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yanglbmeyanglbme
authoredDec 19, 2023
feat: add solutions to lc problems: No.1781+ (doocs#2125)
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‎solution/1700-1799/1781.Sum of Beauty of All Substrings/README.md

+151
Original file line numberDiff line numberDiff line change
@@ -67,6 +67,30 @@ class Solution:
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return ans
6868
```
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```python
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class Solution:
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def beautySum(self, s: str) -> int:
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ans, n = 0, len(s)
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for i in range(n):
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cnt = Counter()
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freq = Counter()
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mi = mx = 1
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for j in range(i, n):
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freq[cnt[s[j]]] -= 1
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cnt[s[j]] += 1
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freq[cnt[s[j]]] += 1
82+
83+
if cnt[s[j]] == 1:
84+
mi = 1
85+
if freq[mi] == 0:
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mi += 1
87+
if cnt[s[j]] > mx:
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mx = cnt[s[j]]
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90+
ans += mx - mi
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return ans
92+
```
93+
7094
### **Java**
7195

7296
<!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -95,6 +119,38 @@ class Solution {
95119
}
96120
```
97121

122+
```java
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class Solution {
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public int beautySum(String s) {
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int n = s.length();
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int ans = 0;
127+
for (int i = 0; i < n; ++i) {
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int[] cnt = new int[26];
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Map<Integer, Integer> freq = new HashMap<>();
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int mi = 1, mx = 1;
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for (int j = i; j < n; ++j) {
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int k = s.charAt(j) - 'a';
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freq.merge(cnt[k], -1, Integer::sum);
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++cnt[k];
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freq.merge(cnt[k], 1, Integer::sum);
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137+
if (cnt[k] == 1) {
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mi = 1;
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}
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if (freq.getOrDefault(mi, 0) == 0) {
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++mi;
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}
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if (cnt[k] > mx) {
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mx = cnt[k];
145+
}
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ans += mx - mi;
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}
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}
149+
return ans;
150+
}
151+
}
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```
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98154
### **C++**
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100156
```cpp
@@ -123,6 +179,39 @@ public:
123179
};
124180
```
125181
182+
```cpp
183+
class Solution {
184+
public:
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int beautySum(string s) {
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int n = s.size();
187+
int ans = 0;
188+
for (int i = 0; i < n; ++i) {
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int cnt[26]{};
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unordered_map<int, int> freq;
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int mi = 1, mx = 1;
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for (int j = i; j < n; ++j) {
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int k = s[j] - 'a';
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--freq[cnt[k]];
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++cnt[k];
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++freq[cnt[k]];
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198+
if (cnt[k] == 1) {
199+
mi = 1;
200+
}
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if (freq[mi] == 0) {
202+
++mi;
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}
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if (cnt[k] > mx) {
205+
mx = cnt[k];
206+
}
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ans += mx - mi;
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}
209+
}
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return ans;
211+
}
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};
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```
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126215
### **Go**
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128217
```go
@@ -149,6 +238,35 @@ func beautySum(s string) (ans int) {
149238
}
150239
```
151240

241+
```go
242+
func beautySum(s string) (ans int) {
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n := len(s)
244+
for i := 0; i < n; i++ {
245+
cnt := [26]int{}
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freq := map[int]int{}
247+
mi, mx := 1, 1
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for j := i; j < n; j++ {
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k := int(s[j] - 'a')
250+
freq[cnt[k]]--
251+
cnt[k]++
252+
freq[cnt[k]]++
253+
254+
if cnt[k] == 1 {
255+
mi = 1
256+
}
257+
if freq[mi] == 0 {
258+
mi++
259+
}
260+
if cnt[k] > mx {
261+
mx = cnt[k]
262+
}
263+
ans += mx - mi
264+
}
265+
}
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return
267+
}
268+
```
269+
152270
### **JavaScript**
153271

154272
```js
@@ -170,6 +288,39 @@ var beautySum = function (s) {
170288
};
171289
```
172290

291+
```js
292+
/**
293+
* @param {string} s
294+
* @return {number}
295+
*/
296+
var beautySum = function (s) {
297+
const n = s.length;
298+
let ans = 0;
299+
for (let i = 0; i < n; ++i) {
300+
const cnt = Array(26).fill(0);
301+
const freq = new Map();
302+
let [mi, mx] = [1, 1];
303+
for (let j = i; j < n; ++j) {
304+
const k = s[j].charCodeAt() - 97;
305+
freq.set(cnt[k], (freq.get(cnt[k]) || 0) - 1);
306+
++cnt[k];
307+
freq.set(cnt[k], (freq.get(cnt[k]) || 0) + 1);
308+
if (cnt[k] === 1) {
309+
mi = 1;
310+
}
311+
if (freq.get(mi) === 0) {
312+
++mi;
313+
}
314+
if (cnt[k] > mx) {
315+
mx = cnt[k];
316+
}
317+
ans += mx - mi;
318+
}
319+
}
320+
return ans;
321+
};
322+
```
323+
173324
### **...**
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175326
```

‎solution/1700-1799/1781.Sum of Beauty of All Substrings/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -37,6 +37,12 @@
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## Solutions
3939

