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feat: add weekly contest 471
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‎solution/3700-3799/3712.Sum of Elements With Frequency Divisible by K/README.md‎

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---
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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/3700-3799/3712.Sum%20of%20Elements%20With%20Frequency%20Divisible%20by%20K/README.md
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---
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<!-- problem:start -->
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# [3712. 出现次数能被 K 整除的元素总和](https://leetcode.cn/problems/sum-of-elements-with-frequency-divisible-by-k)
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[English Version](/solution/3700-3799/3712.Sum%20of%20Elements%20With%20Frequency%20Divisible%20by%20K/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> 和一个整数 <code>k</code>。</p>
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<p>请返回一个整数，表示 <code>nums</code> 中所有其 <strong>出现次数</strong> 能被 <code>k</code> 整除的元素的<strong>总和</strong>；如果没有这样的元素，则返回 0 。</p>
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<p><strong>注意：</strong> 若某个元素在数组中的总出现次数能被 <code>k</code> 整除，则它在求和中会被计算 <strong>恰好</strong> 与其出现次数相同的次数。</p>
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<p>元素 <code>x</code> 的&nbsp;<strong>出现次数&nbsp;</strong>指它在数组中出现的次数。</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 = [1,2,2,3,3,3,3,4], k = 2</span></p>
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<p><strong>输出：</strong> <span class="example-io">16</span></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>数字 1 出现 1 次（奇数次）。</li>
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<li>数字 2 出现 2 次（偶数次）。</li>
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<li>数字 3 出现 4 次（偶数次）。</li>
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<li>数字 4 出现 1 次（奇数次）。</li>
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</ul>
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<p>因此总和为 <code>2 + 2 + 3 + 3 + 3 + 3 = 16</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 = [1,2,3,4,5], k = 2</span></p>
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<p><strong>输出：</strong> <span class="example-io">0</span></p>
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<p><strong>解释：</strong></p>
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<p>没有元素出现偶数次，因此总和为 0。</p>
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</div>
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<p><strong class="example">示例 3：</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong> <span class="example-io">nums = [4,4,4,1,2,3], k = 3</span></p>
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<p><strong>输出：</strong> <span class="example-io">12</span></p>
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<p><strong>解释：</strong></p>
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<ul>
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<li>数字 1 出现 1 次。</li>
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<li>数字 2 出现 1 次。</li>
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<li>数字 3 出现 1 次。</li>
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<li>数字 4 出现 3 次。</li>
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</ul>
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<p>因此总和为 <code>4 + 4 + 4 = 12</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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<li><code>1 &lt;= k &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：计数
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我们用一个哈希表 $\textit{cnt}$ 来记录每个数字的出现次数。遍历数组 $\textit{nums}$，对于每个数字 $x$，我们将 $\textit{cnt}[x]$ 增加 $1$。
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然后，我们遍历哈希表 $\textit{cnt}$，对于每个元素 $x$，如果它的出现次数 $\textit{cnt}[x]$ 能被 $k$ 整除，就将 $x$ 乘以它的出现次数加到结果中。
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时间复杂度为 $O(n)$，其中 $n$ 是数组的长度。空间复杂度为 $O(m)$，其中 $m$ 是数组中不同元素的数量。
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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 sumDivisibleByK(self, nums: List[int], k: int) -> int:
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cnt = Counter(nums)
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return sum(x * v for x, v in cnt.items() if v % k == 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 sumDivisibleByK(int[] nums, int k) {
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Map<Integer, Integer> cnt = new HashMap<>();
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for (int x : nums) {
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cnt.merge(x, 1, Integer::sum);
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}
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int ans = 0;
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for (var e : cnt.entrySet()) {
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int x = e.getKey(), v = e.getValue();
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if (v % k == 0) {
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ans += x * v;
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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 sumDivisibleByK(vector<int>& nums, int k) {
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unordered_map<int, int> cnt;
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for (int x : nums) {
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++cnt[x];
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}
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int ans = 0;
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for (auto& [x, v] : cnt) {
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if (v % k == 0) {
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ans += x * v;
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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 sumDivisibleByK(nums []int, k int) (ans int) {
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cnt := map[int]int{}
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for _, x := range nums {
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cnt[x]++
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}
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for x, v := range cnt {
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if v%k == 0 {
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ans += x * v
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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 sumDivisibleByK(nums: number[], k: number): number {
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const cnt = new Map();
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for (const x of nums) {
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cnt.set(x, (cnt.get(x) || 0) + 1);
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}
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let ans = 0;
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for (const [x, v] of cnt.entries()) {
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if (v % k === 0) {
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ans += x * v;
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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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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

