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feat: add solutions to lc problems: No.2965~2968 (#2112)
* No.2965.Find Missing and Repeated Values * No.2966.Divide Array Into Arrays With Max Difference * No.2968.Apply Operations to Maximize Frequency Score
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‎solution/2900-2999/2965.Find Missing and Repeated Values/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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我们创建一个长度为 $n^2 + 1$ 的数组 $cnt$，统计矩阵中每个数字出现的次数。
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接下来遍历 $i \in [1, n^2]$，如果 $cnt[i] = 2$，则 $i$ 是重复的数字，我们将答案的第一个元素设为 $i$；如果 $cnt[i] = 0$，则 $i$ 是缺失的数字，我们将答案的第二个元素设为 $i$。
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时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 是矩阵的边长。
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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 findMissingAndRepeatedValues(self, grid: List[List[int]]) -> List[int]:
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n = len(grid)
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cnt = [0] * (n * n + 1)
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for row in grid:
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for v in row:
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cnt[v] += 1
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ans = [0] * 2
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for i in range(1, n * n + 1):
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if cnt[i] == 2:
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ans[0] = i
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if cnt[i] == 0:
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ans[1] = i
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return ans
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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[] findMissingAndRepeatedValues(int[][] grid) {
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int n = grid.length;
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int[] cnt = new int[n * n + 1];
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int[] ans = new int[2];
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for (int[] row : grid) {
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for (int x : row) {
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if (++cnt[x] == 2) {
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ans[0] = x;
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}
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}
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}
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for (int x = 1;; ++x) {
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if (cnt[x] == 0) {
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ans[1] = x;
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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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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> findMissingAndRepeatedValues(vector<vector<int>>& grid) {
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int n = grid.size();
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vector<int> cnt(n * n + 1);
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vector<int> ans(2);
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for (auto& row : grid) {
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for (int x : row) {
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if (++cnt[x] == 2) {
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ans[0] = x;
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}
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}
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}
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for (int x = 1;; ++x) {
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if (cnt[x] == 0) {
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ans[1] = x;
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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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### **Go**
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```go
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func findMissingAndRepeatedValues(grid [][]int) []int {
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n := len(grid)
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ans := make([]int, 2)
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cnt := make([]int, n*n+1)
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for _, row := range grid {
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for _, x := range row {
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cnt[x]++
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if cnt[x] == 2 {
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ans[0] = x
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}
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}
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}
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for x := 1; ; x++ {
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if cnt[x] == 0 {
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ans[1] = x
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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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### **TypeScript**
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```ts
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function findMissingAndRepeatedValues(grid: number[][]): number[] {
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const n = grid.length;
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const cnt: number[] = Array(n * n + 1).fill(0);
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const ans: number[] = Array(2).fill(0);
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for (const row of grid) {
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for (const x of row) {
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if (++cnt[x] === 2) {
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ans[0] = x;
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}
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}
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}
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for (let x = 1; ; ++x) {
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if (cnt[x] === 0) {
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ans[1] = x;
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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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### **...**

