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feat: add biweekly contest 169
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‎solution/2500-2599/2528.Maximize the Minimum Powered City/README_EN.md‎

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<pre>
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<strong>Input:</strong> stations = [1,2,4,5,0], r = 1, k = 2
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<strong>Output:</strong> 5
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<strong>Explanation:</strong>
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One of the optimal ways is to install both the power stations at city 1.
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<strong>Explanation:</strong>
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One of the optimal ways is to install both the power stations at city 1.
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So stations will become [1,4,4,5,0].
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- City 0 is provided by 1 + 4 = 5 power stations.
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- City 1 is provided by 1 + 4 + 4 = 9 power stations.
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<pre>
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<strong>Input:</strong> stations = [4,4,4,4], r = 0, k = 3
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<strong>Output:</strong> 4
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<strong>Explanation:</strong>
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<strong>Explanation:</strong>
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It can be proved that we cannot make the minimum power of a city greater than 4.
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</pre>
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‎solution/3700-3799/3731.Find Missing Elements/README.md‎

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comments: true
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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3731.Find%20Missing%20Elements/README.md
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rating: 1217
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source: 第 474 场周赛 Q1
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---
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‎solution/3700-3799/3731.Find Missing Elements/README_EN.md‎

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comments: true
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difficulty: Easy
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3731.Find%20Missing%20Elements/README_EN.md
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rating: 1217
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source: Weekly Contest 474 Q1
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‎solution/3700-3799/3732.Maximum Product of Three Elements After One Replacement/README.md‎

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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3732.Maximum%20Product%20of%20Three%20Elements%20After%20One%20Replacement/README.md
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rating: 1529
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source: 第 474 场周赛 Q2
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‎solution/3700-3799/3732.Maximum Product of Three Elements After One Replacement/README_EN.md‎

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comments: true
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difficulty: Medium
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3732.Maximum%20Product%20of%20Three%20Elements%20After%20One%20Replacement/README_EN.md
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rating: 1529
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source: Weekly Contest 474 Q2
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‎solution/3700-3799/3733.Minimum Time to Complete All Deliveries/README.md‎

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comments: true
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difficulty: 中等
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3733.Minimum%20Time%20to%20Complete%20All%20Deliveries/README.md
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rating: 1972
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source: 第 474 场周赛 Q3
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‎solution/3700-3799/3733.Minimum Time to Complete All Deliveries/README_EN.md‎

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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3733.Minimum%20Time%20to%20Complete%20All%20Deliveries/README_EN.md
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rating: 1972
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source: Weekly Contest 474 Q3
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‎solution/3700-3799/3734.Lexicographically Smallest Palindromic Permutation Greater Than Target/README.md‎

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difficulty: 困难
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3734.Lexicographically%20Smallest%20Palindromic%20Permutation%20Greater%20Than%20Target/README.md
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rating: 2330
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source: 第 474 场周赛 Q4
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‎solution/3700-3799/3734.Lexicographically Smallest Palindromic Permutation Greater Than Target/README_EN.md‎

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difficulty: Hard
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3734.Lexicographically%20Smallest%20Palindromic%20Permutation%20Greater%20Than%20Target/README_EN.md
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rating: 2330
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source: Weekly Contest 474 Q4
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‎solution/3700-3799/3736.Minimum Moves to Equal Array Elements III/README.md‎

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difficulty: 简单
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edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3736.Minimum%20Moves%20to%20Equal%20Array%20Elements%20III/README.md
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---
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<!-- problem:start -->
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# [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)
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[English Version](/solution/3700-3799/3736.Minimum%20Moves%20to%20Equal%20Array%20Elements%20III/README_EN.md)
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## 题目描述
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<!-- description:start -->
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<p>给你一个整数数组 <code>nums</code>。</p>
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<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>
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<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>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1:</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong> <span class="example-io">nums = [2,1,3]</span></p>
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<p><strong>输出:</strong> <span class="example-io">3</span></p>
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<p><strong>解释:</strong></p>
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<p>使所有元素相等的操作如下:</p>
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<ul>
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<li>将 <code>nums[0] = 2</code> 增加 1, 变为 3。</li>
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<li>将 <code>nums[1] = 1</code> 增加 1, 变为 2。</li>
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<li>将 <code>nums[1] = 2</code> 增加 1, 变为 3。</li>
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</ul>
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<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>
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</div>
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<p><strong class="example">示例 2:</strong></p>
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<div class="example-block">
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<p><strong>输入:</strong> <span class="example-io">nums = [4,4,5]</span></p>
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<p><strong>输出:</strong> <span class="example-io">2</span></p>
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<p><strong>解释:</strong></p>
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<p>使所有元素相等的操作如下:</p>
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<ul>
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<li>将 <code>nums[0] = 4</code> 增加 1, 变为 5。</li>
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<li>将 <code>nums[1] = 4</code> 增加 1, 变为 5。</li>
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</ul>
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<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>
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</div>
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<p>&nbsp;</p>
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<p><strong>提示:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 100</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
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</ul>
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<!-- description:end -->
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## 解法
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<!-- solution:start -->
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### 方法一： 计算总和与最大值
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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。
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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。
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时间复杂度 $O(n)$，其中 $n$ 是数组的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def minMoves(self, nums: List[int]) -> int:
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n = len(nums)
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mx = max(nums)
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s = sum(nums)
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return mx * n - s
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```
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#### Java
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```java
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class Solution {
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public int minMoves(int[] nums) {
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int n = nums.length;
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int mx = 0;
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int s = 0;
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for (int x : nums) {
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mx = Math.max(mx, x);
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s += x;
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}
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return mx * n - s;
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}
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}
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```
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#### C++
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```cpp
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class Solution {
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public:
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int minMoves(vector<int>& nums) {
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int n = nums.size();
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int mx = 0;
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int s = 0;
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for (int x : nums) {
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mx = max(mx, x);
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s += x;
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}
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return mx * n - s;
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}
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};
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```
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#### Go
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```go
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func minMoves(nums []int) int {
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mx, s := 0, 0
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for _, x := range nums {
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mx = max(mx, x)
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s += x
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}
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return mx*len(nums) - s
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}
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```
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#### TypeScript
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```ts
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function minMoves(nums: number[]): number {
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const n = nums.length;
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const mx = Math.max(...nums);
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const s = nums.reduce((a, b) => a + b, 0);
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return mx * n - s;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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