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feat: add solutions to lc problem: No.1933
No.1933.Check if String Is Decomposable Into Value-Equal Substrings
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‎solution/0800-0899/0870.Advantage Shuffle/README.md

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Original file line numberDiff line numberDiff line change
@@ -56,8 +56,7 @@
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class Solution:
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def advantageCount(self, nums1: List[int], nums2: List[int]) -> List[int]:
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nums1.sort()
59-
t = [(v, i) for i, v in enumerate(nums2)]
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t.sort()
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t = sorted((v, i) for i, v in enumerate(nums2))
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n = len(nums2)
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ans = [0] * n
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i, j = 0, n - 1

‎solution/0800-0899/0870.Advantage Shuffle/README_EN.md

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class Solution:
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def advantageCount(self, nums1: List[int], nums2: List[int]) -> List[int]:
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nums1.sort()
38-
t = [(v, i) for i, v in enumerate(nums2)]
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t.sort()
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t = sorted((v, i) for i, v in enumerate(nums2))
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n = len(nums2)
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ans = [0] * n
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i, j = 0, n - 1

‎solution/0800-0899/0870.Advantage Shuffle/Solution.py

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@@ -1,8 +1,7 @@
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class Solution:
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def advantageCount(self, nums1: List[int], nums2: List[int]) -> List[int]:
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nums1.sort()
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t = [(v, i) for i, v in enumerate(nums2)]
5-
t.sort()
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t = sorted((v, i) for i, v in enumerate(nums2))
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n = len(nums2)
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ans = [0] * n
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i, j = 0, n - 1

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

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@@ -55,22 +55,109 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：双指针**
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遍历字符串 $s$，用双指针 $i$ 和 $j$ 统计每个等值子字符串的长度。若长度模 $3$ 余 $1$，说明该子字符串长度不符合要求，返回 `false`；若长度模 $3$ 余 $2$，说明出现了长度为 $2$ 的子字符串，若此前已经出现过长度为 $2$ 的子字符串，返回 `false`，否则将 $j$ 的值赋给 $i$，继续遍历。
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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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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。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
65-
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class Solution:
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def isDecomposable(self, s: str) -> bool:
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i, n = 0, len(s)
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cnt2 = 0
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while i < n:
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j = i
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while j < n and s[j] == s[i]:
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j += 1
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if (j - i) % 3 == 1:
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return False
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cnt2 += (j - i) % 3 == 2
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if cnt2 > 1:
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return False
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i = j
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return cnt2 == 1
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public boolean isDecomposable(String s) {
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int i = 0, n = s.length();
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int cnt2 = 0;
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while (i < n) {
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int j = i;
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while (j < n && s.charAt(j) == s.charAt(i)) {
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++j;
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}
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if ((j - i) % 3 == 1) {
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return false;
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}
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if ((j - i) % 3 == 2 && ++cnt2 > 1) {
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return false;
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}
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i = j;
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}
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return cnt2 == 1;
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}
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}
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```
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### **C++**
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```cpp
120+
class Solution {
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public:
122+
bool isDecomposable(string s) {
123+
int i = 0, n = s.size();
124+
int cnt2 = 0;
125+
while (i < n) {
126+
int j = i;
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while (j < n && s[j] == s[i]) ++j;
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if ((j - i) % 3 == 1) return false;
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if ((j - i) % 3 == 2 && ++cnt2 > 1) return false;
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i = j;
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}
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return cnt2 == 1;
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}
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};
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```
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### **Go**
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```go
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func isDecomposable(s string) bool {
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i, n := 0, len(s)
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cnt2 := 0
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for i < n {
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j := i
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for j < n && s[j] == s[i] {
146+
j++
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}
148+
if (j-i)%3 == 1 {
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return false
150+
}
151+
if (j-i)%3 == 2 {
152+
cnt2++
153+
if cnt2 > 1 {
154+
return false
155+
}
156+
}
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i = j
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}
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return cnt2 == 1
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}
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```
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### **...**

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

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@@ -57,13 +57,92 @@
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### **Python3**
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```python
60-
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class Solution:
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def isDecomposable(self, s: str) -> bool:
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i, n = 0, len(s)
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cnt2 = 0
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while i < n:
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j = i
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while j < n and s[j] == s[i]:
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j += 1
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if (j - i) % 3 == 1:
69+
return False
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cnt2 += (j - i) % 3 == 2
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if cnt2 > 1:
72+
return False
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i = j
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return cnt2 == 1
6175
```
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### **Java**
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```java
80+
class Solution {
81+
public boolean isDecomposable(String s) {
82+
int i = 0, n = s.length();
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int cnt2 = 0;
84+
while (i < n) {
85+
int j = i;
86+
while (j < n && s.charAt(j) == s.charAt(i)) {
87+
++j;
88+
}
89+
if ((j - i) % 3 == 1) {
90+
return false;
91+
}
92+
if ((j - i) % 3 == 2 && ++cnt2 > 1) {
93+
return false;
94+
}
95+
i = j;
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}
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return cnt2 == 1;
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}
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}
100+
```
101+
102+
### **C++**
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```cpp
105+
class Solution {
106+
public:
107+
bool isDecomposable(string s) {
108+
int i = 0, n = s.size();
109+
int cnt2 = 0;
110+
while (i < n) {
111+
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;
115+
i = j;
116+
}
117+
return cnt2 == 1;
118+
}
119+
};
120+
```
66121
122+
### **Go**
123+
124+
```go
125+
func isDecomposable(s string) bool {
126+
i, n := 0, len(s)
127+
cnt2 := 0
128+
for i < n {
129+
j := i
130+
for j < n && s[j] == s[i] {
131+
j++
132+
}
133+
if (j-i)%3 == 1 {
134+
return false
135+
}
136+
if (j-i)%3 == 2 {
137+
cnt2++
138+
if cnt2 > 1 {
139+
return false
140+
}
141+
}
142+
i = j
143+
}
144+
return cnt2 == 1
145+
}
67146
```
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69148
### **...**

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

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
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+
}
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};

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

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,21 @@
1+
func isDecomposable(s string) bool {
2+
i, n := 0, len(s)
3+
cnt2 := 0
4+
for i < n {
5+
j := i
6+
for 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 {
13+
cnt2++
14+
if cnt2 > 1 {
15+
return false
16+
}
17+
}
18+
i = j
19+
}
20+
return cnt2 == 1
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}

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

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@@ -0,0 +1,20 @@
1+
class Solution {
2+
public boolean isDecomposable(String s) {
3+
int i = 0, n = s.length();
4+
int cnt2 = 0;
5+
while (i < n) {
6+
int j = i;
7+
while (j < n && s.charAt(j) == s.charAt(i)) {
8+
++j;
9+
}
10+
if ((j - i) % 3 == 1) {
11+
return false;
12+
}
13+
if ((j - i) % 3 == 2 && ++cnt2 > 1) {
14+
return false;
15+
}
16+
i = j;
17+
}
18+
return cnt2 == 1;
19+
}
20+
}

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

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
1+
class Solution:
2+
def isDecomposable(self, s: str) -> bool:
3+
i, n = 0, len(s)
4+
cnt2 = 0
5+
while i < n:
6+
j = i
7+
while j < n and s[j] == s[i]:
8+
j += 1
9+
if (j - i) % 3 == 1:
10+
return False
11+
cnt2 += (j - i) % 3 == 2
12+
if cnt2 > 1:
13+
return False
14+
i = j
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return cnt2 == 1

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