‎solution/3700-3799/3712.Sum of Elements With Frequency Divisible by K/README_EN.md‎

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---
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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/3700-3799/3712.Sum%20of%20Elements%20With%20Frequency%20Divisible%20by%20K/README_EN.md
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---
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<!-- problem:start -->
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# [3712. Sum of Elements With Frequency Divisible by K](https://leetcode.com/problems/sum-of-elements-with-frequency-divisible-by-k)
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[中文文档](/solution/3700-3799/3712.Sum%20of%20Elements%20With%20Frequency%20Divisible%20by%20K/README.md)
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## Description
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<!-- description:start -->
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<p>You are given an integer array <code>nums</code> and an integer <code>k</code>.</p>
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<p>Return an integer denoting the <strong>sum</strong> of all elements in <code>nums</code> whose <strong>frequency</strong> is divisible by <code>k</code>, or 0 if there are no such elements.</p>
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<p><strong>Note:</strong> An element is included in the sum <strong>exactly</strong> as many times as it appears in the array if its total frequency is divisible by <code>k</code>.</p>
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<p>The <strong>frequency</strong> of an element <code>x</code> is the number of times it occurs in the array.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums = [1,2,2,3,3,3,3,4], k = 2</span></p>
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<p><strong>Output:</strong> <span class="example-io">16</span></p>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>The number 1 appears once (odd frequency).</li>
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<li>The number 2 appears twice (even frequency).</li>
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<li>The number 3 appears four times (even frequency).</li>
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<li>The number 4 appears once (odd frequency).</li>
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</ul>
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<p>So, the total sum is <code>2 + 2 + 3 + 3 + 3 + 3 = 16</code>.</p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3,4,5], k = 2</span></p>
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<p><strong>Output:</strong> <span class="example-io">0</span></p>
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<p><strong>Explanation:</strong></p>
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<p>There are no elements that appear an even number of times, so the total sum is 0.</p>
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</div>
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<p><strong class="example">Example 3:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io">nums = [4,4,4,1,2,3], k = 3</span></p>
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<p><strong>Output:</strong> <span class="example-io">12</span></p>
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<p><strong>Explanation:</strong></p>
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<ul>
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<li>The number 1 appears once.</li>
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<li>The number 2 appears once.</li>
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<li>The number 3 appears once.</li>
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<li>The number 4 appears three times.</li>
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</ul>
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<p>So, the total sum is <code>4 + 4 + 4 = 12</code>.</p>
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</div>
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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 &lt;= 100</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 100</code></li>
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<li><code>1 &lt;= k &lt;= 100</code></li>
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</ul>
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<!-- description:end -->
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## Solutions
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<!-- solution:start -->
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### Solution 1: Counting
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We use a hash table $\textit{cnt}$ to record the frequency of each number. We traverse the array $\textit{nums}$, and for each number $x$, we increment $\textit{cnt}[x]$ by $1$.
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Then, we traverse the hash table $\textit{cnt}$. For each element $x$, if its frequency $\textit{cnt}[x]$ is divisible by $k$, we add $x$ multiplied by its frequency to the result.
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The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(m)$, where $m$ is the number of distinct elements in the array.
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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 sumDivisibleByK(self, nums: List[int], k: int) -> int:
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cnt = Counter(nums)
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return sum(x * v for x, v in cnt.items() if v % k == 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 sumDivisibleByK(int[] nums, int k) {
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Map<Integer, Integer> cnt = new HashMap<>();
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for (int x : nums) {
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cnt.merge(x, 1, Integer::sum);
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}
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int ans = 0;
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for (var e : cnt.entrySet()) {
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int x = e.getKey(), v = e.getValue();
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if (v % k == 0) {
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ans += x * v;
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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 sumDivisibleByK(vector<int>& nums, int k) {
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unordered_map<int, int> cnt;
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for (int x : nums) {
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++cnt[x];
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}
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int ans = 0;
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for (auto& [x, v] : cnt) {
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if (v % k == 0) {
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ans += x * v;
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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 sumDivisibleByK(nums []int, k int) (ans int) {
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cnt := map[int]int{}
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for _, x := range nums {
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cnt[x]++
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}
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for x, v := range cnt {
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if v%k == 0 {
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ans += x * v
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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 sumDivisibleByK(nums: number[], k: number): number {
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const cnt = new Map();
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for (const x of nums) {
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cnt.set(x, (cnt.get(x) || 0) + 1);
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}
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let ans = 0;
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for (const [x, v] of cnt.entries()) {
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if (v % k === 0) {
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ans += x * v;
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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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<!-- tabs:end -->
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<!-- solution:end -->
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<!-- problem:end -->

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