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 <= 104s 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 $s$.
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| 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)
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| 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();
}
}
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| 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;
}
};
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| 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)
}
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Solution 2#
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| 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)
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| 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)
}
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