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Formatted question description: https://leetcode.ca/all/2088.html

# 2088. Count Fertile Pyramids in a Land (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:

- The number of cells in the set has to be
**greater than**`1`

and all cells must be**fertile**. - The
**apex**of a pyramid is the**topmost**cell of the pyramid. The**height**of a pyramid is the number of rows it covers. Let`(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:

- The number of cells in the set has to be
**greater than**`1`

and all cells must be**fertile**. - The
**apex**of an inverse pyramid is the**bottommost**cell of the inverse pyramid. The**height**of an inverse pyramid is the number of rows it covers. Let`(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:2Explanation: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:2Explanation: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 = [[1,0,1],[0,0,0],[1,0,1]]Output:0Explanation:There are no pyramidal or inverse pyramidal plots in the grid.

**Example 4:**

Input:grid = [[1,1,1,1,0],[1,1,1,1,1],[1,1,1,1,1],[0,1,0,0,1]]Output:13Explanation: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 <= 10`

^{5}`grid[i][j]`

is either`0`

or`1`

.

**Similar Questions**:

## Solution 1. Prefix Sum

```
// OJ: https://leetcode.com/problems/count-fertile-pyramids-in-a-land/
// Time: O(M^2 * N)
// Space: O(MN) can be reduced to O(1) if we calculate prefix sums in-place
class Solution {
public:
int countPyramids(vector<vector<int>>& A) {
int M = A.size(), N = A[0].size(), ans = 0;
vector<vector<int>> sum(M + 1, vector<int>(N + 1));
for (int i = 0; i < M; ++i) partial_sum(begin(A[i]), end(A[i]), begin(sum[i]) + 1);
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (A[i][j] == 0) continue;
int h = 1;
while (i + h - 1 < M && j + h - 1 < N && j - h + 1 >= 0 && sum[i + h - 1][j + h] - sum[i + h - 1][j - h + 1] == 2 * h - 1) ++h;
ans += max(0, h - 2);
h = 1;
while (i - h + 1 >= 0 && j + h - 1 < N && j - h + 1 >= 0 && sum[i - h + 1][j + h] - sum[i - h + 1][j - h + 1] == 2 * h - 1) ++h;
ans += max(0, h - 2);
}
}
return ans;
}
};
```