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feat: add solutions to lc problems: No.2778~2781 (#1225)
* No.2778.Sum of Squares of Special Elements * No.2779.Maximum Beauty of an Array After Applying Operation * No.2780.Minimum Index of a Valid Split * No.2781.Length of the Longest Valid Substring
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‎solution/2700-2799/2778.Sum of Squares of Special Elements/README.md

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# [2778. 特殊元素平方和](https://leetcode.cn/problems/sum-of-squares-of-special-elements)
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[English Version](/solution/2700-2799/2778.Sum%20of%20Squares%20of%20Special%20Elements/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个下标从 <strong>1</strong> 开始、长度为 <code>n</code> 的整数数组 <code>nums</code> 。</p>
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<p>对 <code>nums</code> 中的元素 <code>nums[i]</code> 而言，如果 <code>n</code> 能够被 <code>i</code> 整除，即 <code>n % i == 0</code> ，则认为 <code>num[i]</code> 是一个 <strong>特殊元素</strong> 。</p>
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<p>返回 <code>nums</code> 中所有 <strong>特殊元素</strong> 的 <strong>平方和</strong> 。</p>
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<p>&nbsp;</p>
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<p><strong>示例 1：</strong></p>
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<pre><strong>输入：</strong>nums = [1,2,3,4]
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<strong>输出：</strong>21
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<strong>解释：</strong>nums 中共有 3 个特殊元素：nums[1] ，因为 4 被 1 整除；nums[2] ，因为 4 被 2 整除；以及 nums[4] ，因为 4 被 4 整除。
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因此，nums 中所有元素的平方和等于 nums[1] * nums[1] + nums[2] * nums[2] + nums[4] * nums[4] = 1 * 1 + 2 * 2 + 4 * 4 = 21 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><strong>输入：</strong>nums = [2,7,1,19,18,3]
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<strong>输出：</strong>63
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<strong>解释：</strong>nums 中共有 4 个特殊元素：nums[1] ，因为 6 被 1 整除；nums[2] ，因为 6 被 2 整除；nums[3] ，因为 6 被 3 整除；以及 nums[6] ，因为 6 被 6 整除。
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因此，nums 中所有元素的平方和等于 nums[1] * nums[1] + nums[2] * nums[2] + nums[3] * nums[3] + nums[6] * nums[6] = 2 * 2 + 7 * 7 + 1 * 1 + 3 * 3 = 63 。 </pre>
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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 == n &lt;= 50</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 50</code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举**
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我们可以枚举数组中的每个元素，判断其是否为特殊元素，如果是则将其平方加入答案中。
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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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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def sumOfSquares(self, nums: List[int]) -> int:
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n = len(nums)
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return sum(x * x for i, x in enumerate(nums, 1) if n % i == 0)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int sumOfSquares(int[] nums) {
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int n = nums.length;
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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if (n % i == 0) {
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ans += nums[i - 1] * nums[i - 1];
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}
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}
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return ans;
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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 sumOfSquares(vector<int>& nums) {
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int n = nums.size();
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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if (n % i == 0) {
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ans += nums[i - 1] * nums[i - 1];
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}
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}
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return ans;
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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 sumOfSquares(nums []int) (ans int) {
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n := len(nums)
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for i, x := range nums {
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if n%(i+1) == 0 {
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ans += x * x
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function sumOfSquares(nums: number[]): number {
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const n = nums.length;
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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if (n % (i + 1) === 0) {
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ans += nums[i] * nums[i];
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}
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}
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return ans;
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2700-2799/2778.Sum of Squares of Special Elements/README_EN.md

