Skip to content

Navigation Menu

Sign in
Appearance settings

Search code, repositories, users, issues, pull requests...

Provide feedback

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly
Sign up
Appearance settings

Commit ad5dea2

Browse files
feat: add TypeScript/Rust solutions to lc problem: No.0828 (doocs#2016)
No.0828.Count Unique Characters of All Substrings of a Given String
1 parent cc5d541 commit ad5dea2

File tree

5 files changed

+139
-11
lines changed

5 files changed

+139
-11
lines changed

‎solution/0800-0899/0828.Count Unique Characters of All Substrings of a Given String/README.md‎

Lines changed: 51 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -56,9 +56,15 @@
5656

5757
**方法一:计算每个字符的贡献**
5858

59-
对于字符串 `s` 的每个字符 $c_i,ドル当它在某个子字符串中仅出现一次时,它会对这个子字符串统计唯一字符时有贡献。只需对每个字符 $c_i,ドル计算有多少子字符串仅包含该字符一次即可。
59+
对于字符串 $s$ 的每个字符 $c_i,ドル当它在某个子字符串中仅出现一次时,它会对这个子字符串统计唯一字符时有贡献。
6060

61-
时间复杂度 $O(n),ドル其中 $n$ 为字符串 `s` 的长度。
61+
因此,我们只需要对每个字符 $c_i,ドル计算有多少子字符串仅包含该字符一次即可。
62+
63+
我们用一个哈希表或者长度为 26ドル$ 的数组 $d,ドル按照下标顺序存储每个字符在 $s$ 中所有出现的位置。
64+
65+
对于每个字符 $c_i,ドル我们遍历 $d[c_i]$ 中的每个位置 $p,ドル找出左侧相邻的位置 $l$ 和右侧相邻的位置 $r,ドル那么从位置 $p$ 向左右两边扩散,满足要求的子字符串的数量就是 $(p - l) \times (r - p)$。我们对每个字符都进行这样的操作,累加所有字符的贡献,即可得到答案。
66+
67+
时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
6268

6369
<!-- tabs:start -->
6470

@@ -132,22 +138,62 @@ public:
132138
### **Go**
133139
134140
```go
135-
func uniqueLetterString(s string) int {
141+
func uniqueLetterString(s string) (ans int) {
136142
d := make([][]int, 26)
137143
for i := range d {
138144
d[i] = []int{-1}
139145
}
140146
for i, c := range s {
141147
d[c-'A'] = append(d[c-'A'], i)
142148
}
143-
ans := 0
144149
for _, v := range d {
145150
v = append(v, len(s))
146151
for i := 1; i < len(v)-1; i++ {
147152
ans += (v[i] - v[i-1]) * (v[i+1] - v[i])
148153
}
149154
}
150-
return ans
155+
return
156+
}
157+
```
158+
159+
### **TypeScript**
160+
161+
```ts
162+
function uniqueLetterString(s: string): number {
163+
const d: number[][] = Array.from({ length: 26 }, () => [-1]);
164+
for (let i = 0; i < s.length; ++i) {
165+
d[s.charCodeAt(i) - 'A'.charCodeAt(0)].push(i);
166+
}
167+
let ans = 0;
168+
for (const v of d) {
169+
v.push(s.length);
170+
171+
for (let i = 1; i < v.length - 1; ++i) {
172+
ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]);
173+
}
174+
}
175+
return ans;
176+
}
177+
```
178+
179+
### **Rust**
180+
181+
```rust
182+
impl Solution {
183+
pub fn unique_letter_string(s: String) -> i32 {
184+
let mut d: Vec<Vec<i32>> = vec![vec![-1; 1]; 26];
185+
for (i, c) in s.chars().enumerate() {
186+
d[(c as usize) - ('A' as usize)].push(i as i32);
187+
}
188+
let mut ans = 0;
189+
for v in d.iter_mut() {
190+
v.push(s.len() as i32);
191+
for i in 1..v.len() - 1 {
192+
ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]);
193+
}
194+
}
195+
ans as i32
196+
}
151197
}
152198
```
153199

‎solution/0800-0899/0828.Count Unique Characters of All Substrings of a Given String/README_EN.md‎

Lines changed: 55 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -50,6 +50,18 @@ Sum of lengths of all substring is 1 +たす 1 +たす 1 +たす 2 +たす 2 +たす 3 =わ 10
5050

