Suppose we have a two-dimensional binary matrix, representing a rectangular chess board, here 0 is for empty cell and 1 is for a knight. The knight is able to move two squares away horizontally and one square vertically, or two squares vertically and one square horizontally (like chess board knight). We have to check whether any two knights are attacking each other or not.
So, if the input is like
0 | 0 | 0 | 0 | 0 |
0 | 1 | 0 | 0 | 0 |
0 | 0 | 0 | 1 | 0 |
Then the output will be True
To solve this, we will follow these steps −
- row, col := row count of matrix, column count of matrix
- for r in range 0 to row-1, do
- for c in range 0 to col-1, do
- if A[r][c] is non-zero, then
- for each nr, nc in [(r+1, c-2), (r+1, c+2), (r+2, c-1), (r+2, c+1)], do
- if nr is in range of row and nc is in range of col and A[nr, nc] is non-zero, then
- return True
- if nr is in range of row and nc is in range of col and A[nr, nc] is non-zero, then
- for each nr, nc in [(r+1, c-2), (r+1, c+2), (r+2, c-1), (r+2, c+1)], do
- if A[r][c] is non-zero, then
- for c in range 0 to col-1, do
- return False
Let us see the following implementation to get better understanding −
Example
class Solution: def solve(self, A): row, col = len(A), len(A[0]) for r in range(row): for c in range(col): if A[r][c]: for nr, nc in ((r+1, c-2), (r+1, c+2), (r+2, c-1), (r+2, c+1)): if 0 <= nr < row and 0 <= nc <col and A[nr][nc]: return True return False ob = Solution() mat = [[0,0,0,0,0], [0,1,0,0,0], [0,0,0,1,0]] print(ob.solve(mat))
Input
[[0,0,0,0,0], [0,1,0,0,0], [0,0,0,1,0]]
Output
True