LeetCode 1020: Number of Enclaves Solution

Master LeetCode problem 1020 (Number of Enclaves), 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.

1020. Number of Enclaves

Medium

Array Depth-First Search Breadth-First Search Union Find Matrix

Problem Explanation

Explanation

To solve this problem, we can perform a depth-first search (DFS) from the boundary cells of the grid and mark all connected land cells. After completing the DFS, we can iterate through the grid and count the remaining land cells that were not marked.

Perform DFS from boundary cells to mark all connected cells.
Count remaining unmarked land cells.

Time complexity: O(m * n) where m is the number of rows and n is the number of columns in the grid. Space complexity: O(m * n) for the recursion stack.

Solution Code

class Solution {
    public int numEnclaves(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;

        for (int i = 0; i < m; i++) {
            dfs(grid, i, 0);
            dfs(grid, i, n - 1);
        }

        for (int j = 0; j < n; j++) {
            dfs(grid, 0, j);
            dfs(grid, m - 1, j);
        }

        int count = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    count++;
                }
            }
        }

        return count;
    }

    private void dfs(int[][] grid, int i, int j) {
        if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length || grid[i][j] == 0) {
            return;
        }

        grid[i][j] = 0;

        dfs(grid, i + 1, j);
        dfs(grid, i - 1, j);
        dfs(grid, i, j + 1);
        dfs(grid, i, j - 1);
    }
}

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