feat: add solutions to lc problems: No.1295,1299

solution/1200-1299/1295.Find Numbers with Even Number of Digits/README.md

@@ -41,7 +41,7 @@ tags:
 
 <pre>
 <strong>输入：</strong>nums = [555,901,482,1771]
-<strong>输出：</strong>1 
+<strong>输出：</strong>1
 <strong>解释： </strong>
 只有 1771 是位数为偶数的数字。
 </pre>
@@ -61,11 +61,13 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一：枚举
+### 方法一：模拟
 
-枚举数组 `nums` 中的每个元素，将其转换为字符串，判断字符串长度是否为偶数，是则答案加一。
+我们遍历数组 $\textit{nums}$ 中的每个元素，对于当前遍历到的元素 $x$，我们直接将其转换为字符串，然后判断其长度是否为偶数即可。若是则将答案加一。
 
-时间复杂度 $O(n \times \log_{10} m)$，空间复杂度 $O(\log_{10} m)$，其中 $n$ 和 $m$ 分别为数组 `nums` 的长度以及数组 `nums` 中的最大元素。
+遍历结束后，我们返回答案即可。
+
+时间复杂度 $O(n \times \log M)$，空间复杂度 $O(\log M)$。其中 $n$ 是数组 $\textit{nums}$ 的长度，而 $M$ 是数组 $\textit{nums}$ 中的元素的最大值。
 
 <!-- tabs:start -->
 
@@ -74,7 +76,7 @@ tags:
 ```python
 class Solution:
     def findNumbers(self, nums: List[int]) -> int:
-        return sum(len(str(v)) % 2 == 0 for v in nums)
+        return sum(len(str(x)) % 2 == 0 for x in nums)
 ```
 
 #### Java
@@ -83,8 +85,8 @@ class Solution:
 class Solution {
     public int findNumbers(int[] nums) {
         int ans = 0;
-        for (int v : nums) {
-            if (String.valueOf(v).length() % 2 == 0) {
+        for (int x : nums) {
+            if (String.valueOf(x).length() % 2 == 0) {
                 ++ans;
             }
         }
@@ -100,8 +102,8 @@ class Solution {
 public:
     int findNumbers(vector<int>& nums) {
         int ans = 0;
-        for (int& v : nums) {
-            ans += to_string(v).size() % 2 == 0;
+        for (int& x : nums) {
+            ans += to_string(x).size() % 2 == 0;
         }
         return ans;
     }
@@ -112,15 +114,23 @@ public:
 
 ```go
 func findNumbers(nums []int) (ans int) {
-	for _, v := range nums {
-		if len(strconv.Itoa(v))%2 == 0 {
+	for _, x := range nums {
+		if len(strconv.Itoa(x))%2 == 0 {
 			ans++
 		}
 	}
 	return
 }
 ```
 
+#### TypeScript
+
+```ts
+function findNumbers(nums: number[]): number {
+    return nums.filter(x => x.toString().length % 2 === 0).length;
+}
+```
+
 #### JavaScript
 
 ```js
@@ -129,11 +139,7 @@ func findNumbers(nums []int) (ans int) {
  * @return {number}
  */
 var findNumbers = function (nums) {
-    let ans = 0;
-    for (const v of nums) {
-        ans += String(v).length % 2 == 0;
-    }
-    return ans;
+    return nums.filter(x => x.toString().length % 2 === 0).length;
 };
 ```
 

solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md

@@ -27,7 +27,7 @@ tags:
 <pre>
 <strong>Input:</strong> nums = [12,345,2,6,7896]
 <strong>Output:</strong> 2
-<strong>Explanation: 
+<strong>Explanation:
 </strong>12 contains 2 digits (even number of digits).&nbsp;
 345 contains 3 digits (odd number of digits).&nbsp;
 2 contains 1 digit (odd number of digits).&nbsp;
@@ -40,7 +40,7 @@ Therefore only 12 and 7896 contain an even number of digits.
 
