feat: update solutions to lc problems: No.1929,1933 (#3599)

* No.1929.Concatenation of Array
* No.1933.Check if String Is Decomposable Into Value-Equal Substrings

Author: yanglbme

solution/1900-1999/1929.Concatenation of Array/README.md

@@ -67,7 +67,11 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：模拟
+
+我们直接根据题目描述模拟，将 $\textit{nums}$ 中的元素依次添加到答案数组中，然后再将 $\textit{nums}$ 中的元素再次添加到答案数组中。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -142,9 +146,7 @@ impl Solution {
  * @return {number[]}
  */
 var getConcatenation = function (nums) {
-    let ans = nums.slice();
-    ans.splice(nums.length, 0, ...nums);
-    return ans;
+    return [...nums, ...nums];
 };
 ```
 

solution/1900-1999/1929.Concatenation of Array/README_EN.md

@@ -60,7 +60,11 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+We directly simulate according to the problem description by adding the elements of $\textit{nums}$ to the answer array one by one, and then adding the elements of $\textit{nums}$ to the answer array again.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
 
 <!-- tabs:start -->
 
@@ -135,9 +139,7 @@ impl Solution {
  * @return {number[]}
  */
 var getConcatenation = function (nums) {
-    let ans = nums.slice();
-    ans.splice(nums.length, 0, ...nums);
-    return ans;
+    return [...nums, ...nums];
 };
 ```
 

solution/1900-1999/1929.Concatenation of Array/Solution.js

@@ -3,7 +3,5 @@
  * @return {number[]}
  */
 var getConcatenation = function (nums) {
-    let ans = nums.slice();
-    ans.splice(nums.length, 0, ...nums);
-    return ans;
+    return [...nums, ...nums];
 };

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

@@ -82,18 +82,14 @@ tags:
 ```python
 class Solution:
     def isDecomposable(self, s: str) -> bool:
-        i, n = 0, len(s)
         cnt2 = 0
-        while i < n:
-            j = i
-            while j < n and s[j] == s[i]:
-                j += 1
-            if (j - i) % 3 == 1:
+        for _, g in groupby(s):
+            m = len(list(g))
+            if m % 3 == 1:
                 return False
-            cnt2 += (j - i) % 3 == 2
+            cnt2 += m % 3 == 2
             if cnt2 > 1:
                 return False
-            i = j
         return cnt2 == 1
 ```
 
@@ -201,30 +197,4 @@ function isDecomposable(s: string): boolean {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```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
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

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

@@ -83,18 +83,14 @@ The time complexity is $O(n)$, where $n$ is the length of the string $s$. The sp
 ```python
 class Solution:
     def isDecomposable(self, s: str) -> bool:
-        i, n = 0, len(s)
         cnt2 = 0
-        while i < n:
-            j = i
-            while j < n and s[j] == s[i]:
-                j += 1
-            if (j - i) % 3 == 1:
+        for _, g in groupby(s):
+            m = len(list(g))
+            if m % 3 == 1:
                 return False
-            cnt2 += (j - i) % 3 == 2
+            cnt2 += m % 3 == 2
             if cnt2 > 1:
                 return False
-            i = j
         return cnt2 == 1
 ```
 
@@ -202,30 +198,4 @@ function isDecomposable(s: string): boolean {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Python3
-
-```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
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->