5151
## Solutions
5252

53+
**Solution 1: Calculate the Contribution of Each Character**
54+
55+
For each character $c_i$ in the string $s,ドル when it appears only once in a substring, it contributes to the count of unique characters in that substring.
56+
57+
Therefore, we only need to calculate for each character $c_i,ドル how many substrings contain this character only once.
58+
59+
We use a hash table or an array $d$ of length 26ドル,ドル to store the positions of each character in $s$ in order of index.
60+
61+
For each character $c_i,ドル we iterate through each position $p$ in $d[c_i],ドル find the adjacent positions $l$ on the left and $r$ on the right, then the number of substrings that meet the requirements by expanding from position $p$ to both sides is $(p - l) \times (r - p)$. We perform this operation for each character, add up the contributions of all characters, and get the answer.
62+
63+
The time complexity is $O(n),ドル and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.
64+
5365
<!-- tabs:start -->
5466

5567
### **Python3**
@@ -118,22 +130,62 @@ public:
118130
### **Go**
119131
120132
```go
121-
func uniqueLetterString(s string) int {
133+
func uniqueLetterString(s string) (ans int) {
122134
d := make([][]int, 26)
123135
for i := range d {
124136
d[i] = []int{-1}
125137
}
126138
for i, c := range s {
127139
d[c-'A'] = append(d[c-'A'], i)
128140
}
129-
ans := 0
130141
for _, v := range d {
131142
v = append(v, len(s))
132143
for i := 1; i < len(v)-1; i++ {
133144
ans += (v[i] - v[i-1]) * (v[i+1] - v[i])
134145
}
135146
}
136-
return ans
147+
return
148+
}
149+
```
150+
151+
### **TypeScript**
152+
153+
```ts
154+
function uniqueLetterString(s: string): number {
155+
const d: number[][] = Array.from({ length: 26 }, () => [-1]);
156+
for (let i = 0; i < s.length; ++i) {
157+
d[s.charCodeAt(i) - 'A'.charCodeAt(0)].push(i);
158+
}
159+
let ans = 0;
160+
for (const v of d) {
161+
v.push(s.length);
162+
163+
for (let i = 1; i < v.length - 1; ++i) {
164+
ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]);
165+
}
166+
}
167+
return ans;
168+
}
169+
```
170+
171+
### **Rust**
172+
173+
```rust
174+
impl Solution {
175+
pub fn unique_letter_string(s: String) -> i32 {
176+
let mut d: Vec<Vec<i32>> = vec![vec![-1; 1]; 26];
177+
for (i, c) in s.chars().enumerate() {
178+
d[(c as usize) - ('A' as usize)].push(i as i32);
179+
}
180+
let mut ans = 0;
181+
for v in d.iter_mut() {
182+
v.push(s.len() as i32);
183+
for i in 1..v.len() - 1 {
184+
ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]);
185+
}
186+
}
187+
ans as i32
188+
}
137189
}
138190
```
139191

Lines changed: 2 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -1,17 +1,16 @@
1-
func uniqueLetterString(s string) int {
1+
func uniqueLetterString(s string) (ansint) {
22
d := make([][]int, 26)
33
for i := range d {
44
d[i] = []int{-1}
55
}
66
for i, c := range s {
77
d[c-'A'] = append(d[c-'A'], i)
88
}
9-
ans := 0
109
for _, v := range d {
1110
v = append(v, len(s))
1211
for i := 1; i < len(v)-1; i++ {
1312
ans += (v[i] - v[i-1]) * (v[i+1] - v[i])
1413
}
1514
}
16-
returnans
15+
return
1716
}
Lines changed: 16 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,16 @@
1+
impl Solution {
2+
pub fn unique_letter_string(s: String) -> i32 {
3+
let mut d: Vec<Vec<i32>> = vec![vec![-1; 1]; 26];
4+
for (i, c) in s.chars().enumerate() {
5+
d[(c as usize) - ('A' as usize)].push(i as i32);
6+
}
7+
let mut ans = 0;
8+
for v in d.iter_mut() {
9+
v.push(s.len() as i32);
10+
for i in 1..v.len() - 1 {
11+
ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]);
12+
}
13+
}
14+
ans as i32
15+
}
16+
}
Lines changed: 15 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
1+
function uniqueLetterString(s: string): number {
2+
const d: number[][] = Array.from({ length: 26 }, () => [-1]);
3+
for (let i = 0; i < s.length; ++i) {
4+
d[s.charCodeAt(i) - 'A'.charCodeAt(0)].push(i);
5+
}
6+
let ans = 0;
7+
for (const v of d) {
8+
v.push(s.length);
9+
10+
for (let i = 1; i < v.length - 1; ++i) {
11+
ans += (v[i] - v[i - 1]) * (v[i + 1] - v[i]);
12+
}
13+
}
14+
return ans;
15+
}

0 commit comments

Comments
(0)

AltStyle によって変換されたページ (->オリジナル) /