diff --git a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README.md b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README.md
index d48ceda73de2d..1f9d17d12be2f 100644
--- a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README.md
+++ b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README.md
@@ -61,7 +61,7 @@
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`。
-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。
+时间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。
<!-- tabs:start -->
@@ -87,6 +87,20 @@ class Solution:
return cnt2 == 1
```
+```python
+class Solution:
+ def isDecomposable(self, s: str) -> bool:
+ cnt2 = 0
+ for _, g in groupby(s):
+ m = len(list(g))
+ if m % 3 == 1:
+ return False
+ cnt2 += m % 3 == 2
+ if cnt2 > 1:
+ return False
+ return cnt2 == 1
+```
+
### **Java**
<!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -120,13 +134,19 @@ class Solution {
class Solution {
public:
bool isDecomposable(string s) {
- int i = 0, n = s.size();
int cnt2 = 0;
- while (i < n) {
+ for (int i = 0, n = s.size(); i < n;) {
int j = i;
- while (j < n && s[j] == s[i]) ++j;
- if ((j - i) % 3 == 1) return false;
- if ((j - i) % 3 == 2 && ++cnt2 > 1) return false;
+ while (j < n && s[j] == s[i]) {
+ ++j;
+ }
+ if ((j - i) % 3 == 1) {
+ return false;
+ }
+ cnt2 += (j - i) % 3 == 2;
+ if (cnt2 > 1) {
+ return false;
+ }
i = j;
}
return cnt2 == 1;
@@ -160,6 +180,29 @@ func isDecomposable(s string) bool {
}
```
+### **TypeScript**
+
+```ts
+function isDecomposable(s: string): boolean {
+ const n = s.length;
+ let cnt2 = 0;
+ for (let i = 0; i < n; ) {
+ let j = i;
+ while (j < n && s[j] === s[i]) {
+ ++j;
+ }
+ if ((j - i) % 3 === 1) {
+ return false;
+ }
+ if ((j - i) % 3 === 2 && ++cnt2 > 1) {
+ return false;
+ }
+ i = j;
+ }
+ return cnt2 === 1;
+}
+```
+
### **...**
```
diff --git a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README_EN.md b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README_EN.md
index 410416d65025f..463da18783da9 100644
--- a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README_EN.md
+++ b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README_EN.md
@@ -52,6 +52,14 @@
## Solutions
+**Solution 1: Two Pointers**
+
+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.
+
+After the traversal, check whether a substring of length $2$ has appeared. If not, return `false`, otherwise return `true`.
+
+The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.
+
<!-- tabs:start -->
### **Python3**
@@ -74,6 +82,20 @@ class Solution:
return cnt2 == 1
```
+```python
+class Solution:
+ def isDecomposable(self, s: str) -> bool:
+ cnt2 = 0
+ for _, g in groupby(s):
+ m = len(list(g))
+ if m % 3 == 1:
+ return False
+ cnt2 += m % 3 == 2
+ if cnt2 > 1:
+ return False
+ return cnt2 == 1
+```
+
### **Java**
```java
@@ -105,13 +127,19 @@ class Solution {
class Solution {
public:
bool isDecomposable(string s) {
- int i = 0, n = s.size();
int cnt2 = 0;
- while (i < n) {
+ for (int i = 0, n = s.size(); i < n;) {
int j = i;
- while (j < n && s[j] == s[i]) ++j;
- if ((j - i) % 3 == 1) return false;
- if ((j - i) % 3 == 2 && ++cnt2 > 1) return false;
+ while (j < n && s[j] == s[i]) {
+ ++j;
+ }
+ if ((j - i) % 3 == 1) {
+ return false;
+ }
+ cnt2 += (j - i) % 3 == 2;
+ if (cnt2 > 1) {
+ return false;
+ }
i = j;
}
return cnt2 == 1;
@@ -145,6 +173,29 @@ func isDecomposable(s string) bool {
}
```
+### **TypeScript**
+
+```ts
+function isDecomposable(s: string): boolean {
+ const n = s.length;
+ let cnt2 = 0;
+ for (let i = 0; i < n; ) {
+ let j = i;
+ while (j < n && s[j] === s[i]) {
+ ++j;
+ }
+ if ((j - i) % 3 === 1) {
+ return false;
+ }
+ if ((j - i) % 3 === 2 && ++cnt2 > 1) {
+ return false;
+ }
+ i = j;
+ }
+ return cnt2 === 1;
+}
+```
+
### **...**
```
diff --git a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution.cpp b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution.cpp
index a36d706241461..9f160c786fe02 100644
--- a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution.cpp
+++ b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution.cpp
@@ -1,15 +1,21 @@
-class Solution {
-public:
- bool isDecomposable(string s) {
- int i = 0, n = s.size();
- int cnt2 = 0;
- while (i < n) {
- int j = i;
- while (j < n && s[j] == s[i]) ++j;
- if ((j - i) % 3 == 1) return false;
- if ((j - i) % 3 == 2 && ++cnt2 > 1) return false;
- i = j;
- }
- return cnt2 == 1;
- }
+class Solution {
+public:
+ bool isDecomposable(string s) {
+ int cnt2 = 0;
+ for (int i = 0, n = s.size(); i < n;) {
+ int j = i;
+ while (j < n && s[j] == s[i]) {
+ ++j;
+ }
+ if ((j - i) % 3 == 1) {
+ return false;
+ }
+ cnt2 += (j - i) % 3 == 2;
+ if (cnt2 > 1) {
+ return false;
+ }
+ i = j;
+ }
+ return cnt2 == 1;
+ }
};
\ No newline at end of file
diff --git a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution.ts b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution.ts
new file mode 100644
index 0000000000000..126b5670d3a1a
--- /dev/null
+++ b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution.ts
@@ -0,0 +1,18 @@
+function isDecomposable(s: string): boolean {
+ const n = s.length;
+ let cnt2 = 0;
+ for (let i = 0; i < n; ) {
+ let j = i;
+ while (j < n && s[j] === s[i]) {
+ ++j;
+ }
+ if ((j - i) % 3 === 1) {
+ return false;
+ }
+ if ((j - i) % 3 === 2 && ++cnt2 > 1) {
+ return false;
+ }
+ i = j;
+ }
+ return cnt2 === 1;
+}