feat: add solutions to lc problems: No.4024~4029

solution/4000-4099/4014.Minimum Total Price After Applying Discounts/README.md

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 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4014.Minimum%20Total%20Price%20After%20Applying%20Discounts/README.md
+rating: 1192
+source: 第 514 场周赛 Q1
 ---
 
 <!-- problem:start -->

solution/4000-4099/4014.Minimum Total Price After Applying Discounts/README_EN.md

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 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4014.Minimum%20Total%20Price%20After%20Applying%20Discounts/README_EN.md
+rating: 1192
+source: Weekly Contest 514 Q1
 ---
 
 <!-- problem:start -->

solution/4000-4099/4015.Weighted Sum of a Tree/README.md

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 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4015.Weighted%20Sum%20of%20a%20Tree/README.md
+rating: 1534
+source: 第 514 场周赛 Q2
 ---
 
 <!-- problem:start -->

solution/4000-4099/4015.Weighted Sum of a Tree/README_EN.md

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 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4015.Weighted%20Sum%20of%20a%20Tree/README_EN.md
+rating: 1534
+source: Weekly Contest 514 Q2
 ---
 
 <!-- problem:start -->

solution/4000-4099/4016.Maximum Area of Two Non-Overlapping Square Submatrices/README.md

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 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4016.Maximum%20Area%20of%20Two%20Non-Overlapping%20Square%20Submatrices/README.md
+rating: 1958
+source: 第 514 场周赛 Q3
 ---
 
 <!-- problem:start -->

solution/4000-4099/4016.Maximum Area of Two Non-Overlapping Square Submatrices/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4016.Maximum%20Area%20of%20Two%20Non-Overlapping%20Square%20Submatrices/README_EN.md
+rating: 1958
+source: Weekly Contest 514 Q3
 ---
 
 <!-- problem:start -->

solution/4000-4099/4017.Peaks in Array II/README.md

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 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4017.Peaks%20in%20Array%20II/README.md
+rating: 2515
+source: 第 514 场周赛 Q4
 ---
 
 <!-- problem:start -->

solution/4000-4099/4017.Peaks in Array II/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4017.Peaks%20in%20Array%20II/README_EN.md
+rating: 2515
+source: Weekly Contest 514 Q4
 ---
 
