feat: update solutions to lc problems: No.1209,1217,1218

solution/1200-1299/1209.Remove All Adjacent Duplicates in String II/README.md

@@ -39,7 +39,7 @@ tags:
 
 <pre><strong>输入：</strong>s = &quot;deeedbbcccbdaa&quot;, k = 3
 <strong>输出：</strong>&quot;aa&quot;
-<strong>解释： 
+<strong>解释：
 </strong>先删除 &quot;eee&quot; 和 &quot;ccc&quot;，得到 &quot;ddbbbdaa&quot;
 再删除 &quot;bbb&quot;，得到 &quot;dddaa&quot;
 最后删除 &quot;ddd&quot;，得到 &quot;aa&quot;</pre>
@@ -81,22 +81,15 @@ tags:
 ```python
 class Solution:
     def removeDuplicates(self, s: str, k: int) -> str:
-        t = []
-        i, n = 0, len(s)
-        while i < n:
-            j = i
-            while j < n and s[j] == s[i]:
-                j += 1
-            cnt = j - i
-            cnt %= k
-            if t and t[-1][0] == s[i]:
-                t[-1][1] = (t[-1][1] + cnt) % k
-                if t[-1][1] == 0:
-                    t.pop()
-            elif cnt:
-                t.append([s[i], cnt])
-            i = j
-        ans = [c * v for c, v in t]
+        stk = []
+        for c in s:
+            if stk and stk[-1][0] == c:
+                stk[-1][1] = (stk[-1][1] + 1) % k
+                if stk[-1][1] == 0:
+                    stk.pop()
+            else:
+                stk.append([c, 1])
+        ans = [c * v for c, v in stk]
         return "".join(ans)
 ```
 
@@ -190,31 +183,4 @@ type pair struct {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def removeDuplicates(self, s: str, k: int) -> str:
-        stk = []
-        for c in s:
-            if stk and stk[-1][0] == c:
-                stk[-1][1] = (stk[-1][1] + 1) % k
-                if stk[-1][1] == 0:
-                    stk.pop()
-            else:
-                stk.append([c, 1])
-        ans = [c * v for c, v in stk]
-        return "".join(ans)
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/1200-1299/1209.Remove All Adjacent Duplicates in String II/README_EN.md

@@ -38,7 +38,7 @@ tags:
 <pre>
 <strong>Input:</strong> s = &quot;deeedbbcccbdaa&quot;, k = 3
 <strong>Output:</strong> &quot;aa&quot;
-<strong>Explanation: 
+<strong>Explanation:
 </strong>First delete &quot;eee&quot; and &quot;ccc&quot;, get &quot;ddbbbdaa&quot;
 Then delete &quot;bbb&quot;, get &quot;dddaa&quot;
 Finally delete &quot;ddd&quot;, get &quot;aa&quot;</pre>
@@ -80,22 +80,15 @@ The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is
 ```python
 class Solution:
     def removeDuplicates(self, s: str, k: int) -> str:
-        t = []
-        i, n = 0, len(s)
-        while i < n:
-            j = i
-            while j < n and s[j] == s[i]:
-                j += 1
-            cnt = j - i
-            cnt %= k
-            if t and t[-1][0] == s[i]:
-                t[-1][1] = (t[-1][1] + cnt) % k
-                if t[-1][1] == 0:
-                    t.pop()
-            elif cnt:
-                t.append([s[i], cnt])
-            i = j
-        ans = [c * v for c, v in t]
+        stk = []
+        for c in s:
+            if stk and stk[-1][0] == c:
+                stk[-1][1] = (stk[-1][1] + 1) % k
+                if stk[-1][1] == 0:
+                    stk.pop()
+            else:
+                stk.append([c, 1])
+        ans = [c * v for c, v in stk]
         return "".join(ans)
 ```
 
@@ -189,31 +182,4 @@ type pair struct {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def removeDuplicates(self, s: str, k: int) -> str:
-        stk = []
-        for c in s:
-            if stk and stk[-1][0] == c:
-                stk[-1][1] = (stk[-1][1] + 1) % k
-                if stk[-1][1] == 0:
-                    stk.pop()
-            else:
-                stk.append([c, 1])
-        ans = [c * v for c, v in stk]
-        return "".join(ans)
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/1200-1299/1217.Minimum Cost to Move Chips to The Same Position/README.md

@@ -83,7 +83,7 @@ tags:
 
 将所有偶数下标的芯片移动到 0 号位置，所有奇数下标的芯片移动到 1 号位置，所有的代价为 0，接下来只需要在 0/1 号位置中选择其中一个较小数量的芯片，移动到另一个位置。所需的最小代价就是那个较小的数量。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为芯片的数量。
+时间复杂度 $O(n)$，其中 $n$ 为芯片的数量。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 

solution/1200-1299/1218.Longest Arithmetic Subsequence of Given Difference/README.md

@@ -66,7 +66,13 @@ tags:
 
 ### 方法一：动态规划
 
-时间复杂度 $O(n)$。
+我们可以使用哈希表 $f$ 来存储以 $x$ 结尾的最长等差子序列的长度。
+
+遍历数组 $\textit{arr}$，对于每个元素 $x$，我们更新 $f[x]$ 为 $f[x - \textit{difference}] + 1$。
+
+遍历结束后，我们返回 $f$ 中的最大值作为答案返回即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{arr}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -127,6 +133,25 @@ func longestSubsequence(arr []int, difference int) (ans int) {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn longest_subsequence(arr: Vec<i32>, difference: i32) -> i32 {
+        let mut f = HashMap::new();
+        let mut ans = 0;
+        for &x in &arr {
+            let count = f.get(&(x - difference)).unwrap_or(&0) + 1;
+            f.insert(x, count);
+            ans = ans.max(count);
+        }
+        ans
+    }
+}
+```
+
 #### TypeScript
 
 ```ts

solution/1200-1299/1218.Longest Arithmetic Subsequence of Given Difference/README_EN.md

@@ -62,7 +62,15 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Dynamic Programming
+
+We can use a hash table $f$ to store the length of the longest arithmetic subsequence ending with $x$.
+
+Traverse the array $\textit{arr}$, and for each element $x$, update $f[x]$ to be $f[x - \textit{difference}] + 1$.
+
+After the traversal, return the maximum value in $f$ as the answer.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{arr}$.
 
 <!-- tabs:start -->
 
@@ -135,6 +143,25 @@ function longestSubsequence(arr: number[], difference: number): number {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn longest_subsequence(arr: Vec<i32>, difference: i32) -> i32 {
+        let mut f = HashMap::new();
+        let mut ans = 0;
+        for &x in &arr {
+            let count = f.get(&(x - difference)).unwrap_or(&0) + 1;
+            f.insert(x, count);
+            ans = ans.max(count);
+        }
+        ans
+    }
+}
+```
+
 #### JavaScript
 
 ```js

solution/1200-1299/1218.Longest Arithmetic Subsequence of Given Difference/Solution.rs

@@ -0,0 +1,14 @@
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn longest_subsequence(arr: Vec<i32>, difference: i32) -> i32 {
+        let mut f = HashMap::new();
+        let mut ans = 0;
+        for &x in &arr {
+            let count = f.get(&(x - difference)).unwrap_or(&0) + 1;
+            f.insert(x, count);
+            ans = ans.max(count);
+        }
+        ans
+    }
+}