diff --git a/docs/interview/experiences/16easyHandwriting.md b/docs/interview/experiences/16hotHandwriting.md
similarity index 93%
rename from docs/interview/experiences/16easyHandwriting.md
rename to docs/interview/experiences/16hotHandwriting.md
index c4c253f..fc7eeae 100644
--- a/docs/interview/experiences/16easyHandwriting.md
+++ b/docs/interview/experiences/16hotHandwriting.md
@@ -16,7 +16,7 @@ nav:
 
 > 练习的不是手写，而是对于这份工作的热爱
 
-> base 50 middle 30 hard 20
+> base 50 middle 30 hard 20 total 100
 
 # 最热的手写,没有之一,没有准备好，就不要出去面试
 
@@ -643,7 +643,7 @@ lodashGet(test, 'other[0].name');
 function lodashSet(obj, path, val) {}
 ```
 
-## 23.场景应用题:红绿灯问题
+## 23.场景应用题, 红绿灯问题
 
 ```js
 function red() {
@@ -1074,7 +1074,7 @@ Number.prototype.minus = function (n) {
 (5).add(3).minus(2);
 ```
 
-## 39.要求设计 LazyMan 类，实现以下功能
+## 39.场景应用题,要求设计 LazyMan 类，实现以下功能
 
 > [要求设计 LazyMan 类，实现以下功能](https://github.com/Advanced-Frontend/Daily-Interview-Question/issues/98)
 
@@ -1202,14 +1202,94 @@ swr('cache', MockPromise, 1000);
 
 ## 42.如何自定义一个事件，使某一个对象能够捕获到？
 
-## 43.
+```js
+
+```
+
+## 43.实现一个 node 异步函数的 promisify
+
+```js
+function promisify(fn) {
+  return (...arr) =>
+    new Promise((resolve, reject) => {
+      arr.push((err, ...val) => {
+        if (err) {
+          reject(err);
+        } else {
+          resolve(val);
+        }
+      });
+      fn.apply(this, arr);
+    });
+}
+```
+
+## 44.实现字符串压缩
 
-## 44.
+```js
+// abbccccaaa->a1b2c4a3
+function compressWord(str) {
+  let res = '';
+  let arr = [];
+  for (let i = 0; i < str.length; i++) {
+    let cur = str[i];
+    const index = arr.indexOf(cur);
+    if (arr.length && index < 0) {
+      res += arr[0] + arr.length;
+      arr = [];
+    }
+    arr.push(cur);
+  }
+  if (arr.length) {
+    res += arr[0] + arr.length;
+  }
+  return res;
+}
+compressWord('abbccccaaa');
+```
 
 ## 45.
 
-## 46.
+## 46.TS 练习嵌套 Awaited<T>
+
+```ts
+type MockAwaited<T extends Promise<unknown>> = T extends Promise<infer P>
+  ? P extends Promise<unknown>
+    ? MockAwaited<P>
+    : P
+  : T;
+```
+
+## 47.TS 练习根据 xx 类型提取类型的对象 FunctionKeys<T, Function>
+
+> 关键词 as 映射
+
+```ts
+type TTP1 = {
+  name: string;
+  getName: () => void;
+  age: number;
+};
+type FunctionKeys<T, F> = {
+  [P in keyof T as T[P] extends F ? P : never]: T[P];
+};
+type Fun1 = FunctionKeys<TTP1, number>;
+```
+
+## 48.TS 练习索引类型转联合类型：IndexToUnion
+
+```ts
+
+```
+
+## 49.TS 练习独一无二类型：Unique
 
-## 47.
+```ts
 
-## 48.
+```
+
+## 50.TS 练习联合类型转交叉类型：UnionToIntersection
+
+```ts
+
+```
diff --git a/docs/interview/experiences/17hotHandwriting.md b/docs/interview/experiences/17middleHandwriting.md
similarity index 100%
rename from docs/interview/experiences/17hotHandwriting.md
rename to docs/interview/experiences/17middleHandwriting.md
diff --git a/docs/interview/experiences/18middleHandwriting.md b/docs/interview/experiences/18hardHandwriting.md
similarity index 100%
rename from docs/interview/experiences/18middleHandwriting.md
rename to docs/interview/experiences/18hardHandwriting.md
diff --git a/docs/interview/experiences/19endHandwriting.md b/docs/interview/experiences/19endHandwriting.md
new file mode 100644
index 0000000..7f364d3
--- /dev/null
+++ b/docs/interview/experiences/19endHandwriting.md
@@ -0,0 +1,13 @@
+---
+title: 连续这么多手写,需要复盘和回归
+order: 19
+group:
+  order: 0
+  title: interview
+nav:
+  order: 3
+  title: 'interview'
+  path: /interview
+---
+
+连续翻越前面几座大山，有什么收获吗？其实并没有
diff --git a/docs/interview/experiences/19hotHandwriting.md b/docs/interview/experiences/19hotHandwriting.md
deleted file mode 100644
index b2a1e2c..0000000
--- a/docs/interview/experiences/19hotHandwriting.md
+++ /dev/null
@@ -1,11 +0,0 @@
----
-title: hotHandwriting2 最热手写
-order: 19
-group:
-  order: 0
-  title: interview
-nav:
-  order: 3
-  title: 'interview'
-  path: /interview
----
diff --git a/docs/interview/experiences/22leetcodeB.md b/docs/interview/experiences/22leetcodeB.md
index 3494176..f3b2092 100644
--- a/docs/interview/experiences/22leetcodeB.md
+++ b/docs/interview/experiences/22leetcodeB.md
@@ -348,6 +348,8 @@ var deleteNode = function (head, val) {
 };
 ```
 
