Merge branch 'doocs:main' into main

Author: taoyq1988

solution/0400-0499/0413.Arithmetic Slices/README.md

@@ -5,6 +5,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0413.Ar
 tags:
     - 数组
     - 动态规划
+    - 滑动窗口
 ---
 
 <!-- problem:start -->

solution/0400-0499/0413.Arithmetic Slices/README_EN.md

@@ -5,6 +5,7 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0413.Ar
 tags:
     - Array
     - Dynamic Programming
+    - Sliding Window
 ---
 
 <!-- problem:start -->

solution/0600-0699/0621.Task Scheduler/README.md

@@ -53,9 +53,9 @@ tags:
 
 <p><strong>示例 3：</strong></p>
 
-<div class="example-block"><strong>输入：</strong>tasks = ["A","A","A","B","B","B"], n = 0</div>
+<div class="example-block"><strong>输入：</strong>tasks = ["A","A","A","B","B","B"], n = 3</div>
 
-<div class="example-block"><strong>输出：</strong>6</div>
+<div class="example-block"><strong>输出：</strong>10</div>
 
 <div class="example-block"><strong>解释：</strong>一种可能的序列为：A -&gt; B -&gt; idle -&gt; idle -&gt; A -&gt; B -&gt; idle -&gt; idle -&gt; A -&gt; B。</div>
 

solution/0700-0799/0731.My Calendar II/README.md

@@ -716,4 +716,74 @@ func (this *MyCalendarTwo) Book(start int, end int) bool {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### Solution 3: Line Sweep
+
+<!-- tabs:start -->
+
+#### TypeScript
+
+```ts
+class MyCalendarTwo {
+    #OVERLAPS = 2;
+    #cnt: Record<number, number> = {};
+
+    book(start: number, end: number): boolean {
+        this.#cnt[start] = (this.#cnt[start] ?? 0) + 1;
+        this.#cnt[end] = (this.#cnt[end] ?? 0) - 1;
+
+        let sum = 0;
+        for (const v of Object.values(this.#cnt)) {
+            sum += v;
+            if (sum > this.#OVERLAPS) {
+                this.#cnt[start]--;
+                this.#cnt[end]++;
+
+                if (!this.#cnt[start]) delete this.#cnt[start];
+                if (!this.#cnt[end]) delete this.#cnt[end];
+
+                return false;
+            }
+        }
+
+        return true;
+    }
+}
+```
+
+#### JavaScript
+
+```js
+class MyCalendarTwo {
+    #OVERLAPS = 2;
+    #cnt = {};
+
+    book(start, end) {
+        this.#cnt[start] = (this.#cnt[start] ?? 0) + 1;
+        this.#cnt[end] = (this.#cnt[end] ?? 0) - 1;
+
+        let sum = 0;
+        for (const v of Object.values(this.#cnt)) {
+            sum += v;
+            if (sum > this.#OVERLAPS) {
+                this.#cnt[start]--;
+                this.#cnt[end]++;
+
+                if (!this.#cnt[start]) delete this.#cnt[start];
+                if (!this.#cnt[end]) delete this.#cnt[end];
+
+                return false;
+            }
+        }
+
+        return true;
+    }
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->

solution/0700-0799/0731.My Calendar II/README_EN.md

@@ -698,4 +698,74 @@ func (this *MyCalendarTwo) Book(start int, end int) bool {
 
 <!-- solution:end -->
 
+<!-- solution:start -->
+
+### Solution 3: Line Sweep
+
+<!-- tabs:start -->
+
+#### TypeScript
+
+```ts
+class MyCalendarTwo {
+    #OVERLAPS = 2;
+    #cnt: Record<number, number> = {};
+
+    book(start: number, end: number): boolean {
+        this.#cnt[start] = (this.#cnt[start] ?? 0) + 1;
+        this.#cnt[end] = (this.#cnt[end] ?? 0) - 1;
+
+        let sum = 0;
+        for (const v of Object.values(this.#cnt)) {
+            sum += v;
+            if (sum > this.#OVERLAPS) {
+                this.#cnt[start]--;
+                this.#cnt[end]++;
+
+                if (!this.#cnt[start]) delete this.#cnt[start];
+                if (!this.#cnt[end]) delete this.#cnt[end];
+
+                return false;
+            }
+        }
+
+        return true;
+    }
+}
+```
+
+#### JavaScript
+
+```js
+class MyCalendarTwo {
+    #OVERLAPS = 2;
+    #cnt = {};
+
+    book(start, end) {
+        this.#cnt[start] = (this.#cnt[start] ?? 0) + 1;
+        this.#cnt[end] = (this.#cnt[end] ?? 0) - 1;
+
+        let sum = 0;
+        for (const v of Object.values(this.#cnt)) {
+            sum += v;
+            if (sum > this.#OVERLAPS) {
+                this.#cnt[start]--;
+                this.#cnt[end]++;
+
+                if (!this.#cnt[start]) delete this.#cnt[start];
+                if (!this.#cnt[end]) delete this.#cnt[end];
+
+                return false;
+            }
+        }
+
+        return true;
+    }
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
 <!-- problem:end -->