‎solution/2900-2999/2965.Find Missing and Repeated Values/README_EN.md

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## Solutions
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**Solution 1: Counting**
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We create an array $cnt$ of length $n^2 + 1$ to count the frequency of each number in the matrix.
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Next, we traverse $i \in [1, n^2]$. If $cnt[i] = 2$, then $i$ is the duplicated number, and we set the first element of the answer to $i$. If $cnt[i] = 0$, then $i$ is the missing number, and we set the second element of the answer to $i$.
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The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the side length of the matrix.
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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 findMissingAndRepeatedValues(self, grid: List[List[int]]) -> List[int]:
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n = len(grid)
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cnt = [0] * (n * n + 1)
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for row in grid:
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for v in row:
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cnt[v] += 1
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ans = [0] * 2
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for i in range(1, n * n + 1):
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if cnt[i] == 2:
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ans[0] = i
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if cnt[i] == 0:
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ans[1] = i
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return ans
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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[] findMissingAndRepeatedValues(int[][] grid) {
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int n = grid.length;
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int[] cnt = new int[n * n + 1];
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int[] ans = new int[2];
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for (int[] row : grid) {
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for (int x : row) {
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if (++cnt[x] == 2) {
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ans[0] = x;
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}
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}
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}
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for (int x = 1;; ++x) {
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if (cnt[x] == 0) {
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ans[1] = x;
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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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### **C++**
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```cpp
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class Solution {
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public:
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vector<int> findMissingAndRepeatedValues(vector<vector<int>>& grid) {
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int n = grid.size();
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vector<int> cnt(n * n + 1);
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vector<int> ans(2);
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for (auto& row : grid) {
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for (int x : row) {
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if (++cnt[x] == 2) {
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ans[0] = x;
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}
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}
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}
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for (int x = 1;; ++x) {
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if (cnt[x] == 0) {
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ans[1] = x;
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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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### **Go**
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```go
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func findMissingAndRepeatedValues(grid [][]int) []int {
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n := len(grid)
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ans := make([]int, 2)
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cnt := make([]int, n*n+1)
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for _, row := range grid {
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for _, x := range row {
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cnt[x]++
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if cnt[x] == 2 {
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ans[0] = x
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}
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}
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}
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for x := 1; ; x++ {
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if cnt[x] == 0 {
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ans[1] = x
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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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### **TypeScript**
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```ts
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function findMissingAndRepeatedValues(grid: number[][]): number[] {
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const n = grid.length;
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const cnt: number[] = Array(n * n + 1).fill(0);
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const ans: number[] = Array(2).fill(0);
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for (const row of grid) {
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for (const x of row) {
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if (++cnt[x] === 2) {
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ans[0] = x;
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}
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}
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}
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for (let x = 1; ; ++x) {
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if (cnt[x] === 0) {
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ans[1] = x;
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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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### **...**

‎solution/2900-2999/2965.Find Missing and Repeated Values/Solution.cpp

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class Solution {
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public:
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vector<int> findMissingAndRepeatedValues(vector<vector<int>>& grid) {
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int n = grid.size();
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vector<int> cnt(n * n + 1);
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vector<int> ans(2);
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for (auto& row : grid) {
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for (int x : row) {
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if (++cnt[x] == 2) {
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ans[0] = x;
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}
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}
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}
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for (int x = 1;; ++x) {
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if (cnt[x] == 0) {
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ans[1] = x;
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return ans;
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}
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}
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}
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};

‎solution/2900-2999/2965.Find Missing and Repeated Values/Solution.go

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func findMissingAndRepeatedValues(grid [][]int) []int {
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n := len(grid)
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ans := make([]int, 2)
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cnt := make([]int, n*n+1)
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for _, row := range grid {
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for _, x := range row {
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cnt[x]++
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if cnt[x] == 2 {
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ans[0] = x
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}
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}
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}
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for x := 1; ; x++ {
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if cnt[x] == 0 {
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ans[1] = x
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return ans
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}
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}
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}

‎solution/2900-2999/2965.Find Missing and Repeated Values/Solution.java

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class Solution {
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public int[] findMissingAndRepeatedValues(int[][] grid) {
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int n = grid.length;
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int[] cnt = new int[n * n + 1];
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int[] ans = new int[2];
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for (int[] row : grid) {
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for (int x : row) {
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if (++cnt[x] == 2) {
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ans[0] = x;
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}
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}
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}
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for (int x = 1;; ++x) {
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if (cnt[x] == 0) {
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ans[1] = x;
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return ans;
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}
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}
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}
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}

‎solution/2900-2999/2965.Find Missing and Repeated Values/Solution.py

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class Solution:
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def findMissingAndRepeatedValues(self, grid: List[List[int]]) -> List[int]:
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n = len(grid)
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cnt = [0] * (n * n + 1)
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for row in grid:
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for v in row:
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cnt[v] += 1
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ans = [0] * 2
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for i in range(1, n * n + 1):
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if cnt[i] == 2:
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ans[0] = i
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if cnt[i] == 0:
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ans[1] = i
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return ans

‎solution/2900-2999/2965.Find Missing and Repeated Values/Solution.ts

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function findMissingAndRepeatedValues(grid: number[][]): number[] {
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const n = grid.length;
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const cnt: number[] = Array(n * n + 1).fill(0);
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const ans: number[] = Array(2).fill(0);
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for (const row of grid) {
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for (const x of row) {
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if (++cnt[x] === 2) {
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ans[0] = x;
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}
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}
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}
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for (let x = 1; ; ++x) {
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if (cnt[x] === 0) {
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ans[1] = x;
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return ans;
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}
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}
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}

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