 <pre>
 <strong>Input:</strong> nums = [555,901,482,1771]
-<strong>Output:</strong> 1 
+<strong>Output:</strong> 1
 <strong>Explanation: </strong>
 Only 1771 contains an even number of digits.
 </pre>
@@ -59,7 +59,13 @@ Only 1771 contains an even number of digits.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Simulation
+
+We traverse each element $x$ in the array $\textit{nums}$. For the current element $x$, we directly convert it to a string and then check if its length is even. If it is, we increment the answer by one.
+
+After the traversal is complete, we return the answer.
+
+The time complexity is $O(n \times \log M)$, and the space complexity is $O(\log M)$. Here, $n$ is the length of the array $\textit{nums}$, and $M$ is the maximum value of the elements in the array $\textit{nums}$.
 
 <!-- tabs:start -->
 
@@ -68,7 +74,7 @@ Only 1771 contains an even number of digits.
 ```python
 class Solution:
     def findNumbers(self, nums: List[int]) -> int:
-        return sum(len(str(v)) % 2 == 0 for v in nums)
+        return sum(len(str(x)) % 2 == 0 for x in nums)
 ```
 
 #### Java
@@ -77,8 +83,8 @@ class Solution:
 class Solution {
     public int findNumbers(int[] nums) {
         int ans = 0;
-        for (int v : nums) {
-            if (String.valueOf(v).length() % 2 == 0) {
+        for (int x : nums) {
+            if (String.valueOf(x).length() % 2 == 0) {
                 ++ans;
             }
         }
@@ -94,8 +100,8 @@ class Solution {
 public:
     int findNumbers(vector<int>& nums) {
         int ans = 0;
-        for (int& v : nums) {
-            ans += to_string(v).size() % 2 == 0;
+        for (int& x : nums) {
+            ans += to_string(x).size() % 2 == 0;
         }
         return ans;
     }
@@ -106,15 +112,23 @@ public:
 
 ```go
 func findNumbers(nums []int) (ans int) {
-	for _, v := range nums {
-		if len(strconv.Itoa(v))%2 == 0 {
+	for _, x := range nums {
+		if len(strconv.Itoa(x))%2 == 0 {
 			ans++
 		}
 	}
 	return
 }
 ```
 
+#### TypeScript
+
+```ts
+function findNumbers(nums: number[]): number {
+    return nums.filter(x => x.toString().length % 2 === 0).length;
+}
+```
+
 #### JavaScript
 
 ```js
@@ -123,11 +137,7 @@ func findNumbers(nums []int) (ans int) {
  * @return {number}
  */
 var findNumbers = function (nums) {
-    let ans = 0;
-    for (const v of nums) {
-        ans += String(v).length % 2 == 0;
-    }
-    return ans;
+    return nums.filter(x => x.toString().length % 2 === 0).length;
 };
 ```
 

solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.cpp

@@ -2,9 +2,9 @@ class Solution {
 public:
     int findNumbers(vector<int>& nums) {
         int ans = 0;
-        for (int& v : nums) {
-            ans += to_string(v).size() % 2 == 0;
+        for (int& x : nums) {
+            ans += to_string(x).size() % 2 == 0;
         }
         return ans;
     }
-};
+};

solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.go

@@ -1,8 +1,8 @@
 func findNumbers(nums []int) (ans int) {
-	for _, v := range nums {
-		if len(strconv.Itoa(v))%2 == 0 {
+	for _, x := range nums {
+		if len(strconv.Itoa(x))%2 == 0 {
 			ans++
 		}
 	}
 	return
-}
+}

solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.java

@@ -1,11 +1,11 @@
 class Solution {
     public int findNumbers(int[] nums) {
         int ans = 0;
-        for (int v : nums) {
-            if (String.valueOf(v).length() % 2 == 0) {
+        for (int x : nums) {
+            if (String.valueOf(x).length() % 2 == 0) {
                 ++ans;
             }
         }
         return ans;
     }
-}
+}

solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.js

@@ -3,9 +3,5 @@
  * @return {number}
  */
 var findNumbers = function (nums) {
-    let ans = 0;
-    for (const v of nums) {
-        ans += String(v).length % 2 == 0;
-    }
-    return ans;
+    return nums.filter(x => x.toString().length % 2 === 0).length;
 };

solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.py

@@ -1,3 +1,3 @@
 class Solution:
     def findNumbers(self, nums: List[int]) -> int:
-        return sum(len(str(v)) % 2 == 0 for v in nums)
+        return sum(len(str(x)) % 2 == 0 for x in nums)

solution/1200-1299/1295.Find Numbers with Even Number of Digits/Solution.ts

@@ -0,0 +1,3 @@
+function findNumbers(nums: number[]): number {
+    return nums.filter(x => x.toString().length % 2 === 0).length;
+}

solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README.md

@@ -75,9 +75,9 @@ tags:
 
