How do you solve Find All Groups of Farmland on LeetCode?

Find All Groups of Farmland (LeetCode #1992) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each group of farmland is rectangular and non-adjacent, which simplifies the search for boundaries.

What is the brute force solution for Find All Groups of Farmland?

The brute force approach for Find All Groups of Farmland is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves scanning the entire matrix to find each group of farmland. For each '1' found, we can determine the boundaries of the rectangle by expanding downwards and rightwards until we hit a '0'. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find All Groups of Farmland?

Find All Groups of Farmland uses the following concepts: Array, Depth-First Search, Breadth-First Search, Matrix. The recommended patterns to study are: Depth-First Search (DFS), Breadth-First Search (BFS), Matrix Traversal.

What are the interview tips for Find All Groups of Farmland?

Clarify the problem statement and ask about edge cases. Think about how to optimize your brute force solution and explain your thought process. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Find All Groups of Farmland?

Not marking cells as visited can lead to incorrect results. Failing to handle edge cases, such as when the matrix is empty.

#1992

Find All Groups of Farmland

Medium

ArrayDepth-First SearchBreadth-First SearchMatrixDepth-First Search (DFS)Breadth-First Search (BFS)Matrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution leverages the fact that each group of farmland is rectangular and non-adjacent. We can traverse the matrix and directly calculate the boundaries of each rectangle without needing to mark cells as visited.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty result list to store the coordinates of groups.
2Step 2: Loop through each cell in the matrix. When a '1' is found, record its position as the top-left corner.
3Step 3: Use a nested loop to find the bottom-right corner by checking the extent of the rectangle.
4Step 4: Add the coordinates of the rectangle to the result list.
5Step 5: Skip the cells that are part of the current rectangle to avoid redundant checks.

solution.py20 lines

1# Full working Python code
2
3def findFarmland(land):
4    result = []
5    m, n = len(land), len(land[0])
6    for i in range(m):
7        for j in range(n):
8            if land[i][j] == 1:
9                r1, c1 = i, j
10                r2, c2 = i, j
11                while r2 + 1 < m and land[r2 + 1][j] == 1:
12                    r2 += 1
13                while c2 + 1 < n and land[i][c2 + 1] == 1:
14                    c2 += 1
15                result.append([r1, c1, r2, c2])
16                # Skip the cells that are part of the current rectangle
17                for x in range(r1, r2 + 1):
18                    for y in range(c1, c2 + 1):
19                        land[x][y] = 0  # Mark as visited
20    return result

ℹ

Complexity note: The time complexity is O(n) because we only traverse each cell once. The space complexity is O(n) due to the result list storing the coordinates of the rectangles.

1Each group of farmland is rectangular and non-adjacent, which simplifies the search for boundaries.
2Marking cells as visited helps avoid counting the same group multiple times.

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