README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3100-3199/3142.Check%20if%20Grid%20Satisfies%20Conditions/README_EN.md	1303	Biweekly Contest 130 Q1	Array Matrix

3142. Check if Grid Satisfies Conditions

中文文档

Description

You are given a 2D matrix grid of size m x n. You need to check if each cell grid[i][j] is:

• Equal to the cell below it, i.e. grid[i][j] == grid[i + 1][j] (if it exists).
• Different from the cell to its right, i.e. grid[i][j] != grid[i][j + 1] (if it exists).

Return true if all the cells satisfy these conditions, otherwise, return false.

 

Example 1:

Input: grid = [[1,0,2],[1,0,2]]

Output: true

Explanation:

All the cells in the grid satisfy the conditions.

Example 2:

Input: grid = [[1,1,1],[0,0,0]]

Output: false

Explanation:

All cells in the first row are equal.

Example 3:

Input: grid = [[1],[2],[3]]

Output: false

Explanation:

Cells in the first column have different values.

 

Constraints:

• 1 <= n, m <= 10
• 0 <= grid[i][j] <= 9

Solutions

Solution 1: Simulation

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.

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)`.

Python3

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

Java

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

C++

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

Go

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

TypeScript

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