How do you solve Count Fertile Pyramids in a Land on LeetCode?

Count Fertile Pyramids in a Land (LeetCode #2088) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m * n) time complexity and O(m * n) space complexity. Key insight: Dynamic programming can significantly reduce redundant calculations.

What is the brute force solution for Count Fertile Pyramids in a Land?

The brute force approach for Count Fertile Pyramids in a Land is Brute Force, with O(m * n * max(m, n)) time complexity and O(1) space complexity. The brute force approach checks every possible apex cell and calculates the maximum height of both pyramids and inverse pyramids by expanding downwards. This method is straightforward but inefficient as it examines all potential heights for each cell. The optimal Optimal Solution approach improves this to O(m * n).

What algorithm or data structure is used to solve Count Fertile Pyramids in a Land?

Count Fertile Pyramids in a Land uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Matrix Traversal.

What are the interview tips for Count Fertile Pyramids in a Land?

Always explain your thought process clearly; it helps the interviewer understand your approach. Don't forget to consider edge cases and discuss them with the interviewer. Practice dynamic programming problems to get comfortable with the concept.

What are common mistakes when solving Count Fertile Pyramids in a Land?

Failing to correctly handle edge cases where the grid is too small. Not initializing DP arrays properly, leading to incorrect counts.

#2088

Count Fertile Pyramids in a Land

Hard

ArrayDynamic ProgrammingMatrixDynamic ProgrammingMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(m * n * max(m, n))	O(m * n)
Space	O(1)	O(m * n)

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Intuition

Time O(m * n)Space O(m * n)

We can use dynamic programming to precompute the maximum height of pyramids and inverse pyramids for each cell, allowing us to count valid plots in linear time. This reduces redundant calculations and speeds up the process significantly.

⚙️

Algorithm

3 steps

1Step 1: Create two DP arrays: dpPyramid and dpInverse to store maximum heights for pyramids and inverse pyramids respectively.
2Step 2: Iterate through the grid to fill dpPyramid from top to bottom and dpInverse from bottom to top.
3Step 3: For each cell, calculate the total number of pyramids and inverse pyramids based on the heights stored in the DP arrays.

solution.py23 lines

1def countPyramids(grid):
2    m, n = len(grid), len(grid[0])
3    dpPyramid = [[0] * n for _ in range(m)]
4    dpInverse = [[0] * n for _ in range(m)]
5    count = 0
6
7    for r in range(m):
8        for c in range(n):
9            if grid[r][c] == 1:
10                dpPyramid[r][c] = 1
11                if r > 0:
12                    dpPyramid[r][c] += min(dpPyramid[r-1][c-1], dpPyramid[r-1][c])
13                count += dpPyramid[r][c] - 1
14
15    for r in range(m-1, -1, -1):
16        for c in range(n):
17            if grid[r][c] == 1:
18                dpInverse[r][c] = 1
19                if r < m - 1:
20                    dpInverse[r][c] += min(dpInverse[r+1][c-1], dpInverse[r+1][c])
21                count += dpInverse[r][c] - 1
22
23    return count

ℹ

Complexity note: The complexity is linear because we only traverse the grid a fixed number of times to compute the heights, making it efficient compared to the brute force approach.

1Dynamic programming can significantly reduce redundant calculations.
2Understanding the structure of pyramids and inverse pyramids is crucial for efficient counting.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.