feat: add solutions to lc problem: No.1933

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;
+}
+```
+
 ### **...**
 
 ```

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;
+}
+```
+
 ### **...**
 
 ```

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;
+    }
 };

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;
+}