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feat: add solutions to lc problem: No.1930
No.1930.Unique Length-3 Palindromic Subsequences
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‎solution/1900-1999/1930.Unique Length-3 Palindromic Subsequences/README.md

+72-56
Original file line numberDiff line numberDiff line change
@@ -87,7 +87,7 @@ tags:
8787

8888
枚举结束后，即可得到答案。
8989

90-
时间复杂度 $O(n \times C)$，空间复杂度 $O(C)$，其中 $n$ 为字符串长度，而 $C$ 为字符集大小。本题中 $C = 26$。
90+
时间复杂度 $O(n \times |\Sigma|)$，其中 $n$ 为字符串长度，而 $\Sigma$ 为字符集大小，本题中 $|\Sigma| = 26$。空间复杂度 $O(|\Sigma|)$ 或 $O(1)$。
9191

9292
<!-- tabs:start -->
9393

@@ -112,11 +112,14 @@ class Solution {
112112
int ans = 0;
113113
for (char c = 'a'; c <= 'z'; ++c) {
114114
int l = s.indexOf(c), r = s.lastIndexOf(c);
115-
Set<Character> cs = new HashSet<>();
115+
int mask = 0;
116116
for (int i = l + 1; i < r; ++i) {
117-
cs.add(s.charAt(i));
117+
int j = s.charAt(i) - 'a';
118+
if ((mask >> j & 1) == 0) {
119+
mask |= 1 << j;
120+
++ans;
121+
}
118122
}
119-
ans += cs.size();
120123
}
121124
return ans;
122125
}
@@ -132,9 +135,14 @@ public:
132135
int ans = 0;
133136
for (char c = 'a'; c <= 'z'; ++c) {
134137
int l = s.find_first_of(c), r = s.find_last_of(c);
135-
unordered_set<char> cs;
136-
for (int i = l + 1; i < r; ++i) cs.insert(s[i]);
137-
ans += cs.size();
138+
int mask = 0;
139+
for (int i = l + 1; i < r; ++i) {
140+
int j = s[i] - 'a';
141+
if (mask >> j & 1 ^ 1) {
142+
mask |= 1 << j;
143+
++ans;
144+
}
145+
}
138146
}
139147
return ans;
140148
}
@@ -147,80 +155,88 @@ public:
147155
func countPalindromicSubsequence(s string) (ans int) {
148156
for c := 'a'; c <= 'z'; c++ {
149157
l, r := strings.Index(s, string(c)), strings.LastIndex(s, string(c))
150-
cs := map[byte]struct{}{}
158+
mask := 0
151159
for i := l + 1; i < r; i++ {
152-
cs[s[i]] = struct{}{}
160+
j := int(s[i] - 'a')
161+
if mask>>j&1 == 0 {
162+
mask |= 1 << j
163+
ans++
164+
}
153165
}
154-
ans += len(cs)
155166
}
156167
return
157168
}
158169
```
159170

160-
#### C#
161-
162-
```cs
163-
public class Solution {
164-
public int CountPalindromicSubsequence(string s) {
165-
int ans = 0;
166-
for (char c = 'a'; c <= 'z'; ++c) {
167-
int l = s.IndexOf(c), r = s.LastIndexOf(c);
168-
HashSet<char> cs = new HashSet<char>();
169-
for (int i = l + 1; i < r; ++i) {
170-
cs.Add(s[i]);
171-
}
172-
ans += cs.Count;
173-
}
174-
return ans;
175-
}
176-
}
177-
```
178-
179171
#### TypeScript
180172

181173
```ts
182-
export function countPalindromicSubsequence(s: string): number {
183-
const cnt = new Map<string, [number, number]>();
184-
const n = s.length;
174+
function countPalindromicSubsequence(s: string): number {
185175
let ans = 0;
186-
187-
for (let i = 0; i < n; i++) {
188-
const ch = s[i];
189-
if (cnt.has(ch)) cnt.get(ch)![1] = i;
190-
else cnt.set(ch, [i, i]);
191-
}
192-
193-
for (const [_, [i, j]] of cnt) {
194-
if (i !== j) {
195-
ans += new Set(s.slice(i + 1, j)).size;
176+
const a = 'a'.charCodeAt(0);
177+
for (let ch = 0; ch < 26; ++ch) {
178+
const c = String.fromCharCode(ch + a);
179+
const l = s.indexOf(c);
180+
const r = s.lastIndexOf(c);
181+
let mask = 0;
182+
for (let i = l + 1; i < r; ++i) {
183+
const j = s.charCodeAt(i) - a;
184+
if (((mask >> j) & 1) ^ 1) {
185+
mask |= 1 << j;
186+
++ans;
187+
}
196188
}
197189
}
198-
199190
return ans;
200191
}
201192
```
202193

203194
#### JavaScript
204195

205196
```js
206-
export function countPalindromicSubsequence(s) {
207-
const cnt = new Map();
208-
const n = s.length;
197+
/**
198+
* @param {string} s
199+
* @return {number}
200+
*/
201+
var countPalindromicSubsequence = function (s) {
