feat: add weekly contest 455

Author: yanglbme

solution/3500-3599/3591.Check if Any Element Has Prime Frequency/README.md

@@ -0,0 +1,230 @@
+---
+comments: true
+difficulty: 简单
+edit_url: https://github.com/doocs/leetcode/edit/main/solution/3500-3599/3591.Check%20if%20Any%20Element%20Has%20Prime%20Frequency/README.md
+---
+
+<!-- problem:start -->
+
+# [3591. 检查元素频次是否为质数](https://leetcode.cn/problems/check-if-any-element-has-prime-frequency)
+
+[English Version](/solution/3500-3599/3591.Check%20if%20Any%20Element%20Has%20Prime%20Frequency/README_EN.md)
+
+## 题目描述
+
+<!-- description:start -->
+
+<p>给你一个整数数组 <code>nums</code>。</p>
+
+<p>如果数组中任一元素的&nbsp;<strong>频次&nbsp;</strong>是&nbsp;<strong>质数</strong>，返回 <code>true</code>；否则，返回 <code>false</code>。</p>
+
+<p>元素 <code>x</code> 的&nbsp;<strong>频次&nbsp;</strong>是它在数组中出现的次数。</p>
+
+<p>质数是一个大于 1 的自然数，并且只有两个因数：1 和它本身。</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 = [1,2,3,4,5,4]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">true</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>数字 4 的频次是 2，而 2 是质数。</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 = [1,2,3,4,5]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">false</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>所有元素的频次都是 1。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">nums = [2,2,2,4,4]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">true</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>数字 2 和 4 的频次都是质数。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
+	<li><code>0 &lt;= nums[i] &lt;= 100</code></li>
+</ul>
+
+<!-- description:end -->
+
+## 解法
+
+<!-- solution:start -->
+
+### 方法一：计数 + 判断质数
+
+我们用一个哈希表 $\text{cnt}$ 统计每个元素的频次。然后遍历 $\text{cnt}$ 中的值，判断是否有质数，如果有则返回 `true`，否则返回 `false`。
+
+时间复杂度 $O(n \times \sqrt{M})$，空间复杂度 $O(n)$。其中 $n$ 是数组 $\text{nums}$ 的长度，而 $M$ 是 $\text{cnt}$ 中的最大值。
+
+<!-- tabs:start -->
+
+#### Python3
+
+```python
+class Solution:
+    def checkPrimeFrequency(self, nums: List[int]) -> bool:
+        def is_prime(x: int) -> bool:
+            if x < 2:
+                return False
+            return all(x % i for i in range(2, int(sqrt(x)) + 1))
+
+        cnt = Counter(nums)
+        return any(is_prime(x) for x in cnt.values())
+```
+
+#### Java
+
+```java
+import java.util.*;
+
+class Solution {
+    public boolean checkPrimeFrequency(int[] nums) {
+        Map<Integer, Integer> cnt = new HashMap<>();
+        for (int x : nums) {
+            cnt.merge(x, 1, Integer::sum);
+        }
+
+        for (int x : cnt.values()) {
+            if (isPrime(x)) {
+                return true;
+            }
+        }
+        return false;
+    }
+
+    private boolean isPrime(int x) {
+        if (x < 2) {
+            return false;
+        }
+        for (int i = 2; i <= x / i; i++) {
+            if (x % i == 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+}
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    bool checkPrimeFrequency(vector<int>& nums) {
+        unordered_map<int, int> cnt;
+        for (int x : nums) {
+            ++cnt[x];
+        }
+
+        for (auto& [_, x] : cnt) {
+            if (isPrime(x)) {
+                return true;
+            }
+        }
+        return false;
+    }
+
+private:
+    bool isPrime(int x) {
+        if (x < 2) {
+            return false;
+        }
+        for (int i = 2; i <= x / i; ++i) {
+            if (x % i == 0) {
+                return false;
+            }
+        }
+        return true;
+    }
+};
+```
+
+#### Go
+
+```go
+func checkPrimeFrequency(nums []int) bool {
+	cnt := make(map[int]int)
+	for _, x := range nums {
+		cnt[x]++
+	}
+	for _, x := range cnt {
+		if isPrime(x) {
+			return true
+		}
+	}
+	return false
+}
+
+func isPrime(x int) bool {
+	if x < 2 {
+		return false
+	}
+	for i := 2; i*i <= x; i++ {
+		if x%i == 0 {
+			return false
+		}
+	}
+	return true
+}
+```
+
+#### TypeScript
+
+```ts
+function checkPrimeFrequency(nums: number[]): boolean {
+    const cnt: Record<number, number> = {};
+    for (const x of nums) {
+        cnt[x] = (cnt[x] || 0) + 1;
+    }
+    for (const x of Object.values(cnt)) {
+        if (isPrime(x)) {
+            return true;
+        }
+    }
+    return false;
+}
+
+function isPrime(x: number): boolean {
+    if (x < 2) {
+        return false;
+    }
+    for (let i = 2; i * i <= x; i++) {
+        if (x % i === 0) {
+            return false;
+        }
+    }
+    return true;
+}
+```
+
+<!-- tabs:end -->
+
+<!-- solution:end -->
+
+<!-- problem:end -->