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feat: add solutions to lc problem: No.1933 (#2000)
No.1933.Check if String Is Decomposable Into Value-Equal Substrings
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‎solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README.md‎

Lines changed: 49 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -61,7 +61,7 @@
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遍历结束后，判断是否出现过长度为 $2$ 的子字符串，若没有，返回 `false`，否则返回 `true`。
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64-
时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串 $s$ 的长度。
64+
时间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。
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6666
<!-- tabs:start -->
6767

@@ -87,6 +87,20 @@ class Solution:
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return cnt2 == 1
8888
```
8989

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```python
91+
class Solution:
92+
def isDecomposable(self, s: str) -> bool:
93+
cnt2 = 0
94+
for _, g in groupby(s):
95+
m = len(list(g))
96+
if m % 3 == 1:
97+
return False
98+
cnt2 += m % 3 == 2
99+
if cnt2 > 1:
100+
return False
101+
return cnt2 == 1
102+
```
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90104
### **Java**
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92106
<!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -120,13 +134,19 @@ class Solution {
120134
class Solution {
121135
public:
122136
bool isDecomposable(string s) {
123-
int i = 0, n = s.size();
124137
int cnt2 = 0;
125-
while (i < n) {
138+
for (int i = 0, n = s.size(); i < n;) {
126139
int j = i;
127-
while (j < n && s[j] == s[i]) ++j;
128-
if ((j - i) % 3 == 1) return false;
129-
if ((j - i) % 3 == 2 && ++cnt2 > 1) return false;
140+
while (j < n && s[j] == s[i]) {
141+
++j;
142+
}
143+
if ((j - i) % 3 == 1) {
144+
return false;
145+
}
146+
cnt2 += (j - i) % 3 == 2;
147+
if (cnt2 > 1) {
148+
return false;
149+
}
130150
i = j;
131151
}
132152
return cnt2 == 1;
@@ -160,6 +180,29 @@ func isDecomposable(s string) bool {
160180
}
161181
```
162182

183+
### **TypeScript**
184+
185+
```ts
186+
function isDecomposable(s: string): boolean {
187+
const n = s.length;
188+
let cnt2 = 0;
189+
for (let i = 0; i < n; ) {
190+
let j = i;
191+
while (j < n && s[j] === s[i]) {
192+
++j;
193+
}
194+
if ((j - i) % 3 === 1) {
195+
return false;
196+
}
197+
if ((j - i) % 3 === 2 && ++cnt2 > 1) {
198+
return false;
199+
}
200+
i = j;
201+
}
202+
return cnt2 === 1;
203+
}
204+
```
205+
163206
### **...**
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165208
```

‎solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README_EN.md‎

Lines changed: 56 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -52,6 +52,14 @@
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## Solutions
5454

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**Solution 1: Two Pointers**
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We traverse the string $s$, using two pointers $i$ and $j$ to count the length of each equal substring. If the length modulo $3$ is $1$, it means that the length of this substring does not meet the requirements, so we return `false`. If the length modulo $3$ is $2$, it means that a substring of length $2$ has appeared. If a substring of length $2$ has appeared before, return `false`, otherwise assign the value of $j$ to $i$ and continue to traverse.
58+
59+
After the traversal, check whether a substring of length $2$ has appeared. If not, return `false`, otherwise return `true`.
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61+
The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
@@ -74,6 +82,20 @@ class Solution:
7482
return cnt2 == 1
7583
```
7684

85+
```python
86+
class Solution:
87+
def isDecomposable(self, s: str) -> bool:
88+
cnt2 = 0
89+
for _, g in groupby(s):
90+
m = len(list(g))
91+
if m % 3 == 1:
92+
return False
93+
cnt2 += m % 3 == 2
94+
if cnt2 > 1:
95+
return False
96+
return cnt2 == 1
97+
```
98+
7799
### **Java**
78100

79101
```java
@@ -105,13 +127,19 @@ class Solution {
105127
class Solution {
106128
public:
107129
bool isDecomposable(string s) {
108-
int i = 0, n = s.size();
109130
int cnt2 = 0;
110-
while (i < n) {
131+
for (int i = 0, n = s.size(); i < n;) {
111132
int j = i;
112-
while (j < n && s[j] == s[i]) ++j;
113-
if ((j - i) % 3 == 1) return false;
114-
if ((j - i) % 3 == 2 && ++cnt2 > 1) return false;
133+
while (j < n && s[j] == s[i]) {
134+
++j;
135+
}
136+
if ((j - i) % 3 == 1) {
137+
return false;
138+
}
139+
cnt2 += (j - i) % 3 == 2;
140+
if (cnt2 > 1) {
141+
return false;
142+
}
115143
i = j;
116144
}
117145
return cnt2 == 1;
@@ -145,6 +173,29 @@ func isDecomposable(s string) bool {
145173
}
146174
```
147175

176+
### **TypeScript**
177+
178+
```ts
179+
function isDecomposable(s: string): boolean {
180+
const n = s.length;
181+
let cnt2 = 0;
182+
for (let i = 0; i < n; ) {
183+
let j = i;
184+
while (j < n && s[j] === s[i]) {
185+
++j;
186+
}
187+
if ((j - i) % 3 === 1) {
188+
return false;
189+
}
190+
if ((j - i) % 3 === 2 && ++cnt2 > 1) {
191+
return false;
192+
}
193+
i = j;
194+
}
195+
return cnt2 === 1;
196+
}
197+
```
198+
148199
### **...**
149200

150201
```

‎solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution.cpp‎

Lines changed: 20 additions & 14 deletions
Original file line numberDiff line numberDiff line change
@@ -1,15 +1,21 @@
1-
class Solution {
2-
public:
3-
bool isDecomposable(string s) {
4-
int i = 0, n = s.size();
5-
int cnt2 = 0;
6-
while (i < n) {
7-
int j = i;
8-
while (j < n && s[j] == s[i]) ++j;
9-
if ((j - i) % 3 == 1) return false;
10-
if ((j - i) % 3 == 2 && ++cnt2 > 1) return false;
11-
i = j;
12-
}
13-
return cnt2 == 1;
14-
}
1+
class Solution {
2+
public:
3+
bool isDecomposable(string s) {
4+
int cnt2 = 0;
5+
for (int i = 0, n = s.size(); i < n;) {
6+
int j = i;
7+
while (j < n && s[j] == s[i]) {
8+
++j;
9+
}
10+
if ((j - i) % 3 == 1) {
11+
return false;
12+
}
13+
cnt2 += (j - i) % 3 == 2;
14+
if (cnt2 > 1) {
15+
return false;
16+
}
17+
i = j;
18+
}
19+
return cnt2 == 1;
20+
}
1521
};

‎solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution.ts‎

Lines changed: 18 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,18 @@
1+
function isDecomposable(s: string): boolean {
2+
const n = s.length;
3+
let cnt2 = 0;
4+
for (let i = 0; i < n; ) {
5+
let j = i;
6+
while (j < n && s[j] === s[i]) {
7+
++j;
8+
}
9+
if ((j - i) % 3 === 1) {
10+
return false;
11+
}
12+
if ((j - i) % 3 === 2 && ++cnt2 > 1) {
13+
return false;
14+
}
15+
i = j;
16+
}
17+
return cnt2 === 1;
18+
}

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