209202
let ans = 0;
210-
211-
for (let i = 0; i < n; i++) {
212-
const ch = s[i];
213-
if (cnt.has(ch)) cnt.get(ch)[1] = i;
214-
else cnt.set(ch, [i, i]);
203+
const a = 'a'.charCodeAt(0);
204+
for (let ch = 0; ch < 26; ++ch) {
205+
const c = String.fromCharCode(ch + a);
206+
const l = s.indexOf(c);
207+
const r = s.lastIndexOf(c);
208+
let mask = 0;
209+
for (let i = l + 1; i < r; ++i) {
210+
const j = s.charCodeAt(i) - a;
211+
if (((mask >> j) & 1) ^ 1) {
212+
mask |= 1 << j;
213+
++ans;
214+
}
215+
}
215216
}
217+
return ans;
218+
};
219+
```
216220

217-
for (const [_, [i, j]] of cnt) {
218-
if (i !== j) {
219-
ans += new Set(s.slice(i + 1, j)).size;
221+
#### C#
222+
223+
```cs
224+
public class Solution {
225+
public int CountPalindromicSubsequence(string s) {
226+
int ans = 0;
227+
for (char c = 'a'; c <= 'z'; ++c) {
228+
int l = s.IndexOf(c), r = s.LastIndexOf(c);
229+
int mask = 0;
230+
for (int i = l + 1; i < r; ++i) {
231+
int j = s[i] - 'a';
232+
if ((mask >> j & 1) == 0) {
233+
mask |= 1 << j;
234+
++ans;
235+
}
236+
}
220237
}
238+
return ans;
221239
}
222-
223-
return ans;
224240
}
225241
```
226242

‎solution/1900-1999/1930.Unique Length-3 Palindromic Subsequences/README_EN.md

+78-56
Original file line numberDiff line numberDiff line change
@@ -79,7 +79,13 @@ tags:
7979

8080
<!-- solution:start -->
8181

82-
### Solution 1
82+
### Solution 1: Enumerate Both End Characters + Hash Table
83+
84+
Since the string contains only lowercase letters, we can directly enumerate all pairs of end characters. For each pair of end characters $c$, we find their first and last occurrence positions $l$ and $r$ in the string. If $r - l > 1$, it means we have found a palindromic subsequence that meets the conditions. We then count the number of unique characters between $[l+1,..r-1]$, which gives the number of palindromic subsequences with $c$ as the end characters, and add it to the answer.
85+
86+
After enumerating all pairs, we get the answer.
87+
88+
The time complexity is $O(n \times |\Sigma|)$, where $n$ is the length of the string and $\Sigma$ is the size of the character set. In this problem, $|\Sigma| = 26$. The space complexity is $O(|\Sigma|)$ or $O(1)$.
8389

8490
<!-- tabs:start -->
8591

@@ -104,11 +110,14 @@ class Solution {
104110
int ans = 0;
105111
for (char c = 'a'; c <= 'z'; ++c) {
106112
int l = s.indexOf(c), r = s.lastIndexOf(c);
107-
Set<Character> cs = new HashSet<>();
113+
int mask = 0;
108114
for (int i = l + 1; i < r; ++i) {
109-
cs.add(s.charAt(i));
115+
int j = s.charAt(i) - 'a';
116+
if ((mask >> j & 1) == 0) {
117+
mask |= 1 << j;
118+
++ans;
119+
}
110120
}
111-
ans += cs.size();
112121
}
113122
return ans;
114123
}
@@ -124,9 +133,14 @@ public:
124133
int ans = 0;
125134
for (char c = 'a'; c <= 'z'; ++c) {
126135
int l = s.find_first_of(c), r = s.find_last_of(c);
127-
unordered_set<char> cs;
128-
for (int i = l + 1; i < r; ++i) cs.insert(s[i]);
129-
ans += cs.size();
136+
int mask = 0;
137+
for (int i = l + 1; i < r; ++i) {
138+
int j = s[i] - 'a';
139+
if (mask >> j & 1 ^ 1) {
140+
mask |= 1 << j;
141+
++ans;
142+
}
143+
}
130144
}
131145
return ans;
132146
}
@@ -139,80 +153,88 @@ public:
139153
func countPalindromicSubsequence(s string) (ans int) {
140154
for c := 'a'; c <= 'z'; c++ {
141155
l, r := strings.Index(s, string(c)), strings.LastIndex(s, string(c))
142-
cs := map[byte]struct{}{}
156+
mask := 0
143157
for i := l + 1; i < r; i++ {
144-
cs[s[i]] = struct{}{}
158+
j := int(s[i] - 'a')
159+
if mask>>j&1 == 0 {
160+
mask |= 1 << j
161+
ans++
162+
}
145163
}
146-
ans += len(cs)
147164
}
148165
return
149166
}
150167
```
151168