 ### 方法一：哈希表 + 枚举
 
-根据题目描述，如果一个长串满足条件，那么这个长串的子串（长度至少为 `minSize`）也一定满足条件。因此，我们只需要枚举 $s$ 中所有长度为 `minSize` 的子串，然后利用哈希表记录所有子串的出现次数，找出最大的次数作为答案即可。
+根据题目描述，如果一个长串满足条件，那么这个长串的长度为 $\textit{minSize}$ 的子串也一定满足条件。因此，我们只需要枚举 $s$ 中所有长度为 $\textit{minSize}$ 的子串，然后利用哈希表记录所有子串的出现次数，找出最大的次数作为答案即可。
 
-时间复杂度 $O(n \times m)$，空间复杂度 $O(n \times m)$。其中 $n$ 和 $m$ 分别为字符串 $s$ 的长度以及 `minSize` 的大小。本题中 $m$ 不超过 $26$。
+时间复杂度 $O(n \times m)$，空间复杂度 $O(n \times m)$。其中 $n$ 和 $m$ 分别为字符串 $s$ 的长度以及 $\textit{minSize}$ 的大小。本题中 $m$ 不超过 $26$。
 
 <!-- tabs:start -->
 
@@ -162,6 +162,24 @@ func maxFreq(s string, maxLetters int, minSize int, maxSize int) (ans int) {
 }
 ```
 
+#### TypeScript
+
+```ts
+function maxFreq(s: string, maxLetters: number, minSize: number, maxSize: number): number {
+    const cnt = new Map<string, number>();
+    let ans = 0;
+    for (let i = 0; i < s.length - minSize + 1; ++i) {
+        const t = s.slice(i, i + minSize);
+        const ss = new Set(t.split(''));
+        if (ss.size <= maxLetters) {
+            cnt.set(t, (cnt.get(t) || 0) + 1);
+            ans = Math.max(ans, cnt.get(t)!);
+        }
+    }
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/README_EN.md

@@ -61,7 +61,11 @@ It satisfies the conditions, 2 unique letters and size 3 (between minSize and ma
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Hash Table + Enumeration
+
+According to the problem description, if a long string meets the condition, then its substring of length $\textit{minSize}$ must also meet the condition. Therefore, we only need to enumerate all substrings of length $\textit{minSize}$ in $s$, then use a hash table to record the occurrence frequency of all substrings, and find the maximum frequency as the answer.
+
+The time complexity is $O(n \times m)$, and the space complexity is $O(n \times m)$. Here, $n$ and $m$ are the lengths of the string $s$ and $\textit{minSize}$, respectively. In this problem, $m$ does not exceed $26$.
 
 <!-- tabs:start -->
 
@@ -146,6 +150,24 @@ func maxFreq(s string, maxLetters int, minSize int, maxSize int) (ans int) {
 }
 ```
 
+#### TypeScript
+
+```ts
+function maxFreq(s: string, maxLetters: number, minSize: number, maxSize: number): number {
+    const cnt = new Map<string, number>();
+    let ans = 0;
+    for (let i = 0; i < s.length - minSize + 1; ++i) {
+        const t = s.slice(i, i + minSize);
+        const ss = new Set(t.split(''));
+        if (ss.size <= maxLetters) {
+            cnt.set(t, (cnt.get(t) || 0) + 1);
+            ans = Math.max(ans, cnt.get(t)!);
+        }
+    }
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1200-1299/1297.Maximum Number of Occurrences of a Substring/Solution.ts

@@ -0,0 +1,13 @@
+function maxFreq(s: string, maxLetters: number, minSize: number, maxSize: number): number {
+    const cnt = new Map<string, number>();
+    let ans = 0;
+    for (let i = 0; i < s.length - minSize + 1; ++i) {
+        const t = s.slice(i, i + minSize);
+        const ss = new Set(t.split(''));
+        if (ss.size <= maxLetters) {
+            cnt.set(t, (cnt.get(t) || 0) + 1);
+            ans = Math.max(ans, cnt.get(t)!);
+        }
+    }
+    return ans;
+}