+## 10.括号生成
+
 ## [11.轮转数组](https://leetcode.cn/problems/rotate-array/?envType=study-plan-v2&envId=top-100-liked)
 
 ### 第一种方式
@@ -500,7 +502,46 @@ lengthOfLongestSubstring('abcabcbb');
 ## [15.乘积最大子数组](https://leetcode.cn/problems/maximum-product-subarray/)
 
 ```js
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+var maxProduct = function (nums) {
+  let max = Number.MIN_SAFE_INTEGER;
+  let imax = 1,
+    imin = 1;
+  for (let num of nums) {
+    if (num < 0) {
+      let tmp = imax;
+      imax = imin;
+      imin = tmp;
+    }
+    imax = Math.max(imax * num, num);
+    imin = Math.min(imin * num, num);
+    max = Math.max(max, imax);
+  }
+  return max;
+};
+```
 
+### second
+
+```js
+var maxProduct = function (nums) {
+  let len = nums.length;
+  let dpMax = new Array(len).fill(1);
+  dpMax[0] = nums[0];
+  let dpMin = new Array(len).fill(1);
+  dpMin[0] = nums[0];
+  let max = nums[0];
+  for (let i = 1; i < len; i++) {
+    let cur = nums[i];
+    dpMax[i] = Math.max(dpMin[i - 1] * cur, Math.max(dpMax[i - 1] * cur, cur));
+    dpMin[i] = Math.min(dpMin[i - 1] * cur, Math.min(dpMax[i - 1] * cur, cur));
+    max = Math.max(max, dpMax[i]);
+  }
+  return max;
+};
 ```
 
