Question

316. Remove Duplicate Letters

中文文档

Description

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/

Solutions

Solution 1: Stack

We use an array last to record the last occurrence of each character, a stack to save the result string, and an array vis or an integer variable mask to record whether the current character is in the stack.

Traverse the string $s$, for each character $c$, if $c$ is not in the stack, we need to check whether the top element of the stack is greater than $c$. If it is greater than $c$ and the top element of the stack will appear later, we pop the top element of the stack and push $c$ into the stack.

Finally, concatenate the elements in the stack into a string and return it as the result.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the string $s$.

class Solution:
  def removeDuplicateLetters(self, s: str) -> str:
    last = {c: i for i, c in enumerate(s)}
    stk = []
    vis = set()
    for i, c in enumerate(s):
      if c in vis:
        continue
      while stk and stk[-1] > c and last[stk[-1]] > i:
        vis.remove(stk.pop())
      stk.append(c)
      vis.add(c)
    return ''.join(stk)

class Solution {
  public String removeDuplicateLetters(String s) {
    int n = s.length();
    int[] last = new int[26];
    for (int i = 0; i < n; ++i) {
      last[s.charAt(i) - 'a'] = i;
    }
    Deque<Character> stk = new ArrayDeque<>();
    int mask = 0;
    for (int i = 0; i < n; ++i) {
      char c = s.charAt(i);
      if (((mask >> (c - 'a')) & 1) == 1) {
        continue;
      }
      while (!stk.isEmpty() && stk.peek() > c && last[stk.peek() - 'a'] > i) {
        mask ^= 1 << (stk.pop() - 'a');
      }
      stk.push(c);
      mask |= 1 << (c - 'a');
    }
    StringBuilder ans = new StringBuilder();
    for (char c : stk) {
      ans.append(c);
    }
    return ans.reverse().toString();
  }
}

class Solution {
public:
  string removeDuplicateLetters(string s) {
    int n = s.size();
    int last[26] = {0};
    for (int i = 0; i < n; ++i) {
      last[s[i] - 'a'] = i;
    }
    string ans;
    int mask = 0;
    for (int i = 0; i < n; ++i) {
      char c = s[i];
      if ((mask >> (c - 'a')) & 1) {
        continue;
      }
      while (!ans.empty() && ans.back() > c && last[ans.back() - 'a'] > i) {
        mask ^= 1 << (ans.back() - 'a');
        ans.pop_back();
      }
      ans.push_back(c);
      mask |= 1 << (c - 'a');
    }
    return ans;
  }
};

func removeDuplicateLetters(s string) string {
  last := make([]int, 26)
  for i, c := range s {
    last[c-'a'] = i
  }
  stk := []rune{}
  vis := make([]bool, 128)
  for i, c := range s {
    if vis[c] {
      continue
    }
    for len(stk) > 0 && stk[len(stk)-1] > c && last[stk[len(stk)-1]-'a'] > i {
      vis[stk[len(stk)-1]] = false
      stk = stk[:len(stk)-1]
    }
    stk = append(stk, c)
    vis[c] = true
  }
  return string(stk)
}

Solution 2

class Solution:
  def removeDuplicateLetters(self, s: str) -> str:
    count, in_stack = [0] * 128, [False] * 128
    stack = []
    for c in s:
      count[ord(c)] += 1
    for c in s:
      count[ord(c)] -= 1
      if in_stack[ord(c)]:
        continue
      while len(stack) and stack[-1] > c:
        peek = stack[-1]
        if count[ord(peek)] < 1:
          break
        in_stack[ord(peek)] = False
        stack.pop()
      stack.append(c)
      in_stack[ord(c)] = True
    return ''.join(stack)

func removeDuplicateLetters(s string) string {
  count, in_stack, stack := make([]int, 128), make([]bool, 128), make([]rune, 0)
  for _, c := range s {
    count[c] += 1
  }

  for _, c := range s {
    count[c] -= 1
    if in_stack[c] {
      continue
    }
    for len(stack) > 0 && stack[len(stack)-1] > c && count[stack[len(stack)-1]] > 0 {
      peek := stack[len(stack)-1]
      stack = stack[0 : len(stack)-1]
      in_stack[peek] = false
    }
    stack = append(stack, c)
    in_stack[c] = true
  }
  return string(stack)
}