solution/0700-0799/0731.My Calendar II/Solution3.js

@@ -0,0 +1,25 @@
+class MyCalendarTwo {
+    #OVERLAPS = 2;
+    #cnt = {};
+
+    book(start, end) {
+        this.#cnt[start] = (this.#cnt[start] ?? 0) + 1;
+        this.#cnt[end] = (this.#cnt[end] ?? 0) - 1;
+
+        let sum = 0;
+        for (const v of Object.values(this.#cnt)) {
+            sum += v;
+            if (sum > this.#OVERLAPS) {
+                this.#cnt[start]--;
+                this.#cnt[end]++;
+
+                if (!this.#cnt[start]) delete this.#cnt[start];
+                if (!this.#cnt[end]) delete this.#cnt[end];
+
+                return false;
+            }
+        }
+
+        return true;
+    }
+}

solution/0700-0799/0731.My Calendar II/Solution3.ts

@@ -0,0 +1,25 @@
+class MyCalendarTwo {
+    #OVERLAPS = 2;
+    #cnt: Record<number, number> = {};
+
+    book(start: number, end: number): boolean {
+        this.#cnt[start] = (this.#cnt[start] ?? 0) + 1;
+        this.#cnt[end] = (this.#cnt[end] ?? 0) - 1;
+
+        let sum = 0;
+        for (const v of Object.values(this.#cnt)) {
+            sum += v;
+            if (sum > this.#OVERLAPS) {
+                this.#cnt[start]--;
+                this.#cnt[end]++;
+
+                if (!this.#cnt[start]) delete this.#cnt[start];
+                if (!this.#cnt[end]) delete this.#cnt[end];
+
+                return false;
+            }
+        }
+
+        return true;
+    }
+}

solution/0900-0999/0908.Smallest Range I/README.md

@@ -19,7 +19,7 @@ tags:
 
 <p>给你一个整数数组 <code>nums</code>，和一个整数 <code>k</code> 。</p>
 
-<p>在一个操作中，您可以选择 <code>0 &lt;= i &lt; nums.length</code> 的任何索引 <code>i</code> 。将 <code>nums[i]</code> 改为 <code>nums[i] + x</code> ，其中 <code>x</code> 是一个范围为 <code>[-k, k]</code> 的整数。对于每个索引 <code>i</code> ，最多 <strong>只能 </strong>应用 <strong>一次</strong> 此操作。</p>
+<p>在一个操作中，您可以选择 <code>0 &lt;= i &lt; nums.length</code> 的任何索引 <code>i</code> 。将 <code>nums[i]</code> 改为 <code>nums[i] + x</code> ，其中 <code>x</code> 是一个范围为 <code>[-k, k]</code> 的任意整数。对于每个索引 <code>i</code> ，最多 <strong>只能 </strong>应用 <strong>一次</strong> 此操作。</p>
 
 <p><code>nums</code>&nbsp;的&nbsp;<strong>分数&nbsp;</strong>是&nbsp;<code>nums</code>&nbsp;中最大和最小元素的差值。&nbsp;</p>
 

solution/1500-1599/1503.Last Moment Before All Ants Fall Out of a Plank/README.md

@@ -90,9 +90,9 @@ tags:
 
 题目关键点在于两只蚂蚁相遇，然后分别调转方向的情况，实际上相当于两只蚂蚁继续往原来的方向移动。因此，我们只需要求出所有蚂蚁中最远的那只蚂蚁的移动距离即可。
 
-注意 $left$ 和 $right$ 数组的长度可能为 $0$。
+注意 $\textit{left}$ 和 $\textit{right}$ 数组的长度可能为 $0$。
 
