feat: add biweekly contest 169

solution/2500-2599/2528.Maximize the Minimum Powered City/README_EN.md

@@ -45,8 +45,8 @@ tags:
 <pre>
 <strong>Input:</strong> stations = [1,2,4,5,0], r = 1, k = 2
 <strong>Output:</strong> 5
-<strong>Explanation:</strong>
-One of the optimal ways is to install both the power stations at city 1.
+<strong>Explanation:</strong> 
+One of the optimal ways is to install both the power stations at city 1. 
 So stations will become [1,4,4,5,0].
 - City 0 is provided by 1 + 4 = 5 power stations.
 - City 1 is provided by 1 + 4 + 4 = 9 power stations.
@@ -62,7 +62,7 @@ Since it is not possible to obtain a larger power, we return 5.
 <pre>
 <strong>Input:</strong> stations = [4,4,4,4], r = 0, k = 3
 <strong>Output:</strong> 4
-<strong>Explanation:</strong>
+<strong>Explanation:</strong> 
 It can be proved that we cannot make the minimum power of a city greater than 4.
 </pre>
 

solution/3700-3799/3731.Find Missing Elements/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 简单
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3731.Find%20Missing%20Elements/README.md
+rating: 1217
+source: 第 474 场周赛 Q1
 ---
 
 <!-- problem:start -->

solution/3700-3799/3731.Find Missing Elements/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Easy
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3731.Find%20Missing%20Elements/README_EN.md
+rating: 1217
+source: Weekly Contest 474 Q1
 ---
 
 <!-- problem:start -->

solution/3700-3799/3732.Maximum Product of Three Elements After One Replacement/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3732.Maximum%20Product%20of%20Three%20Elements%20After%20One%20Replacement/README.md
+rating: 1529
+source: 第 474 场周赛 Q2
 ---
 
 <!-- problem:start -->

solution/3700-3799/3732.Maximum Product of Three Elements After One Replacement/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3732.Maximum%20Product%20of%20Three%20Elements%20After%20One%20Replacement/README_EN.md
+rating: 1529
+source: Weekly Contest 474 Q2
 ---
 
 <!-- problem:start -->

solution/3700-3799/3733.Minimum Time to Complete All Deliveries/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 中等
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3733.Minimum%20Time%20to%20Complete%20All%20Deliveries/README.md
+rating: 1972
+source: 第 474 场周赛 Q3
 ---
 
 <!-- problem:start -->

solution/3700-3799/3733.Minimum Time to Complete All Deliveries/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Medium
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3733.Minimum%20Time%20to%20Complete%20All%20Deliveries/README_EN.md
+rating: 1972
+source: Weekly Contest 474 Q3
 ---
 
 <!-- problem:start -->

solution/3700-3799/3734.Lexicographically Smallest Palindromic Permutation Greater Than Target/README.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: 困难
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3734.Lexicographically%20Smallest%20Palindromic%20Permutation%20Greater%20Than%20Target/README.md
+rating: 2330
+source: 第 474 场周赛 Q4
 ---
 
 <!-- problem:start -->

solution/3700-3799/3734.Lexicographically Smallest Palindromic Permutation Greater Than Target/README_EN.md

@@ -2,6 +2,8 @@
 comments: true
 difficulty: Hard
 edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3734.Lexicographically%20Smallest%20Palindromic%20Permutation%20Greater%20Than%20Target/README_EN.md
+rating: 2330
+source: Weekly Contest 474 Q4
 ---
 
 <!-- problem:start -->

solution/3700-3799/3736.Minimum Moves to Equal Array Elements III/README.md

@@ -0,0 +1,163 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3736.Minimum%20Moves%20to%20Equal%20Array%20Elements%20III/README.md
+---
+
+<!-- problem:start -->
+
+# [3736. 最小操作次数使数组元素相等 III](https://leetcode.cn/problems/minimum-moves-to-equal-array-elements-iii)
+
+[English Version](/solution/3700-3799/3736.Minimum%20Moves%20to%20Equal%20Array%20Elements%20III/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个整数数组 <code>nums</code>。</p>
+
+<p>在一步操作中，你可以将任意单个元素 <code>nums[i]</code> 的值&nbsp;<strong>增加</strong> 1。</p>
+
+<p>返回使数组中的所有元素都&nbsp;<strong>相等&nbsp;</strong>所需的&nbsp;<strong>最小总操作次数&nbsp;</strong>。</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">nums = [2,1,3]</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">3</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>使所有元素相等的操作如下:</p>
+
+<ul>
+	<li>将 <code>nums[0] = 2</code> 增加 1, 变为 3。</li>
+	<li>将 <code>nums[1] = 1</code> 增加 1, 变为 2。</li>
+	<li>将 <code>nums[1] = 2</code> 增加 1, 变为 3。</li>
+</ul>
+
+<p>现在，<code>nums</code> 中的所有元素都等于 3。最小总操作次数为 <code>3</code>。</p>
+</div>
+
+<p><strong class="example">示例 2:</strong></p>
+
+<div class="example-block">
+<p><strong>输入:</strong> <span class="example-io">nums = [4,4,5]</span></p>
+
+<p><strong>输出:</strong> <span class="example-io">2</span></p>
+
+<p><strong>解释:</strong></p>
+
+<p>使所有元素相等的操作如下:</p>
+
+<ul>
+	<li>将 <code>nums[0] = 4</code> 增加 1, 变为 5。</li>
+	<li>将 <code>nums[1] = 4</code> 增加 1, 变为 5。</li>
+</ul>
+
+<p>现在，<code>nums</code> 中的所有元素都等于 5。最小总操作次数为 <code>2</code>。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一： 计算总和与最大值
+
+本题要求将数组中的所有元素变为相等，且每次只能将某个元素增加 1。为了使操作次数最少，我们应当将所有元素都变为数组中的最大值。
+
+因此，我们可以先计算数组的最大值 $\textit{mx}$ 和数组元素的总和 $\textit{s}$，那么将所有元素变为 $\textit{mx}$ 所需的操作次数即为 $\textit{mx} \times n - \textit{s}$，其中 $n$ 是数组的长度。
+
+时间复杂度 $O(n)$，其中 $n$ 是数组的长度。空间复杂度 $O(1)$。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def minMoves(self, nums: List[int]) -> int:
+        n = len(nums)
+        mx = max(nums)
+        s = sum(nums)
+        return mx * n - s
+```
+
+#### Java
+
+```java
+class Solution {
+    public int minMoves(int[] nums) {
+        int n = nums.length;
+        int mx = 0;
+        int s = 0;
+        for (int x : nums) {
+            mx = Math.max(mx, x);
+            s += x;
+        }
+        return mx * n - s;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int minMoves(vector<int>& nums) {
+        int n = nums.size();
+        int mx = 0;
+        int s = 0;
+        for (int x : nums) {
+            mx = max(mx, x);
+            s += x;
+        }
+        return mx * n - s;
+    }
+};
+```
+
+#### Go
+
+```go
+func minMoves(nums []int) int {
+	mx, s := 0, 0
+	for _, x := range nums {
+		mx = max(mx, x)
+		s += x
+	}
+	return mx*len(nums) - s
+}
+```
+
+#### TypeScript
+
+```ts
+function minMoves(nums: number[]): number {
+    const n = nums.length;
+    const mx = Math.max(...nums);
+    const s = nums.reduce((a, b) => a + b, 0);
+    return mx * n - s;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->