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feat: add solutions to lc problem: No.2128 (doocs#1288)
No.2128.Remove All Ones With Row and Column Flips
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‎solution/1000-1099/1072.Flip Columns For Maximum Number of Equal Rows/README.md

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时间复杂度 $O(m \times n)$，空间复杂度 $O(m)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。
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相似题目：
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- [2128. 通过翻转行或列来去除所有的 1](/solution/2100-2199/2128.Remove%20All%20Ones%20With%20Row%20and%20Column%20Flips/README.md)
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<!-- tabs:start -->
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### **Python3**

‎solution/2100-2199/2128.Remove All Ones With Row and Column Flips/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：哈希表**
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我们观察发现，如果矩阵中的两行满足以下条件之一，则它们可以通过翻转某些列的方式得到相等的行：
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1. 两行的对应位置元素相等，即如果其中一行元素为 $1,0,0,1$，则另一行元素也为 $1,0,0,1$；
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1. 两行的对应位置元素相反，即如果其中一行元素为 $1,0,0,1$，则另一行元素为 $0,1,1,0$。
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我们称满足以上条件之一的两行元素为“等价行”，那么题目所求的答案即为矩阵中最多包含等价行的行数。
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因此，我们可以遍历矩阵的每一行，将每一行转换成第一个元素为 $0$ 的“等价行”。具体做法如下：
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- 如果当前行的第一个元素为 $0$，那么当前行的元素保持不变；
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- 如果当前行的第一个元素为 $1$，那么我们将当前行的每个元素进行翻转，即 $0$ 变成 $1$, $1$ 变成 $0$。也就是说，我们将以 $1$ 开头的行翻转成以 $0$ 开头的“等价行”。
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这样一来，我们只需要用一个哈希表来统计转换后的每一行，如果最后哈希表只有一个元素，那么说明我们可以通过翻转行或列，将矩阵中所有的 $1$ 去除。
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时间复杂度 $O(m \times n)$，空间复杂度 $O(m)$。其中 $m$ 和 $n$ 分别是矩阵的行数和列数。
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相似题目：
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- [1072. 按列翻转得到最大值等行数](/solution/1000-1099/1072.Flip%20Columns%20For%20Maximum%20Number%20of%20Equal%20Rows/README.md)
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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 removeOnes(self, grid: List[List[int]]) -> bool:
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s = set()
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for row in grid:
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t = tuple(row) if row[0] == grid[0][0] else tuple(x ^ 1 for x in row)
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s.add(t)
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return len(s) == 1
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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 boolean removeOnes(int[][] grid) {
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Set<String> s = new HashSet<>();
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int n = grid[0].length;
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for (var row : grid) {
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var cs = new char[n];
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for (int i = 0; i < n; ++i) {
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cs[i] = (char) (row[0] ^ row[i]);
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}
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s.add(String.valueOf(cs));
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}
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return s.size() == 1;
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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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bool removeOnes(vector<vector<int>>& grid) {
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unordered_set<string> s;
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for (auto& row : grid) {
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string t;
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for (int x : row) {
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t.push_back('0' + (row[0] == 0 ? x : x ^ 1));
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}
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s.insert(t);
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}
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return s.size() == 1;
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}
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};
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```
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### **TypeScript**
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### **Go**
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```go
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func removeOnes(grid [][]int) bool {
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s := map[string]bool{}
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for _, row := range grid {
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t := []byte{}
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for _, x := range row {
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if row[0] == 1 {
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x ^= 1
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}
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t = append(t, byte(x)+'0')
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}
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s[string(t)] = true
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}
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return len(s) == 1
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}
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```
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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### **TypeScript**
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```ts
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function removeOnes(grid: number[][]): boolean {
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const s = new Set<string>();
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for (const row of grid) {
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if (row[0] === 1) {
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for (let i = 0; i < row.length; i++) {
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row[i] ^= 1;
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}
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}
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const t = row.join('');
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s.add(t);
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}
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return s.size === 1;
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}
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```
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### **...**