 <!-- problem:start -->

solution/4000-4099/4024.Nearest Available Drone/README.md

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+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/4000-4099/4024.Nearest%20Available%20Drone/README.md
+---
+
+<!-- problem:start -->
+
+# [4024. 最近的可用无人机](https://leetcode.cn/problems/nearest-available-drone)
+
+[English Version](/solution/4000-4099/4024.Nearest%20Available%20Drone/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个二维整数数组 <code>drones</code>，其中 <code>drones[i] = [x<sub>i</sub>, y<sub>i</sub>, range<sub>i</sub>]</code> 表示第 <code>i<sup>th</sup></code> 架无人机的横坐标、纵坐标和飞行范围。</p>
+
+<p>另给你一个整数数组 <code>target = [tx, ty]</code>，表示目标的坐标。</p>
+
+<p>如果无人机 <code>drones[i]</code> 的坐标与目标坐标之间的<strong>曼哈顿距离</strong><strong>小于或等于</strong>其 <code>range<sub>i</sub></code>，则该无人机能够到达目标。</p>
+
+<p>返回能够到达目标且与目标之间<strong>曼哈顿距离最小</strong>的无人机的<strong>下标</strong>。如果存在多个符合条件的无人机，则返回其中<strong>最小的下标</strong>。如果没有无人机能够到达目标，则返回 -1。</p>
+
+<p>两个坐标 <code>(x<sub>i</sub>, y<sub>i</sub>)</code> 和 <code>(x<sub>j</sub>, y<sub>j</sub>)</code> 之间的<strong>曼哈顿距离</strong>为 <code>|x<sub>i</sub> - x<sub>j</sub>| + |y<sub>i</sub> - y<sub>j</sub>|</code>。</p>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">drones = [[0,0,8],[2,2,9]], target = [3,4]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li><code>drones[0]</code> 与 <code>target</code> 之间的距离为 <code>|0 - 3| + |0 - 4| = 7</code>，没有超出其飞行范围 8。</li>
+	<li><code>drones[1]</code> 与 <code>target</code> 之间的距离为 <code>|2 - 3| + |2 - 4| = 3</code>，没有超出其飞行范围 9。</li>
+	<li>由于 <code>drones[1]</code> 是距离目标最近的无人机，因此答案为 1。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">drones = [[2,1,5],[4,4,5],[6,6,8]], target = [5,5]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li><code>drones[0]</code> 与 <code>target</code> 之间的距离为 <code>|2 - 5| + |1 - 5| = 7</code>，大于其飞行范围 5。</li>
+	<li><code>drones[1]</code> 与 <code>target</code> 之间的距离为 <code>|4 - 5| + |4 - 5| = 2</code>，没有超出其飞行范围 5。</li>
+	<li><code>drones[2]</code> 与 <code>target</code> 之间的距离为 <code>|6 - 5| + |6 - 5| = 2</code>，没有超出其飞行范围 8。</li>
+	<li><code>drones[1]</code> 和 <code>drones[2]</code> 都是距离目标最近的无人机。由于需要返回最小下标，因此答案为 1。</li>
+</ul>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">drones = [[4,4,5]], target = [8,6]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">-1</span></p>
+
+<p><strong>解释：</strong></p>
+
+<ul>
+	<li><code>drones[0]</code> 与 <code>target</code> 之间的距离为 <code>|4 - 8| + |4 - 6| = 6</code>，大于其飞行范围 5。</li>
+	<li>没有无人机能够到达目标，因此答案为 -1。</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= drones.length &lt;= 100</code></li>
+	<li><code>drones[i] = [x<sub>i</sub>, y<sub>i</sub>, range<sub>i</sub>]</code></li>
+	<li><code>target = [tx, ty]</code></li>
+	<li><code>-25 &lt;= x<sub>i</sub>, y<sub>i</sub>, tx, ty &lt;= 25</code></li>
+	<li><code>1 &lt;= range<sub>i</sub> &lt;= 100</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：遍历
+
+我们遍历每一架无人机，计算其与目标的曼哈顿距离 $d = |x_i - t_x| + |y_i - t_y|$。若 $d \le \textit{range}_i$，则该无人机可达。在所有可达无人机中，选择距离最小的一架；若距离相同，由于我们从左到右遍历且仅在距离严格更小时更新答案，因此会自动保留更小的下标。若没有可达无人机，返回 $-1$。
+
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 是无人机的数量。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def nearestDrone(self, drones: list[list[int]], target: list[int]) -> int:
+        ans = -1
+        mn = inf
+        tx, ty = target
+        for i, (x, y, r) in enumerate(drones):
+            d = abs(x - tx) + abs(y - ty)
+            if d <= r and mn > d:
+                ans = i
+                mn = d
+        return ans
+```
+
+#### Java
+
+```java
+class Solution {
+    public int nearestDrone(int[][] drones, int[] target) {
+        int ans = -1;
+        int mn = Integer.MAX_VALUE;
+        int tx = target[0], ty = target[1];
+
+        for (int i = 0; i < drones.length; i++) {
+            int x = drones[i][0];
+            int y = drones[i][1];
+            int r = drones[i][2];
+
+            int d = Math.abs(x - tx) + Math.abs(y - ty);
+
+            if (d <= r && mn > d) {
+                ans = i;
+                mn = d;
+            }
+        }
+
+        return ans;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int nearestDrone(vector<vector<int>>& drones, vector<int>& target) {
+        int ans = -1;
+        int mn = INT_MAX;
+        int tx = target[0], ty = target[1];
+
+        for (int i = 0; i < drones.size(); i++) {
+            int x = drones[i][0];
+            int y = drones[i][1];
+            int r = drones[i][2];
+
+            int d = abs(x - tx) + abs(y - ty);
+
+            if (d <= r && mn > d) {
+                ans = i;
+                mn = d;
+            }
+        }
+
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+func nearestDrone(drones [][]int, target []int) int {
+	ans := -1
+	mn := math.MaxInt32
+	tx, ty := target[0], target[1]
+
+	for i, drone := range drones {
+		x, y, r := drone[0], drone[1], drone[2]
+
+		d := abs(x-tx) + abs(y-ty)
+
+		if d <= r && mn > d {
+			ans = i
+			mn = d
+		}
+	}
+
+	return ans
+}
+
+func abs(x int) int {
+	if x < 0 {
+		return -x
+	}
+	return x
+}
+```
+
+#### TypeScript
+
+```ts
+function nearestDrone(drones: number[][], target: number[]): number {
+    let ans = -1;
+    let mn = Infinity;
+    const [tx, ty] = target;
+
+    for (let i = 0; i < drones.length; i++) {
+        const [x, y, r] = drones[i];
+
+        const d = Math.abs(x - tx) + Math.abs(y - ty);
+
+        if (d <= r && mn > d) {
+            ans = i;
+            mn = d;
+        }
+    }
+
+    return ans;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->