feat: add solutions to lc problems: No.3210~3212 (doocs#3219)

* No.3210.Find the Encrypted String
* No.3211.Generate Binary Strings Without Adjacent Zeros
* No.3212.Count Submatrices With Equal Frequency of X and Y

Author: yanglbme

solution/0000-0099/0017.Letter Combinations of a Phone Number/README.md

@@ -165,13 +165,13 @@ func letterCombinations(digits string) []string {
 
 ```ts
 function letterCombinations(digits: string): string[] {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans: string[] = [''];
     const d = ['abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz'];
     for (const i of digits) {
-        const s = d[parseInt(i) - 2];
+        const s = d[+i - 2];
         const t: string[] = [];
         for (const a of ans) {
             for (const b of s) {
@@ -218,13 +218,13 @@ impl Solution {
  * @return {string[]}
  */
 var letterCombinations = function (digits) {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans = [''];
     const d = ['abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz'];
     for (const i of digits) {
-        const s = d[parseInt(i) - 2];
+        const s = d[+i - 2];
         const t = [];
         for (const a of ans) {
             for (const b of s) {
@@ -392,7 +392,7 @@ func letterCombinations(digits string) (ans []string) {
 
 ```ts
 function letterCombinations(digits: string): string[] {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans: string[] = [];
@@ -403,7 +403,7 @@ function letterCombinations(digits: string): string[] {
             ans.push(t.join(''));
             return;
         }
-        const s = d[parseInt(digits[i]) - 2];
+        const s = d[+digits[i] - 2];
         for (const c of s) {
             t.push(c);
             dfs(i + 1);
@@ -453,7 +453,7 @@ impl Solution {
  * @return {string[]}
  */
 var letterCombinations = function (digits) {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans = [];
@@ -464,7 +464,7 @@ var letterCombinations = function (digits) {
             ans.push(t.join(''));
             return;
         }
-        const s = d[parseInt(digits[i]) - 2];
+        const s = d[+digits[i] - 2];
         for (const c of s) {
             t.push(c);
             dfs(i + 1);

solution/0000-0099/0017.Letter Combinations of a Phone Number/README_EN.md

@@ -161,13 +161,13 @@ func letterCombinations(digits string) []string {
 
 ```ts
 function letterCombinations(digits: string): string[] {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans: string[] = [''];
     const d = ['abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz'];
     for (const i of digits) {
-        const s = d[parseInt(i) - 2];
+        const s = d[+i - 2];
         const t: string[] = [];
         for (const a of ans) {
             for (const b of s) {
@@ -214,13 +214,13 @@ impl Solution {
  * @return {string[]}
  */
 var letterCombinations = function (digits) {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans = [''];
     const d = ['abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz'];
     for (const i of digits) {
-        const s = d[parseInt(i) - 2];
+        const s = d[+i - 2];
         const t = [];
         for (const a of ans) {
             for (const b of s) {
@@ -388,7 +388,7 @@ func letterCombinations(digits string) (ans []string) {
 
 ```ts
 function letterCombinations(digits: string): string[] {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans: string[] = [];
@@ -399,7 +399,7 @@ function letterCombinations(digits: string): string[] {
             ans.push(t.join(''));
             return;
         }
-        const s = d[parseInt(digits[i]) - 2];
+        const s = d[+digits[i] - 2];
         for (const c of s) {
             t.push(c);
             dfs(i + 1);
@@ -449,7 +449,7 @@ impl Solution {
  * @return {string[]}
  */
 var letterCombinations = function (digits) {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans = [];
@@ -460,7 +460,7 @@ var letterCombinations = function (digits) {
             ans.push(t.join(''));
             return;
         }
-        const s = d[parseInt(digits[i]) - 2];
+        const s = d[+digits[i] - 2];
         for (const c of s) {
             t.push(c);
             dfs(i + 1);

solution/0000-0099/0017.Letter Combinations of a Phone Number/Solution.js

@@ -3,13 +3,13 @@
  * @return {string[]}
  */
 var letterCombinations = function (digits) {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans = [''];
     const d = ['abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz'];
     for (const i of digits) {
-        const s = d[parseInt(i) - 2];
+        const s = d[+i - 2];
         const t = [];
         for (const a of ans) {
             for (const b of s) {

solution/0000-0099/0017.Letter Combinations of a Phone Number/Solution.ts

@@ -1,11 +1,11 @@
 function letterCombinations(digits: string): string[] {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans: string[] = [''];
     const d = ['abc', 'def', 'ghi', 'jkl', 'mno', 'pqrs', 'tuv', 'wxyz'];
     for (const i of digits) {
-        const s = d[parseInt(i) - 2];
+        const s = d[+i - 2];
         const t: string[] = [];
         for (const a of ans) {
             for (const b of s) {

solution/0000-0099/0017.Letter Combinations of a Phone Number/Solution2.js

@@ -3,7 +3,7 @@
  * @return {string[]}
  */
 var letterCombinations = function (digits) {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans = [];
@@ -14,7 +14,7 @@ var letterCombinations = function (digits) {
             ans.push(t.join(''));
             return;
         }
-        const s = d[parseInt(digits[i]) - 2];
+        const s = d[+digits[i] - 2];
         for (const c of s) {
             t.push(c);
             dfs(i + 1);

