LeetCode 463: Island Perimeter Solution

Master LeetCode problem 463 (Island Perimeter), a easy 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.

463. Island Perimeter

Easy

Array Depth-First Search Breadth-First Search Matrix

Problem Explanation

Explanation

To solve this problem, we can iterate over each cell in the grid and for each land cell (where grid[i][j] == 1), we can check the adjacent cells to determine the perimeter contribution of that cell. We can count the number of adjacent water cells and add that to the perimeter. The final perimeter will be the sum of all individual cell contributions.

We can iterate through the grid and for each land cell, check its adjacent cells (up, down, left, right) to count the number of water cells. The perimeter contribution for each land cell will be 4 minus the count of water cells around it.

Time complexity: O(row * col)
Space complexity: O(1)

Solution Code

class Solution {
    public int islandPerimeter(int[][] grid) {
        int perimeter = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    perimeter += 4;
                    if (i > 0 && grid[i - 1][j] == 1) perimeter--;
                    if (i < grid.length - 1 && grid[i + 1][j] == 1) perimeter--;
                    if (j > 0 && grid[i][j - 1] == 1) perimeter--;
                    if (j < grid[0].length - 1 && grid[i][j + 1] == 1) perimeter--;
                }
            }
        }
        return perimeter;
    }
}

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