Description
Given a string s, return the lexicographically smallest subsequence of s that contains all the distinct characters of s exactly once.
Example 1:
Input: s = "bcabc"
Output: "abc"
Example 2:
Input: s = "cbacdcbc"
Output: "acdb"
Constraints:
1 <= s.length <= 1000s consists of lowercase English letters.
Note: This question is the same as 316:
https://leetcode.com/problems/remove-duplicate-letters/
Solutions
Solution 1
PythonJavaC++GoTypeScript
class Solution:
def smallestSubsequence(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)(code-box)
class Solution {
public String smallestSubsequence(String text) {
int[] cnt = new int[26];
for (char c : text.toCharArray()) {
++cnt[c - 'a'];
}
boolean[] vis = new boolean[26];
char[] cs = new char[text.length()];
int top = -1;
for (char c : text.toCharArray()) {
--cnt[c - 'a'];
if (!vis[c - 'a']) {
while (top >= 0 && c < cs[top] && cnt[cs[top] - 'a'] > 0) {
vis[cs[top--] - 'a'] = false;
}
cs[++top] = c;
vis[c - 'a'] = true;
}
}
return String.valueOf(cs, 0, top + 1);
}
}(code-box)
class Solution {
public:
string smallestSubsequence(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;
}
};(code-box)
func smallestSubsequence(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)
}(code-box)
function smallestSubsequence(s: string): string {
const f = (c: string): number => c.charCodeAt(0) - 'a'.charCodeAt(0);
const last: number[] = new Array(26).fill(0);
for (const [i, c] of [...s].entries()) {
last[f(c)] = i;
}
const stk: string[] = [];
let mask = 0;
for (const [i, c] of [...s].entries()) {
const x = f(c);
if ((mask >> x) & 1) {
continue;
}
while (stk.length && stk[stk.length - 1] > c && last[f(stk[stk.length - 1])] > i) {
mask ^= 1 << f(stk.pop()!);
}
stk.push(c);
mask |= 1 << x;
}
return stk.join('');
}(code-box)
Solution 2
Java
class Solution {
public String smallestSubsequence(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();
}
}(code-box)
