Leetcode #1992: Find All Groups of Farmland
In this guide, we solve Leetcode #1992 Find All Groups of Farmland 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
You are given a 0-indexed m x n binary matrix land where a 0 represents a hectare of forested land and a 1 represents a hectare of farmland. To keep the land organized, there are designated rectangular areas of hectares that consist entirely of farmland.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Array, Matrix
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: land = [[1,0,0],[0,1,1],[0,1,1]]
Output: [[0,0,0,0],[1,1,2,2]]
Explanation:
The first group has a top left corner at land[0][0] and a bottom right corner at land[0][0].
The second group has a top left corner at land[1][1] and a bottom right corner at land[2][2].
Python Solution
class Solution:
def findFarmland(self, land: List[List[int]]) -> List[List[int]]:
m, n = len(land), len(land[0])
ans = []
for i in range(m):
for j in range(n):
if (
land[i][j] == 0
or (j > 0 and land[i][j - 1] == 1)
or (i > 0 and land[i - 1][j] == 1)
):
continue
x, y = i, j
while x + 1 < m and land[x + 1][j] == 1:
x += 1
while y + 1 < n and land[x][y + 1] == 1:
y += 1
ans.append([i, j, x, y])
return ans
Complexity
The time complexity is O(V+E). The space complexity is O(V).
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.