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This approach uses a Breadth-First Search (BFS) algorithm to explore each group of farmland on the matrix.
Starting from a farmland cell (1), the BFS can be used to explore all connected cells that belong to the same group, thereby finding the bottom-right corner of that group. Use a queue to facilitate the exploration of individual farmlands.
Mark cells as visited as you explore them to ensure each cell is processed only once.
Time Complexity: O(m * n), as each cell is visited at most once.
Space Complexity: O(m * n) in the worst case due to additional space used for storing visited cells and the queue.
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 public int[][] FindFarmland(int[][] land) {
6 int m = land.Length;
7 int n = land[0].Length;
8 bool[,] visited = new bool[m, n];
9 var results = new List<int[]>();
10
11 for (int i = 0; i < m; i++) {
12 for (int j = 0; j < n; j++) {
13 if (land[i][j] == 1 && !visited[i, j]) {
14 int maxRow = i, maxCol = j;
15 var queue = new Queue<(int, int)>();
16 queue.Enqueue((i, j));
17 visited[i, j] = true;
18
19 while (queue.Count > 0) {
20 var (row, col) = queue.Dequeue();
21 maxRow = Math.Max(maxRow, row);
22 maxCol = Math.Max(maxCol, col);
23
24 int[,] directions = { {0, 1}, {1, 0} };
25 for (int d = 0; d < directions.GetLength(0); d++) {
26 int newRow = row + directions[d, 0], newCol = col + directions[d, 1];
27 if (newRow < m && newCol < n && land[newRow][newCol] == 1 && !visited[newRow, newCol]) {
28 visited[newRow, newCol] = true;
29 queue.Enqueue((newRow, newCol));
30 }
31 }
32 }
33 results.Add(new int[] { i, j, maxRow, maxCol });
34 }
35 }
36 }
37 return results.ToArray();
38 }
39}
The C# approach uses BFS to traverse each group of farmland. It checks cells' unvisited status and '1' value for expansion, determining the farmland's boundaries.
This approach implements a Union-Find (also called Disjoint Set Union) data structure to efficiently group farmland cells together. After processing all connections, each cell will belong to a group denoted by its root, and by traversing all cells, we can identify the top-left and bottom-right corners of each group.
Time Complexity: O(m * n * α(n)), with α being the inverse Ackermann function, which grows very slowly, making this almost O(m * n).
Space Complexity: O(m * n) due to storage for parents and ranks of each cell.
using System.Collections.Generic;
public class Solution {
public int[][] FindFarmland(int[][] land) {
int m = land.Length, n = land[0].Length;
int[] parent = new int[m * n];
int[] rank = new int[m * n];
for (int i = 0; i < m * n; i++) {
parent[i] = i;
}
int Find(int x) {
if (parent[x] != x) {
parent[x] = Find(parent[x]);
}
return parent[x];
}
void Union(int x, int y) {
int rootX = Find(x);
int rootY = Find(y);
if (rootX != rootY) {
if (rank[rootX] > rank[rootY]) {
parent[rootY] = rootX;
} else if (rank[rootX] < rank[rootY]) {
parent[rootX] = rootY;
} else {
parent[rootY] = rootX;
rank[rootX]++;
}
}
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (land[i][j] == 1) {
int index = i * n + j;
if (i > 0 && land[i-1][j] == 1) {
Union(index, (i-1) * n + j);
}
if (j > 0 && land[i][j-1] == 1) {
Union(index, i * n + (j-1));
}
}
}
}
var bounds = new Dictionary<int, int[]>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (land[i][j] == 1) {
int root = Find(i * n + j);
if (!bounds.ContainsKey(root)) {
bounds[root] = new int[] { i, j, i, j };
} else {
bounds[root][0] = Math.Min(bounds[root][0], i);
bounds[root][1] = Math.Min(bounds[root][1], j);
bounds[root][2] = Math.Max(bounds[root][2], i);
bounds[root][3] = Math.Max(bounds[root][3], j);
}
}
}
}
var result = new List<int[]>(bounds.Values);
return result.ToArray();
}
}
This C# solution leverages the efficiency of Union-Find for connecting farmland components: post-processing yields the corners per component for output.