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feat: add solutions to lc problem: No.3005 (doocs#2216)
No.3005.Count Elements With Maximum Frequency
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‎solution/3000-3099/3005.Count Elements With Maximum Frequency/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：计数**
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我们可以用一个哈希表或数组 $cnt$ 记录每个元素出现的次数。
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然后我们遍历 $cnt$，找到出现次数最多的元素，记其出现次数为 $mx$，累加出现次数等于 $mx$ 的元素的出现次数，即为答案。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $nums$ 的长度。
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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 maxFrequencyElements(self, nums: List[int]) -> int:
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cnt = Counter(nums)
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mx = max(cnt.values())
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return sum(x for x in cnt.values() if x == mx)
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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 maxFrequencyElements(int[] nums) {
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int[] cnt = new int[101];
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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, mx = -1;
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for (int x : cnt) {
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if (mx < x) {
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mx = x;
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ans = x;
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} else if (mx == x) {
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ans += x;
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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 maxFrequencyElements(vector<int>& nums) {
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int cnt[101]{};
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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, mx = -1;
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for (int x : cnt) {
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if (mx < x) {
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mx = x;
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ans = x;
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} else if (mx == x) {
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ans += x;
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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 maxFrequencyElements(nums []int) (ans int) {
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cnt := [101]int{}
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for _, x := range nums {
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cnt[x]++
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}
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mx := -1
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for _, x := range cnt {
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if mx < x {
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mx, ans = x, x
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} else if mx == x {
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ans += 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 maxFrequencyElements(nums: number[]): number {
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const cnt: number[] = Array(101).fill(0);
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for (const x of nums) {
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++cnt[x];
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}
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let [ans, mx] = [0, -1];
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for (const x of cnt) {
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if (mx < x) {
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mx = x;
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ans = x;
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} else if (mx === x) {
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ans += x;
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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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### **...**

‎solution/3000-3099/3005.Count Elements With Maximum Frequency/README_EN.md

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## Solutions
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**Solution 1: Counting**
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We can use a hash table or array $cnt$ to record the occurrence of each element.
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Then we traverse $cnt$ to find the element with the most occurrences, and let its occurrence be $mx$. We sum up the occurrences of elements that appear $mx$ times, which is the answer.
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The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.
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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 maxFrequencyElements(self, nums: List[int]) -> int:
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cnt = Counter(nums)
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mx = max(cnt.values())
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return sum(x for x in cnt.values() if x == mx)
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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 maxFrequencyElements(int[] nums) {
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int[] cnt = new int[101];
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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, mx = -1;
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for (int x : cnt) {
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if (mx < x) {
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mx = x;
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ans = x;
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} else if (mx == x) {
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ans += x;
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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 maxFrequencyElements(vector<int>& nums) {
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int cnt[101]{};
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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, mx = -1;
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for (int x : cnt) {
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if (mx < x) {
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mx = x;
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ans = x;
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} else if (mx == x) {
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ans += x;
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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 maxFrequencyElements(nums []int) (ans int) {
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cnt := [101]int{}
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for _, x := range nums {
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cnt[x]++
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}
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mx := -1
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for _, x := range cnt {
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if mx < x {
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mx, ans = x, x
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} else if mx == x {
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ans += 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 maxFrequencyElements(nums: number[]): number {
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const cnt: number[] = Array(101).fill(0);
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for (const x of nums) {
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++cnt[x];
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}
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let [ans, mx] = [0, -1];
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for (const x of cnt) {
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if (mx < x) {
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mx = x;
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ans = x;
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} else if (mx === x) {
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ans += x;
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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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### **...**

‎solution/3000-3099/3005.Count Elements With Maximum Frequency/Solution.cpp

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class Solution {
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public:
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int maxFrequencyElements(vector<int>& nums) {
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int cnt[101]{};
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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, mx = -1;
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for (int x : cnt) {
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if (mx < x) {
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mx = x;
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ans = x;
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} else if (mx == x) {
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ans += x;
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}
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}
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return ans;
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}
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};

‎solution/3000-3099/3005.Count Elements With Maximum Frequency/Solution.go

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func maxFrequencyElements(nums []int) (ans int) {
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cnt := [101]int{}
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for _, x := range nums {
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cnt[x]++
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}
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mx := -1
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for _, x := range cnt {
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if mx < x {
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mx, ans = x, x
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} else if mx == x {
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ans += x
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}
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}
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return
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}

‎solution/3000-3099/3005.Count Elements With Maximum Frequency/Solution.java

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class Solution {
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public int maxFrequencyElements(int[] nums) {
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int[] cnt = new int[101];
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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, mx = -1;
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for (int x : cnt) {
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if (mx < x) {
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mx = x;
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ans = x;
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} else if (mx == x) {
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ans += x;
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}
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}
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return ans;
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}
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}

‎solution/3000-3099/3005.Count Elements With Maximum Frequency/Solution.py

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class Solution:
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def maxFrequencyElements(self, nums: List[int]) -> int:
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cnt = Counter(nums)
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mx = max(cnt.values())
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return sum(x for x in cnt.values() if x == mx)

‎solution/3000-3099/3005.Count Elements With Maximum Frequency/Solution.ts

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function maxFrequencyElements(nums: number[]): number {
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const cnt: number[] = Array(101).fill(0);
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for (const x of nums) {
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++cnt[x];
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}
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let [ans, mx] = [0, -1];
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for (const x of cnt) {
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if (mx < x) {
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mx = x;
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ans = x;
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} else if (mx === x) {
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ans += x;
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}
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}
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return ans;
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}

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