LeetCode 1992: Find All Groups of Farmland Solution

Master LeetCode problem 1992 (Find All Groups of Farmland), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1992. Find All Groups of Farmland

Medium

Array Depth-First Search Breadth-First Search Matrix

Problem Explanation

Explanation:

To solve this problem, we can perform a depth-first search (DFS) on the given matrix land. We iterate through each cell in the matrix, and when we encounter a cell with the value 1, we start a DFS from that cell to find the coordinates of the farmland group. During the DFS, we mark visited cells and keep track of the top left and bottom right corners of the group.

Algorithm:

Initialize an empty list to store the result.
Iterate through each cell in the matrix:
- If the cell value is 1 and it has not been visited:
  - Start a DFS from that cell.
  - During the DFS, update the top left and bottom right corners of the group.
Return the list of coordinates of the farmland groups.

Time Complexity:

The time complexity of this approach is O(m * n), where m is the number of rows and n is the number of columns in the matrix.

Space Complexity:

The space complexity is O(m * n) due to the recursion stack during the DFS.

: :

Solution Code

class Solution {
    public int[][] findFarmland(int[][] land) {
        List<int[]> result = new ArrayList<>();
        int m = land.length;
        int n = land[0].length;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (land[i][j] == 1) {
                    int[] corners = new int[]{i, j, i, j};
                    dfs(land, i, j, corners);
                    result.add(corners);
                }
            }
        }

        return result.toArray(new int[0][]);
    }

    private void dfs(int[][] land, int row, int col, int[] corners) {
        if (row < 0 || row >= land.length || col < 0 || col >= land[0].length || land[row][col] == 0) {
            return;
        }

        land[row][col] = 0; // Mark as visited

        corners[2] = Math.max(corners[2], row); // Update bottom right row
        corners[3] = Math.max(corners[3], col); // Update bottom right col

        dfs(land, row + 1, col, corners); // Down
        dfs(land, row - 1, col, corners); // Up
        dfs(land, row, col + 1, corners); // Right
        dfs(land, row, col - 1, corners); // Left
    }
}

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