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feat: add solutions to lc problems: No.3142,3143 (#2796)
* No.3142.Check if Grid Satisfies Conditions * No.3143.Maximum Points Inside the Square
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31 files changed

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‎solution/3100-3199/3142.Check if Grid Satisfies Conditions/README.md

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## 解法
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### 方法一
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### 方法一：模拟
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我们可以遍历每一个格子，判断其是否满足题目条件，如果有一个格子不满足条件，我们就返回 `false`，否则返回 `true`。
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时间复杂度 $O(m \times n)$，其中 $m$ 和 $n$ 分别是矩阵 `grid` 的行数和列数。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def satisfiesConditions(self, grid: List[List[int]]) -> bool:
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m, n = len(grid), len(grid[0])
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for i, row in enumerate(grid):
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for j, x in enumerate(row):
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if i + 1 < m and x != grid[i + 1][j]:
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return False
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if j + 1 < n and x == grid[i][j + 1]:
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return False
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return True
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```
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```java
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class Solution {
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public boolean satisfiesConditions(int[][] grid) {
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int m = grid.length, n = grid[0].length;
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (i + 1 < m && grid[i][j] != grid[i + 1][j]) {
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return false;
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}
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if (j + 1 < n && grid[i][j] == grid[i][j + 1]) {
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return false;
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}
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}
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}
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return true;
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}
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}
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```
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```cpp
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class Solution {
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public:
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bool satisfiesConditions(vector<vector<int>>& grid) {
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int m = grid.size(), n = grid[0].size();
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (i + 1 < m && grid[i][j] != grid[i + 1][j]) {
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return false;
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}
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if (j + 1 < n && grid[i][j] == grid[i][j + 1]) {
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return false;
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}
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}
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}
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return true;
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}
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};
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```
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```go
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func satisfiesConditions(grid [][]int) bool {
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m, n := len(grid), len(grid[0])
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for i, row := range grid {
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for j, x := range row {
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if i+1 < m && x != grid[i+1][j] {
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return false
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}
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if j+1 < n && x == grid[i][j+1] {
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return false
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}
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}
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}
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return true
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}
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```
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```ts
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function satisfiesConditions(grid: number[][]): boolean {
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const [m, n] = [grid.length, grid[0].length];
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for (let i = 0; i < m; ++i) {
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for (let j = 0; j < n; ++j) {
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if (i + 1 < m && grid[i][j] !== grid[i + 1][j]) {
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return false;
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}
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if (j + 1 < n && grid[i][j] === grid[i][j + 1]) {
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return false;
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}
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}
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}
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return true;
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}
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```
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<!-- tabs:end -->

‎solution/3100-3199/3142.Check if Grid Satisfies Conditions/README_EN.md

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## Solutions
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### Solution 1
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### Solution 1: Simulation
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We can iterate through each cell and determine whether it meets the conditions specified in the problem. If there is a cell that does not meet the conditions, we return `false`, otherwise, we return `true`.
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The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns of the matrix `grid` respectively. The space complexity is $O(1)`.
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<!-- tabs:start -->
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```python
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class Solution:
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def satisfiesConditions(self, grid: List[List[int]]) -> bool:
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m, n = len(grid), len(grid[0])
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for i, row in enumerate(grid):
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for j, x in enumerate(row):
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if i + 1 < m and x != grid[i + 1][j]:
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return False
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if j + 1 < n and x == grid[i][j + 1]:
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return False
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return True
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```
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```java
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class Solution {
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public boolean satisfiesConditions(int[][] grid) {
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int m = grid.length, n = grid[0].length;
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (i + 1 < m && grid[i][j] != grid[i + 1][j]) {
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return false;
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}
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if (j + 1 < n && grid[i][j] == grid[i][j + 1]) {
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return false;
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}
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}
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}
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return true;
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}
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}
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```
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```cpp
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class Solution {
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public:
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bool satisfiesConditions(vector<vector<int>>& grid) {
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int m = grid.size(), n = grid[0].size();
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (i + 1 < m && grid[i][j] != grid[i + 1][j]) {
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return false;
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}
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if (j + 1 < n && grid[i][j] == grid[i][j + 1]) {
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return false;
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}
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}
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}
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return true;
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}
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};
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```
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```go
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func satisfiesConditions(grid [][]int) bool {
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m, n := len(grid), len(grid[0])
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for i, row := range grid {
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for j, x := range row {
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if i+1 < m && x != grid[i+1][j] {
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return false
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}
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if j+1 < n && x == grid[i][j+1] {
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return false
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}
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}
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}
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return true
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}
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```
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```ts
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function satisfiesConditions(grid: number[][]): boolean {
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const [m, n] = [grid.length, grid[0].length];
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for (let i = 0; i < m; ++i) {
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for (let j = 0; j < n; ++j) {
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if (i + 1 < m && grid[i][j] !== grid[i + 1][j]) {
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return false;
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}
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if (j + 1 < n && grid[i][j] === grid[i][j + 1]) {
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return false;
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}
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}
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}
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return true;
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}
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```
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<!-- tabs:end -->

‎solution/3100-3199/3142.Check if Grid Satisfies Conditions/Solution.cpp

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class Solution {
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public:
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bool satisfiesConditions(vector<vector<int>>& grid) {
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int m = grid.size(), n = grid[0].size();
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (i + 1 < m && grid[i][j] != grid[i + 1][j]) {
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return false;
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}
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if (j + 1 < n && grid[i][j] == grid[i][j + 1]) {
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return false;
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}
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}
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}
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return true;
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}
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};

‎solution/3100-3199/3142.Check if Grid Satisfies Conditions/Solution.go

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@@ -0,0 +1,14 @@
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func satisfiesConditions(grid [][]int) bool {
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m, n := len(grid), len(grid[0])
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for i, row := range grid {
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for j, x := range row {
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if i+1 < m && x != grid[i+1][j] {
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return false
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}
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if j+1 < n && x == grid[i][j+1] {
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return false
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}
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}
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}
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return true
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}

‎solution/3100-3199/3142.Check if Grid Satisfies Conditions/Solution.java

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class Solution {
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public boolean satisfiesConditions(int[][] grid) {
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int m = grid.length, n = grid[0].length;
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for (int i = 0; i < m; ++i) {
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for (int j = 0; j < n; ++j) {
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if (i + 1 < m && grid[i][j] != grid[i + 1][j]) {
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return false;
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}
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if (j + 1 < n && grid[i][j] == grid[i][j + 1]) {
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return false;
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}
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}
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}
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return true;
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}
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}

‎solution/3100-3199/3142.Check if Grid Satisfies Conditions/Solution.py

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class Solution:
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def satisfiesConditions(self, grid: List[List[int]]) -> bool:
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m, n = len(grid), len(grid[0])
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for i, row in enumerate(grid):
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for j, x in enumerate(row):
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if i + 1 < m and x != grid[i + 1][j]:
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return False
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if j + 1 < n and x == grid[i][j + 1]:
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return False
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return True

‎solution/3100-3199/3142.Check if Grid Satisfies Conditions/Solution.ts

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function satisfiesConditions(grid: number[][]): boolean {
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const [m, n] = [grid.length, grid[0].length];
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for (let i = 0; i < m; ++i) {
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for (let j = 0; j < n; ++j) {
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if (i + 1 < m && grid[i][j] !== grid[i + 1][j]) {
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return false;
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}
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if (j + 1 < n && grid[i][j] === grid[i][j + 1]) {
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return false;
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}
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}
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}
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return true;
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}

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