest subsequence of text that contains all the distinct characters of text exactly once.

Example 1:

Input: "cdadabcc"
Output: "adbc"

Example 2:

Input: "abcd"
Output: "abcd"

Example 3:

Input: "ecbacba"
Output: "eacb"

Example 4:

Input: "leetcode"
Output: "letcod"

 

Constraints:

1. 1 <= text.length <= 1000
2. text consists of lowercase English letters.

 

Key observation: if the current character has not been selected before and is smaller than the previous character(s), then we should try to remove these bigger characters. A special case of this is that a previous character is bigger than the current character but it is the last one of its instances. We can not remove this character because we need each unique character exactly once.

 

Based on this, we have the following algorithm:

1. get character frequency count mapping.

2. go over input string s, for each character, decrease its count by 1. If the current character c has been picked, go to the next character in s. 

3. otherwise, remove all bigger characters that can still be added later from the top of the stack. For each popped character, update its picked boolean to false.

4. add c to the stack and mark its picked boolean to true.

5. after going over s, return all the remaining characters as the final result.

 

 

class Solution {
    public String removeDuplicateLetters(String s) {
        int[] cnt = new int[26];
        boolean[] picked = new boolean[26];
        for(int i = 0; i < s.length(); i++) {
            cnt[s.charAt(i) - 'a']++;
        }
        ArrayDeque<Character> dq = new ArrayDeque<>();
        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            int diff = c - 'a';
            cnt[diff]--;
            if(picked[diff]) {
                continue;
            }
            while(dq.size() > 0 && dq.peekLast() > c && cnt[dq.peekLast() - 'a'] > 0) {
                char rm = dq.pollLast();
                picked[rm - 'a'] = false;
            }
            dq.addLast(c);            
            picked[diff] = true;
        }
        StringBuilder buffer = new StringBuilder();
        while(dq.size() > 0) {
            buffer.append(dq.pollFirst());
        }
        return buffer.toString();
    }
}