40+
**Solution 1: Enumeration + Counting**
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Enumerate the starting position $i$ of each substring, find all substrings with the character at this starting position as the left endpoint, then calculate the beauty value of each substring, and accumulate it to the answer.
43+
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The time complexity is $O(n^2 \times C)$, and the space complexity is $O(C)$. Here, $n$ is the length of the string, and $C$ is the size of the character set. In this problem, $C = 26$.
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4046
<!-- tabs:start -->
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4248
### **Python3**
@@ -53,6 +59,30 @@ class Solution:
5359
return ans
5460
```
5561

62+
```python
63+
class Solution:
64+
def beautySum(self, s: str) -> int:
65+
ans, n = 0, len(s)
66+
for i in range(n):
67+
cnt = Counter()
68+
freq = Counter()
69+
mi = mx = 1
70+
for j in range(i, n):
71+
freq[cnt[s[j]]] -= 1
72+
cnt[s[j]] += 1
73+
freq[cnt[s[j]]] += 1
74+
75+
if cnt[s[j]] == 1:
76+
mi = 1
77+
if freq[mi] == 0:
78+
mi += 1
79+
if cnt[s[j]] > mx:
80+
mx = cnt[s[j]]
81+
82+
ans += mx - mi
83+
return ans
84+
```
85+
5686
### **Java**
5787

5888
```java
@@ -79,6 +109,38 @@ class Solution {
79109
}
80110
```
81111

112+
```java
113+
class Solution {
114+
public int beautySum(String s) {
115+
int n = s.length();
116+
int ans = 0;
117+
for (int i = 0; i < n; ++i) {
118+
int[] cnt = new int[26];
119+
Map<Integer, Integer> freq = new HashMap<>();
120+
int mi = 1, mx = 1;
121+
for (int j = i; j < n; ++j) {
122+
int k = s.charAt(j) - 'a';
123+
freq.merge(cnt[k], -1, Integer::sum);
124+
++cnt[k];
125+
freq.merge(cnt[k], 1, Integer::sum);
126+
127+
if (cnt[k] == 1) {
128+
mi = 1;
129+
}
130+
if (freq.getOrDefault(mi, 0) == 0) {
131+
++mi;
132+
}
133+
if (cnt[k] > mx) {
134+
mx = cnt[k];
135+
}
136+
ans += mx - mi;
137+
}
138+
}
139+
return ans;
140+
}
141+
}
142+
```
143+
82144
### **C++**
83145

84146
```cpp
@@ -107,6 +169,39 @@ public:
107169
};
108170
```
109171
172+
```cpp
173+
class Solution {
174+
public:
175+
int beautySum(string s) {
176+
int n = s.size();
177+
int ans = 0;
178+
for (int i = 0; i < n; ++i) {
179+
int cnt[26]{};
180+
unordered_map<int, int> freq;
181+
int mi = 1, mx = 1;
182+
for (int j = i; j < n; ++j) {
183+
int k = s[j] - 'a';
184+
--freq[cnt[k]];
185+
++cnt[k];
186+
++freq[cnt[k]];
187+
188+
if (cnt[k] == 1) {
189+
mi = 1;
190+
}
191+
if (freq[mi] == 0) {
192+
++mi;
193+
}
194+
if (cnt[k] > mx) {
195+
mx = cnt[k];
196+
}
197+
ans += mx - mi;
198+
}
199+
}
200+
return ans;
201+
}
202+
};
203+
```
204+
110205
### **Go**
111206

112207
```go
@@ -133,6 +228,35 @@ func beautySum(s string) (ans int) {
133228
}
134229
```
135230

231+
```go
232+
func beautySum(s string) (ans int) {
233+
n := len(s)
234+
for i := 0; i < n; i++ {
235+
cnt := [26]int{}
236+
freq := map[int]int{}
237+
mi, mx := 1, 1
238+
for j := i; j < n; j++ {
239+
k := int(s[j] - 'a')
240+
freq[cnt[k]]--
241+
cnt[k]++
242+
freq[cnt[k]]++
243+
244+
if cnt[k] == 1 {
245+
mi = 1
246+
}
247+
if freq[mi] == 0 {
248+
mi++
249+
}
250+
if cnt[k] > mx {
251+
mx = cnt[k]
252+
}
253+
ans += mx - mi
254+
}
255+
}
256+
return
257+
}
258+
```
259+
136260
### **JavaScript**
137261

138262
```js
@@ -154,6 +278,39 @@ var beautySum = function (s) {
154278
};
155279
```
156280

281+
```js
282+
/**
283+
* @param {string} s
284+
* @return {number}
285+
*/
286+
var beautySum = function (s) {
287+
const n = s.length;
288+
let ans = 0;
289+
for (let i = 0; i < n; ++i) {
290+
const cnt = Array(26).fill(0);
291+
const freq = new Map();
292+
let [mi, mx] = [1, 1];
293+
for (let j = i; j < n; ++j) {
294+
const k = s[j].charCodeAt() - 97;
295+
freq.set(cnt[k], (freq.get(cnt[k]) || 0) - 1);
296+
++cnt[k];
297+
freq.set(cnt[k], (freq.get(cnt[k]) || 0) + 1);
298+
if (cnt[k] === 1) {
299+
mi = 1;
300+
}
301+
if (freq.get(mi) === 0) {
302+
++mi;
303+
}
304+
if (cnt[k] > mx) {
305+
mx = cnt[k];
306+
}
307+
ans += mx - mi;
308+
}
309+
}
310+
return ans;
311+
};
312+
```
313+
157314
### **...**
158315

159316
```

‎solution/1700-1799/1784.Check if Binary String Has at Most One Segment of Ones/README.md

+1-1
Original file line numberDiff line numberDiff line change
@@ -50,7 +50,7 @@
5050

5151
因此，只需要判断字符串 $s$ 是否存在 "01" 串即可。
5252

53-
时间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
53+
时间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。
5454

5555
<!-- tabs:start -->
5656

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