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3142. Check if Grid Satisfies Conditions
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)`.
-
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; } }
-
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; } };
-
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
-
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 }
-
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; }