463. Island Perimeter

/**
 * Refer:
 * https://leetcode.com/problems/island-perimeter/discuss/95001/clear-and-easy-java-solution
 *
 * Loop over the matrix and count the number of islands. If the current dot is
 * an island, count if it has any right neighbour or down neighbour. The result
 * is islands*4 - neighbours*2.
 *
 * Time Complexity: O(M * N)
 *
 * Space Complexity: O(1)
 *
 * M = Number of rows in the grid. N = Number of columns in the grid.
 */
class Solution {
    public int islandPerimeter(int[][] grid) {
        if (grid == null || grid.length == 0 || grid[0].length == 0) {
            return 0;
        }

        int islands = 0;
        int neighbours = 0;
        int rows = grid.length;
        int cols = grid[0].length;

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 1) {
                    islands++;
                    if (j < cols - 1 && grid[i][j + 1] == 1) {
                        neighbours++;
                    }
                    if (i < rows - 1 && grid[i + 1][j] == 1) {
                        neighbours++;
                    }
                }
            }
        }

        return islands * 4 - neighbours * 2;
    }
}