solution/0000-0099/0017.Letter Combinations of a Phone Number/Solution2.ts

@@ -1,5 +1,5 @@
 function letterCombinations(digits: string): string[] {
-    if (digits.length == 0) {
+    if (digits.length === 0) {
         return [];
     }
     const ans: string[] = [];
@@ -10,7 +10,7 @@ function letterCombinations(digits: string): string[] {
             ans.push(t.join(''));
             return;
         }
-        const s = d[parseInt(digits[i]) - 2];
+        const s = d[+digits[i] - 2];
         for (const c of s) {
             t.push(c);
             dfs(i + 1);

solution/0700-0799/0724.Find Pivot Index/README.md

@@ -77,13 +77,13 @@ tags:
 
 ### 方法一：前缀和
 
-我们定义变量 $left$ 表示数组 `nums` 中下标 $i$ 左侧元素之和，变量 $right$ 表示数组 `nums` 中下标 $i$ 右侧元素之和。初始时 $left = 0$, $right = \sum_{i = 0}^{n - 1} nums[i]$。
+我们定义变量 $left$ 表示数组 $\textit{nums}$ 中下标 $i$ 左侧元素之和，变量 $right$ 表示数组 $\textit{nums}$ 中下标 $i$ 右侧元素之和。初始时 $left = 0$, $right = \sum_{i = 0}^{n - 1} nums[i]$。
 
-遍历数组 `nums`，对于当前遍历到的数字 $x$，我们更新 $right = right - x$，此时如果 $left=right$，说明当前下标 $i$ 就是中间位置，直接返回即可。否则，我们更新 $left = left + x$，继续遍历下一个数字。
+遍历数组 $\textit{nums}$，对于当前遍历到的数字 $x$，我们更新 $right = right - x$，此时如果 $left=right$，说明当前下标 $i$ 就是中间位置，直接返回即可。否则，我们更新 $left = left + x$，继续遍历下一个数字。
 
 遍历结束，如果没有找到中间位置，返回 $-1$。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 `nums` 的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
 
 相似题目：
 

solution/0700-0799/0724.Find Pivot Index/README_EN.md

@@ -73,7 +73,20 @@ Right sum = nums[1] + nums[2] = 1 + -1 = 0
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Prefix Sum
+
+We define a variable $left$ to represent the sum of elements to the left of index $i$ in the array $\textit{nums}$, and a variable $right$ to represent the sum of elements to the right of index $i$ in the array $\textit{nums}$. Initially, $left = 0$, $right = \sum_{i = 0}^{n - 1} nums[i]$.
+
+We traverse the array $\textit{nums}$. For the current number $x$ being traversed, we update $right = right - x$. At this point, if $left = right$, it indicates that the current index $i$ is the middle position, and we can return it directly. Otherwise, we update $left = left + x$ and continue to traverse the next number.
+
+If the middle position is not found by the end of the traversal, return $-1$.
+
+The time complexity is $O(n)$, and the space complexity is $O(1)$. Here, $n$ is the length of the array $\textit{nums}$.
+
+Similar Problems:
+
+-   [1991. Find the Middle Index in Array](https://github.com/doocs/leetcode/blob/main/solution/1900-1999/1991.Find%20the%20Middle%20Index%20in%20Array/README_EN.md)
+-   [2574. Left and Right Sum Differences](https://github.com/doocs/leetcode/blob/main/solution/2500-2599/2574.Left%20and%20Right%20Sum%20Differences/README_EN.md)
 
 <!-- tabs:start -->
 

solution/3200-3299/3210.Find the Encrypted String/README.md

@@ -69,32 +69,80 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3200-3299/3210.Fi
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：模拟
+
+我们可以使用模拟的方法，对字符串的第 $i$ 个字符，我们将其替换为字符串的第 $(i + k) \bmod n$ 个字符。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是字符串 $s$ 的长度。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def getEncryptedString(self, s: str, k: int) -> str:
+        cs = list(s)
+        n = len(s)
+        for i in range(n):
+            cs[i] = s[(i + k) % n]
+        return "".join(cs)
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public String getEncryptedString(String s, int k) {
+        char[] cs = s.toCharArray();
+        int n = cs.length;
+        for (int i = 0; i < n; ++i) {
+            cs[i] = s.charAt((i + k) % n);
+        }
+        return new String(cs);
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    string getEncryptedString(string s, int k) {
+        int n = s.length();
+        string cs(n, ' ');
+        for (int i = 0; i < n; ++i) {
+            cs[i] = s[(i + k) % n];
+        }
+        return cs;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func getEncryptedString(s string, k int) string {
+	cs := []byte(s)
+	for i := range s {
+		cs[i] = s[(i+k)%len(s)]
+	}
+	return string(cs)
+}
+```
 
+#### TypeScript
+
+```ts
+function getEncryptedString(s: string, k: number): string {
+    const cs: string[] = [];
+    const n = s.length;
+    for (let i = 0; i < n; ++i) {
+        cs[i] = s[(i + k) % n];
+    }
+    return cs.join('');
+}
 ```
 
 <!-- tabs:end -->