Hard
A farmer has a rectangular grid of land with m
rows and n
columns that can be divided into unit cells. Each cell is either fertile (represented by a 1
) or barren (represented by a 0
). All cells outside the grid are considered barren.
A pyramidal plot of land can be defined as a set of cells with the following criteria:
1
and all cells must be fertile.(r, c)
be the apex of the pyramid, and its height be h
. Then, the plot comprises of cells (i, j)
where r <= i <= r + h - 1
and c - (i - r) <= j <= c + (i - r)
.An inverse pyramidal plot of land can be defined as a set of cells with similar criteria:
1
and all cells must be fertile.(r, c)
be the apex of the pyramid, and its height be h
. Then, the plot comprises of cells (i, j)
where r - h + 1 <= i <= r
and c - (r - i) <= j <= c + (r - i)
.Some examples of valid and invalid pyramidal (and inverse pyramidal) plots are shown below. Black cells indicate fertile cells.
Given a 0-indexed m x n
binary matrix grid
representing the farmland, return the total number of pyramidal and inverse pyramidal plots that can be found in grid
.
Example 1:
Input: grid = [[0,1,1,0],[1,1,1,1]]
Output: 2
Explanation: The 2 possible pyramidal plots are shown in blue and red respectively.
There are no inverse pyramidal plots in this grid.
Hence total number of pyramidal and inverse pyramidal plots is 2 + 0 = 2.
Example 2:
Input: grid = [[1,1,1],[1,1,1]]
Output: 2
Explanation: The pyramidal plot is shown in blue, and the inverse pyramidal plot is shown in red.
Hence the total number of plots is 1 + 1 = 2.
Example 3:
Input: grid = [[1,1,1,1,0],[1,1,1,1,1],[1,1,1,1,1],[0,1,0,0,1]]
Output: 13
Explanation: There are 7 pyramidal plots, 3 of which are shown in the 2nd and 3rd figures.
There are 6 inverse pyramidal plots, 2 of which are shown in the last figure.
The total number of plots is 7 + 6 = 13.
Constraints:
m == grid.length
n == grid[i].length
1 <= m, n <= 1000
1 <= m * n <= 105
grid[i][j]
is either 0
or 1
.public class Solution {
public int countPyramids(int[][] grid) {
int m = grid.length;
int n = grid[0].length;
int[][] rev = new int[m][n];
for (int i = 0; i < m; ++i) {
System.arraycopy(grid[i], 0, rev[m - i - 1], 0, n);
}
return cal(grid) + cal(rev);
}
private int cal(int[][] grid) {
int m = grid.length;
int n = grid[0].length;
int res = 0;
for (int i = 1; i < m; ++i) {
int cnt = 0;
for (int j = 0; j < n; ++j) {
if (0 != grid[i][j]) {
cnt++;
} else {
cnt = 0;
}
if (0 == cnt || 0 == j) {
continue;
}
grid[i][j] = Math.min(grid[i - 1][j - 1] + 1, (cnt + 1) >> 1);
res += grid[i][j] - 1;
}
}
return res;
}
}