Leetcode #2088: Count Fertile Pyramids in a Land
In this guide, we solve Leetcode #2088 Count Fertile Pyramids in a Land in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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).
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Dynamic Programming, Matrix
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
class Solution:
def countPyramids(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
f = [[0] * n for _ in range(m)]
ans = 0
for i in range(m - 1, -1, -1):
for j in range(n):
if grid[i][j] == 0:
f[i][j] = -1
elif not (i == m - 1 or j == 0 or j == n - 1):
f[i][j] = min(f[i + 1][j - 1], f[i + 1][j], f[i + 1][j + 1]) + 1
ans += f[i][j]
for i in range(m):
for j in range(n):
if grid[i][j] == 0:
f[i][j] = -1
elif i == 0 or j == 0 or j == n - 1:
f[i][j] = 0
else:
f[i][j] = min(f[i - 1][j - 1], f[i - 1][j], f[i - 1][j + 1]) + 1
ans += f[i][j]
return ans
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.