‎solution/2100-2199/2128.Remove All Ones With Row and Column Flips/README_EN.md

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### **Python3**
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```python
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class Solution:
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def removeOnes(self, grid: List[List[int]]) -> bool:
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s = set()
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for row in grid:
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t = tuple(row) if row[0] == grid[0][0] else tuple(x ^ 1 for x in row)
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s.add(t)
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return len(s) == 1
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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 boolean removeOnes(int[][] grid) {
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Set<String> s = new HashSet<>();
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int n = grid[0].length;
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for (var row : grid) {
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var cs = new char[n];
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for (int i = 0; i < n; ++i) {
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cs[i] = (char) (row[0] ^ row[i]);
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}
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s.add(String.valueOf(cs));
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}
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return s.size() == 1;
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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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bool removeOnes(vector<vector<int>>& grid) {
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unordered_set<string> s;
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for (auto& row : grid) {
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string t;
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for (int x : row) {
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t.push_back('0' + (row[0] == 0 ? x : x ^ 1));
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}
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s.insert(t);
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}
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return s.size() == 1;
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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 removeOnes(grid [][]int) bool {
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s := map[string]bool{}
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for _, row := range grid {
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t := []byte{}
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for _, x := range row {
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if row[0] == 1 {
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x ^= 1
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}
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t = append(t, byte(x)+'0')
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}
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s[string(t)] = true
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}
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return len(s) == 1
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}
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```
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### **TypeScript**
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```ts
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function removeOnes(grid: number[][]): boolean {
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const s = new Set<string>();
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for (const row of grid) {
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if (row[0] === 1) {
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for (let i = 0; i < row.length; i++) {
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row[i] ^= 1;
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}
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}
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const t = row.join('');
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s.add(t);
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}
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return s.size === 1;
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}
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```
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### **...**

‎solution/2100-2199/2128.Remove All Ones With Row and Column Flips/Solution.cpp

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class Solution {
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public:
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bool removeOnes(vector<vector<int>>& grid) {
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unordered_set<string> s;
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for (auto& row : grid) {
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string t;
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for (int x : row) {
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t.push_back('0' + (row[0] == 0 ? x : x ^ 1));
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}
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s.insert(t);
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}
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return s.size() == 1;
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}
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};

‎solution/2100-2199/2128.Remove All Ones With Row and Column Flips/Solution.go

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func removeOnes(grid [][]int) bool {
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s := map[string]bool{}
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for _, row := range grid {
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t := []byte{}
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for _, x := range row {
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if row[0] == 1 {
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x ^= 1
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}
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t = append(t, byte(x)+'0')
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}
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s[string(t)] = true
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}
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return len(s) == 1
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}

‎solution/2100-2199/2128.Remove All Ones With Row and Column Flips/Solution.java

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class Solution {
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public boolean removeOnes(int[][] grid) {
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Set<String> s = new HashSet<>();
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int n = grid[0].length;
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for (var row : grid) {
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var cs = new char[n];
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for (int i = 0; i < n; ++i) {
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cs[i] = (char) (row[0] ^ row[i]);
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}
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s.add(String.valueOf(cs));
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}
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return s.size() == 1;
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}
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}

‎solution/2100-2199/2128.Remove All Ones With Row and Column Flips/Solution.py

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class Solution:
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def removeOnes(self, grid: List[List[int]]) -> bool:
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s = set()
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for row in grid:
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t = tuple(row) if row[0] == grid[0][0] else tuple(x ^ 1 for x in row)
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s.add(t)
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return len(s) == 1

‎solution/2100-2199/2128.Remove All Ones With Row and Column Flips/Solution.ts

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function removeOnes(grid: number[][]): boolean {
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const s = new Set<string>();
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for (const row of grid) {
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if (row[0] === 1) {
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for (let i = 0; i < row.length; i++) {
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row[i] ^= 1;
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}
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}
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const t = row.join('');
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s.add(t);
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}
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return s.size === 1;
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}

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