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 thetotal number of pyramidal and inverse pyramidal plots that can be found in grid.

Examples

Example 1


![](https://assets.leetcode.com/uploads/2021/12/22/1.JPG)

    
    
    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


![](https://assets.leetcode.com/uploads/2021/12/22/2.JPG)

    
    
    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


![](https://assets.leetcode.com/uploads/2021/12/22/3.JPG)

    
    
    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 <= 10^5
grid[i][j] is either 0 or 1.

Solution

Method 1 – Dynamic Programming (DP) for Pyramid Heights

Intuition

A cell can be the apex of a pyramid of height h if the three cells directly below it can be the apex of pyramids of height at least h-1, and the current cell is fertile. We can use DP to compute the maximum height of a pyramid with apex at each cell, both for normal and inverse pyramids.

Approach

For each cell, compute the maximum height of a pyramid with apex at (i, j) (top-down) using DP.
For each cell, compute the maximum height of an inverse pyramid with apex at (i, j) (bottom-up) using DP.
For each cell, add (height - 1) to the answer (since height 1 is not counted as a pyramid).
Return the total count for both normal and inverse pyramids.

Code

[{"language":"C++","code":"class Solution {\npublic:\n    int countPyramids(vector<vector<int>>& grid) {\n        int m = grid.size(), n = grid[0].size(), ans = 0;\n        auto dp = grid;\n        for (int i = 1; i < m; ++i) {\n            for (int j = 1; j < n-1; ++j) {\n                if (grid[i][j] && grid[i-1][j-1] && grid[i-1][j] && grid[i-1][j+1])\n                    dp[i][j] = min({dp[i-1][j-1], dp[i-1][j], dp[i-1][j+1]}) + 1;\n            }\n        }\n        for (int i = 0; i < m; ++i) for (int j = 0; j < n; ++j) ans += dp[i][j] - 1;\n        reverse(grid.begin(), grid.end());\n        dp = grid;\n        for (int i = 1; i < m; ++i) {\n            for (int j = 1; j < n-1; ++j) {\n

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