 ## [16.螺旋矩阵](https://leetcode.cn/problems/spiral-matrix/description/)
@@ -542,6 +583,8 @@ function matrix(nums) {
 }
 ```
 
+## 17.按照版本号对数组排序
+
 ## 18.翻转二叉树，二叉树的左右节点翻转
 
 ## [19.每日温度](https://leetcode.cn/problems/daily-temperatures/)
@@ -1235,6 +1278,8 @@ findLength([1, 2, 3, 2, 1], [3, 2, 1, 4, 7]);
 
 ## [45.最小覆盖子串](https://leetcode.cn/problems/minimum-window-substring/description/)
 
+## 46.回文子串
+
 ## [47.和为 K 的子数组](https://leetcode.cn/problems/subarray-sum-equals-k/)
 
 > (前缀和)
diff --git a/docs/interview/experiences/practise/202404/0401.js b/docs/interview/experiences/practise/202404/0401.js
index ac318be..26bcb72 100644
--- a/docs/interview/experiences/practise/202404/0401.js
+++ b/docs/interview/experiences/practise/202404/0401.js
@@ -10,6 +10,41 @@
  * 8.最长重复子数组
  */
 
+function maxMul(arr) {
+  let len = arr.length;
+  let dpMax = new Array(len).fill(1);
+  dpMax[0] = arr[0];
+  let dpMin = new Array(len).fill(1);
+  dpMin[0] = arr[0];
+  let max = arr[0];
+  for (let i = 1; i < len; i++) {
+    let cur = arr[i];
+    dpMax[i] = Math.max(dpMin[i - 1] * cur, cur, dpMax[i - 1] * cur);
+    dpMin[i] = Math.min(dpMin[i - 1] * cur, cur, dpMax[i - 1] * cur);
+    max = Math.max(dpMax[i], max);
+  }
+  return max;
+}
+
+function maxMul2(arr) {
+  let len = arr.length;
+  let max = 1;
+  let min = 1;
+  let res = Number.MIN_SAFE_INTEGER;
+  for (let i = 0; i < len; i++) {
+    let cur = arr[i];
+    if (cur < 0) {
+      let tmp = max;
+      max = min;
+      min = tmp;
+    }
+    min = Math.min(min * cur, cur);
+    max = Math.max(max * cur, cur);
+    res = Math.max(res, max);
+  }
+  return res;
+}
+
 class Route {
   constructor() {
     this.routes = [];
diff --git a/docs/interview/experiences/practise/202404/0402.js b/docs/interview/experiences/practise/202404/0402.js
index ccd10b8..4c84970 100644
--- a/docs/interview/experiences/practise/202404/0402.js
+++ b/docs/interview/experiences/practise/202404/0402.js
@@ -1,6 +1,55 @@
 /**
  * 1.利用正则筛选只包含大小写字母的字符串
  * 2.Object.assign(target, ...source)
+ * 3.两个数组的交集
+ * 4.字符串转换整数
+ * 5.字符串压缩
+ * 6.单词搜索
+ * 7.版本号进行排序
+ * 8.每日温度
+ * 9.最长连续重复子串
+ * 10.零钱兑换II
  */
 
+function myAssign(target, ...source) {
+  if (typeof target != 'object') {
+    throw new TypeError('error');
+  }
+  source.forEach((item) => {
+    if (Object.keys(item).length) {
+      for (let key in item) {
+        if (item.hasOwnProperty(key)) {
+          target[key] = item[key];
+        }
+      }
+    }
+  });
+  return target;
+}
+Object.assign = myAssign;
+
 // ['Abc', 'DeF', '123', '_ghI'];
+function dataF(arr) {
+  return arr.filter((item) => /^[a-zA-Z]+$/.test(item)).map((item) => item.toUpperCase());
+}
+dataF(['Abc', 'DeF', '123', '_ghI']);
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number[]}
+ */
+var intersection = function (nums1, nums2) {
+  let a = [...new Set(nums1)];
+  let b = [...new Set(nums2)];
+  let m = new Map();
+  let ans = [];
+  for (let a1 of a) {
+    m.set(a1, 1);
+  }
+  for (let b1 of b) {
+    if (m.has(b1)) {
+      ans.push(b1);
+    }
+  }
+  return ans;
+};
diff --git a/docs/interview/experiences/practise/202404/0403.js b/docs/interview/experiences/practise/202404/0403.js
new file mode 100644
index 0000000..6b89001
--- /dev/null
+++ b/docs/interview/experiences/practise/202404/0403.js
@@ -0,0 +1,8 @@
+/**
+ * 1.最小栈
+ * 2.最大子序和/最大乘积子数组
+ * 3.字符串转换整数
+ * 4.判断子序列
+ * 5.跳跃游戏II
+ * 6.单词搜索
+ */