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LeetCode 695: Max Area of Island Solution

Master LeetCode problem 695 (Max Area of Island), 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.

Problem Explanation

Explanation:

To solve this problem, we can use Depth First Search (DFS) to explore each cell in the grid. We will start DFS from each cell with a value of 1 and calculate the area of the island connected to that cell. We will keep track of the maximum area found so far and return it as the result.

  1. Initialize a variable maxArea to store the maximum area found.
  2. Iterate over each cell in the grid:
    • If the cell value is 1, start DFS from that cell to explore the connected island.
    • During DFS, mark the visited cell as 0 to avoid revisiting it.
    • Calculate the area of the island during DFS and update maxArea if a larger area is found.
  3. Return the maxArea as the result.

Time Complexity: O(mn) where m is the number of rows and n is the number of columns in the grid. Space Complexity: O(mn) for the recursive stack and visited array.

:

Solution Code

class Solution {
    public int maxAreaOfIsland(int[][] grid) {
        int maxArea = 0;
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    maxArea = Math.max(maxArea, dfs(grid, i, j));
                }
            }
        }
        return maxArea;
    }
    
    private int dfs(int[][] grid, int i, int j) {
        if (i < 0 || i >= grid.length || j < 0 || j >= grid[0].length || grid[i][j] == 0) {
            return 0;
        }
        grid[i][j] = 0;
        int area = 1;
        area += dfs(grid, i + 1, j);
        area += dfs(grid, i - 1, j);
        area += dfs(grid, i, j + 1);
        area += dfs(grid, i, j - 1);
        return area;
    }
}

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