152-
#### C#
153-
154-
```cs
155-
public class Solution {
156-
public int CountPalindromicSubsequence(string s) {
157-
int ans = 0;
158-
for (char c = 'a'; c <= 'z'; ++c) {
159-
int l = s.IndexOf(c), r = s.LastIndexOf(c);
160-
HashSet<char> cs = new HashSet<char>();
161-
for (int i = l + 1; i < r; ++i) {
162-
cs.Add(s[i]);
163-
}
164-
ans += cs.Count;
165-
}
166-
return ans;
167-
}
168-
}
169-
```
170-
171169
#### TypeScript
172170

173171
```ts
174-
export function countPalindromicSubsequence(s: string): number {
175-
const cnt = new Map<string, [number, number]>();
176-
const n = s.length;
172+
function countPalindromicSubsequence(s: string): number {
177173
let ans = 0;
178-
179-
for (let i = 0; i < n; i++) {
180-
const ch = s[i];
181-
if (cnt.has(ch)) cnt.get(ch)![1] = i;
182-
else cnt.set(ch, [i, i]);
183-
}
184-
185-
for (const [_, [i, j]] of cnt) {
186-
if (i !== j) {
187-
ans += new Set(s.slice(i + 1, j)).size;
174+
const a = 'a'.charCodeAt(0);
175+
for (let ch = 0; ch < 26; ++ch) {
176+
const c = String.fromCharCode(ch + a);
177+
const l = s.indexOf(c);
178+
const r = s.lastIndexOf(c);
179+
let mask = 0;
180+
for (let i = l + 1; i < r; ++i) {
181+
const j = s.charCodeAt(i) - a;
182+
if (((mask >> j) & 1) ^ 1) {
183+
mask |= 1 << j;
184+
++ans;
185+
}
188186
}
189187
}
190-
191188
return ans;
192189
}
193190
```
194191

195192
#### JavaScript
196193

197194
```js
198-
export function countPalindromicSubsequence(s) {
199-
const cnt = new Map();
200-
const n = s.length;
195+
/**
196+
* @param {string} s
197+
* @return {number}
198+
*/
199+
var countPalindromicSubsequence = function (s) {
201200
let ans = 0;
202-
203-
for (let i = 0; i < n; i++) {
204-
const ch = s[i];
205-
if (cnt.has(ch)) cnt.get(ch)[1] = i;
206-
else cnt.set(ch, [i, i]);
201+
const a = 'a'.charCodeAt(0);
202+
for (let ch = 0; ch < 26; ++ch) {
203+
const c = String.fromCharCode(ch + a);
204+
const l = s.indexOf(c);
205+
const r = s.lastIndexOf(c);
206+
let mask = 0;
207+
for (let i = l + 1; i < r; ++i) {
208+
const j = s.charCodeAt(i) - a;
209+
if (((mask >> j) & 1) ^ 1) {
210+
mask |= 1 << j;
211+
++ans;
212+
}
213+
}
207214
}
215+
return ans;
216+
};
217+
```
218+
219+
#### C#
208220

209-
for (const [_, [i, j]] of cnt) {
210-
if (i !== j) {
211-
ans += new Set(s.slice(i + 1, j)).size;
221+
```cs
222+
public class Solution {
223+
public int CountPalindromicSubsequence(string s) {
224+
int ans = 0;
225+
for (char c = 'a'; c <= 'z'; ++c) {
226+
int l = s.IndexOf(c), r = s.LastIndexOf(c);
227+
int mask = 0;
228+
for (int i = l + 1; i < r; ++i) {
229+
int j = s[i] - 'a';
230+
if ((mask >> j & 1) == 0) {
231+
mask |= 1 << j;
232+
++ans;
233+
}
234+
}
212235
}
236+
return ans;
213237
}
214-
215-
return ans;
216238
}
217239
```
218240

‎solution/1900-1999/1930.Unique Length-3 Palindromic Subsequences/Solution.cpp

+9-4
Original file line numberDiff line numberDiff line change
@@ -4,10 +4,15 @@ class Solution {
44
int ans = 0;
55
for (char c = 'a'; c <= 'z'; ++c) {
66
int l = s.find_first_of(c), r = s.find_last_of(c);
7-
unordered_set<char> cs;
8-
for (int i = l + 1; i < r; ++i) cs.insert(s[i]);
9-
ans += cs.size();
7+
int mask = 0;
8+
for (int i = l + 1; i < r; ++i) {
9+
int j = s[i] - 'a';
10+
if (mask >> j & 1 ^ 1) {
11+
mask |= 1 << j;
12+
++ans;
13+
}
14+
}
1015
}
1116
return ans;
1217
}
13-
};
18+
};

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