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# [2778. Sum of Squares of Special Elements](https://leetcode.com/problems/sum-of-squares-of-special-elements)
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[中文文档](/solution/2700-2799/2778.Sum%20of%20Squares%20of%20Special%20Elements/README.md)
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## Description
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<p>You are given a <strong>1-indexed</strong> integer array <code>nums</code> of length <code>n</code>.</p>
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<p>An element <code>nums[i]</code> of <code>nums</code> is called <strong>special</strong> if <code>i</code> divides <code>n</code>, i.e. <code>n % i == 0</code>.</p>
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<p>Return <em>the <strong>sum of the squares</strong> of all <strong>special</strong> elements of </em><code>nums</code>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,3,4]
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<strong>Output:</strong> 21
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<strong>Explanation:</strong> There are exactly 3 special elements in nums: nums[1] since 1 divides 4, nums[2] since 2 divides 4, and nums[4] since 4 divides 4.
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Hence, the sum of the squares of all special elements of nums is nums[1] * nums[1] + nums[2] * nums[2] + nums[4] * nums[4] = 1 * 1 + 2 * 2 + 4 * 4 = 21.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [2,7,1,19,18,3]
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<strong>Output:</strong> 63
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<strong>Explanation:</strong> There are exactly 4 special elements in nums: nums[1] since 1 divides 6, nums[2] since 2 divides 6, nums[3] since 3 divides 6, and nums[6] since 6 divides 6.
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Hence, the sum of the squares of all special elements of nums is nums[1] * nums[1] + nums[2] * nums[2] + nums[3] * nums[3] + nums[6] * nums[6] = 2 * 2 + 7 * 7 + 1 * 1 + 3 * 3 = 63.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length == n &lt;= 50</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 50</code></li>
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</ul>
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## Solutions
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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 sumOfSquares(self, nums: List[int]) -> int:
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n = len(nums)
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return sum(x * x for i, x in enumerate(nums, 1) if n % i == 0)
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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 sumOfSquares(int[] nums) {
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int n = nums.length;
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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if (n % i == 0) {
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ans += nums[i - 1] * nums[i - 1];
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}
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}
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return ans;
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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 sumOfSquares(vector<int>& nums) {
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int n = nums.size();
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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if (n % i == 0) {
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ans += nums[i - 1] * nums[i - 1];
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}
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}
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return ans;
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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 sumOfSquares(nums []int) (ans int) {
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n := len(nums)
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for i, x := range nums {
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if n%(i+1) == 0 {
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ans += x * x
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function sumOfSquares(nums: number[]): number {
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const n = nums.length;
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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if (n % (i + 1) === 0) {
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ans += nums[i] * nums[i];
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}
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}
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return ans;
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}
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/2700-2799/2778.Sum of Squares of Special Elements/Solution.cpp

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class Solution {
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public:
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int sumOfSquares(vector<int>& nums) {
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int n = nums.size();
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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if (n % i == 0) {
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ans += nums[i - 1] * nums[i - 1];
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}
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}
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return ans;
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}
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};

‎solution/2700-2799/2778.Sum of Squares of Special Elements/Solution.go

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func sumOfSquares(nums []int) (ans int) {
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n := len(nums)
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for i, x := range nums {
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if n%(i+1) == 0 {
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ans += x * x
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}
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}
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return
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}

‎solution/2700-2799/2778.Sum of Squares of Special Elements/Solution.java

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class Solution {
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public int sumOfSquares(int[] nums) {
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int n = nums.length;
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int ans = 0;
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for (int i = 1; i <= n; ++i) {
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if (n % i == 0) {
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ans += nums[i - 1] * nums[i - 1];
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}
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}
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return ans;
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}
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}

‎solution/2700-2799/2778.Sum of Squares of Special Elements/Solution.py

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class Solution:
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def sumOfSquares(self, nums: List[int]) -> int:
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n = len(nums)
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return sum(x * x for i, x in enumerate(nums, 1) if n % i == 0)

‎solution/2700-2799/2778.Sum of Squares of Special Elements/Solution.ts

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function sumOfSquares(nums: number[]): number {
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const n = nums.length;
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let ans = 0;
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for (let i = 0; i < n; ++i) {
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if (n % (i + 1) === 0) {
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ans += nums[i] * nums[i];
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}
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}
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return ans;
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}

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