notes/OJ notes/pages/Leetcode Max-Area-of-Island.md

Leetcode Max-Area-of-Island

2022-07-15 10:00

Algorithms:

#algorithm

Data structures:

#DS #vector_2d

Difficulty:

#coding_problem #difficulty-medium

Additional tags:

#leetcode

Revisions:

N/A

Related topics:

Links:

• Link to problem

---

Problem

You are given an m x n binary matrix grid. An island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.) You may assume all four edges of the grid are surrounded by water.

The area of an island is the number of cells with a value 1 in the island.

Return the maximum area of an island in grid. If there is no island, return 0.

Examples

Example 1:

**Input:** grid = [[0,0,1,0,0,0,0,1,0,0,0,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,1,1,0,1,0,0,0,0,0,0,0,0],[0,1,0,0,1,1,0,0,1,0,1,0,0],[0,1,0,0,1,1,0,0,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,0,1,0,0],[0,0,0,0,0,0,0,1,1,1,0,0,0],[0,0,0,0,0,0,0,1,1,0,0,0,0]]
**Output:** 6
**Explanation:** The answer is not 11, because the island must be connected 4-directionally.

Example 2:

**Input:** grid = [[0,0,0,0,0,0,0,0]]
**Output:** 0

Constraints

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 50
• grid[i][j] is either 0 or 1.

Thoughts

[!summary] This is a search problem, can be implemented using #DFS or #BFS

Simple O(mn) solution.

We keep track of pile we visited, using:

• an 2-d bool vector visited(which is the one I'm using)
• change the value of pile to 0

and check for:

• x, y is not OOB
• x, y is in a island -> (value == 1)
• x, y is not visited (see above)

And for each unvisited pile, run DFS or BFS to check the size, and keep track of max size with a variable

Solution

BFS

class Solution {
  int getSize(vector<vector<int>> &grid, int m, int n, int x, int y,
              vector<vector<bool>> &visited) {
    queue<pair<int, int>> todo;
    todo.push({x, y});
    int r, c;
    int size = 0;

    while (!todo.empty()) {
      r = todo.front().first;
      c = todo.front().second;
      todo.pop();

      if (visited[r][c] || grid[r][c] != 1) {
        // Visited this place, or the color is not 1
        continue;
      }

      size++;
      visited[r][c] = true;

      if (r > 0) {
        todo.push({r - 1, c});
      }
      if (r < m - 1) {
        todo.push({r + 1, c});
      }
      if (c > 0) {
        todo.push({r, c - 1});
      }
      if (c < n - 1) {
        todo.push({r, c + 1});
      }
    }

    return size;
  }

public:
  int maxAreaOfIsland(vector<vector<int>> &grid) {
    int m = grid.size(), n = grid[0].size();
    int maxSize = 0;

    vector<vector<bool>> visited;
    for (int i = 0; i < m; i++) {
      visited.push_back({});
      for (int j = 0; j < n; j++) {
        visited.back().push_back(false);
      }
    }

    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        if (grid[i][j] == 1 && !visited[i][j]) {
          maxSize = max(maxSize, getSize(grid, m, n, i, j, visited));
        }
      }
    }
    return maxSize;
  }
};

DFS iterative (simply change from queue to stack in BFS)

class Solution {
  // DFS iterative
  int getSize(vector<vector<int>> &grid, int m, int n, int x, int y,
              vector<vector<bool>> &visited) {
    stack<pair<int, int>> todo;
    todo.push({x, y});
    int r, c;
    int size = 0;

    while (!todo.empty()) {
      r = todo.top().first;
      c = todo.top().second;
      todo.pop();

      if (visited[r][c] || grid[r][c] != 1) {
        // Visited this place, or the color is not 1
        continue;
      }

      size++;
      visited[r][c] = true;

      if (r > 0) {
        todo.push({r - 1, c});
      }
      if (r < m - 1) {
        todo.push({r + 1, c});
      }
      if (c > 0) {
        todo.push({r, c - 1});
      }
      if (c < n - 1) {
        todo.push({r, c + 1});
      }
    }

    return size;
  }

public:
  int maxAreaOfIsland(vector<vector<int>> &grid) {
    int m = grid.size(), n = grid[0].size();
    int maxSize = 0;

    vector<vector<bool>> visited;
    for (int i = 0; i < m; i++) {
      visited.push_back({});
      for (int j = 0; j < n; j++) {
        visited.back().push_back(false);
      }
    }

    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        if (grid[i][j] == 1 && !visited[i][j]) {
          maxSize = max(maxSize, getSize(grid, m, n, i, j, visited));
        }
      }
    }
    return maxSize;
  }
};

DFS recursive

class Solution {
  // DFS recursive
  int getSize(vector<vector<int>> &grid, int m, int n, int x, int y,
              vector<vector<bool>> &visited) {
    if (x >= 0 && x < m && y >= 0 && y < n && !visited[x][y] &&
        grid[x][y] == 1) {
      visited[x][y] = true;
      return 1 + getSize(grid, m, n, x + 1, y, visited) +
             getSize(grid, m, n, x - 1, y, visited) +
             getSize(grid, m, n, x, y + 1, visited) +
             getSize(grid, m, n, x, y - 1, visited);
    } else {
      return 0;
    }
  }

public:
  int maxAreaOfIsland(vector<vector<int>> &grid) {
    int m = grid.size(), n = grid[0].size();
    int maxSize = 0;

    vector<vector<bool>> visited;
    for (int i = 0; i < m; i++) {
      visited.push_back({});
      for (int j = 0; j < n; j++) {
        visited.back().push_back(false);
      }
    }

    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        if (grid[i][j] == 1 && !visited[i][j]) {
          maxSize = max(maxSize, getSize(grid, m, n, i, j, visited));
        }
      }
    }
    return maxSize;
  }
};