-时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为木板的长度。
+时间复杂度 $O(n)$，其中 $n$ 为木板的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 

solution/1500-1599/1503.Last Moment Before All Ants Fall Out of a Plank/README_EN.md

@@ -77,7 +77,13 @@ The last moment when an ant was on the plank is t = 4 seconds. After that, it fa
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Brain Teaser
+
+The key point of the problem is that when two ants meet and then turn around, it is equivalent to the two ants continuing to move in their original directions. Therefore, we only need to find the maximum distance moved by any ant.
+
+Note that the lengths of the $\textit{left}$ and $\textit{right}$ arrays may be $0$.
+
+The time complexity is $O(n)$, where $n$ is the length of the plank. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 

solution/1500-1599/1504.Count Submatrices With All Ones/README.md

@@ -197,6 +197,37 @@ func numSubmat(mat [][]int) (ans int) {
 }
 ```
 
+#### TypeScript
+
+```ts
+function numSubmat(mat: number[][]): number {
+    const m = mat.length;
+    const n = mat[0].length;
+    const g: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
+
+    for (let i = 0; i < m; i++) {
+        for (let j = 0; j < n; j++) {
+            if (mat[i][j]) {
+                g[i][j] = j === 0 ? 1 : 1 + g[i][j - 1];
+            }
+        }
+    }
+
+    let ans = 0;
+    for (let i = 0; i < m; i++) {
+        for (let j = 0; j < n; j++) {
+            let col = Infinity;
+            for (let k = i; k >= 0; k--) {
+                col = Math.min(col, g[k][j]);
+                ans += col;
+            }
+        }
+    }
+
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1500-1599/1504.Count Submatrices With All Ones/README_EN.md

@@ -69,7 +69,13 @@ Total number of rectangles = 8 + 5 + 2 + 4 + 2 + 2 + 1 = 24.
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Enumeration + Prefix Sum
+
+We can enumerate the bottom-right corner $(i, j)$ of the matrix, and then enumerate the first row $k$ upwards. The width of the matrix with $(i, j)$ as the bottom-right corner in each row is $\min_{k \leq i} \textit{g}[k][j]$, where $\textit{g}[k][j]$ represents the width of the matrix with $(k, j)$ as the bottom-right corner in the $k$-th row.
+
+Therefore, we can preprocess a 2D array $g[i][j]$, where $g[i][j]$ represents the number of consecutive $1$s from the $j$-th column to the left in the $i$-th row.
+
+The time complexity is $O(m^2 \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the matrix, respectively.
 
 <!-- tabs:start -->
 
@@ -184,6 +190,37 @@ func numSubmat(mat [][]int) (ans int) {
 }
 ```
 
+#### TypeScript
+
+```ts
+function numSubmat(mat: number[][]): number {
+    const m = mat.length;
+    const n = mat[0].length;
+    const g: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
+
+    for (let i = 0; i < m; i++) {
+        for (let j = 0; j < n; j++) {
+            if (mat[i][j]) {
+                g[i][j] = j === 0 ? 1 : 1 + g[i][j - 1];
+            }
+        }
+    }
+
+    let ans = 0;
+    for (let i = 0; i < m; i++) {
+        for (let j = 0; j < n; j++) {
+            let col = Infinity;
+            for (let k = i; k >= 0; k--) {
+                col = Math.min(col, g[k][j]);
+                ans += col;
+            }
+        }
+    }
+
+    return ans;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1500-1599/1504.Count Submatrices With All Ones/Solution.ts

@@ -0,0 +1,26 @@
+function numSubmat(mat: number[][]): number {
+    const m = mat.length;
+    const n = mat[0].length;
+    const g: number[][] = Array.from({ length: m }, () => Array(n).fill(0));
+
+    for (let i = 0; i < m; i++) {
+        for (let j = 0; j < n; j++) {
+            if (mat[i][j]) {
+                g[i][j] = j === 0 ? 1 : 1 + g[i][j - 1];
+            }
+        }
+    }
+
+    let ans = 0;
+    for (let i = 0; i < m; i++) {
+        for (let j = 0; j < n; j++) {
+            let col = Infinity;
+            for (let k = i; k >= 0; k--) {
+                col = Math.min(col, g[k][j]);
+                ans += col;
+            }
+        }
+    }
+
+    return ans;
+}