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1568. Minimum Number of Days to Disconnect Island 👍

  • Time: $O((mn)^2)$
  • Space: $O(mn)$
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class Solution {
 public:
  int minDays(vector<vector<int>>& grid) {
    if (disconnected(grid))
      return 0;

    // Try to remove 1 land.
    for (int i = 0; i < grid.size(); ++i)
      for (int j = 0; j < grid[0].size(); ++j)
        if (grid[i][j] == 1) {
          grid[i][j] = 0;
          if (disconnected(grid))
            return 1;
          grid[i][j] = 1;
        }

    // Remove 2 lands.
    return 2;
  }

 private:
  const vector<int> dirs{0, 1, 0, -1, 0};

  bool disconnected(const vector<vector<int>>& grid) {
    int islandsCount = 0;
    vector<vector<bool>> seen(grid.size(), vector<bool>(grid[0].size()));
    for (int i = 0; i < grid.size(); ++i)
      for (int j = 0; j < grid[0].size(); ++j) {
        if (grid[i][j] == 0 || seen[i][j])
          continue;
        if (++islandsCount > 1)
          return true;
        dfs(grid, i, j, seen);
      }
    return islandsCount != 1;
  }

  void dfs(const vector<vector<int>>& grid, int i, int j,
           vector<vector<bool>>& seen) {
    seen[i][j] = true;
    for (int k = 0; k < 4; ++k) {
      const int x = i + dirs[k];
      const int y = j + dirs[k + 1];
      if (x < 0 || x == grid.size() || y < 0 || y == grid[0].size())
        continue;
      if (grid[x][y] == 0 || seen[x][y])
        continue;
      dfs(grid, x, y, seen);
    }
  }
};

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class Solution {
  public int minDays(int[][] grid) {
    if (disconnected(grid))
      return 0;

    // Try to remove 1 land.
    for (int i = 0; i < grid.length; ++i)
      for (int j = 0; j < grid[0].length; ++j)
        if (grid[i][j] == 1) {
          grid[i][j] = 0;
          if (disconnected(grid))
            return 1;
          grid[i][j] = 1;
        }

    // Remove 2 lands.
    return 2;
  }

  private final int[] dirs = {0, 1, 0, -1, 0};

  private boolean disconnected(int[][] grid) {
    int islandsCount = 0;
    boolean[][] seen = new boolean[grid.length][grid[0].length];
    for (int i = 0; i < grid.length; ++i)
      for (int j = 0; j < grid[0].length; ++j) {
        if (grid[i][j] == 0 || seen[i][j])
          continue;
        if (++islandsCount > 1)
          return true;
        dfs(grid, i, j, seen);
      }

    return islandsCount != 1;
  }

  private void dfs(int[][] grid, int i, int j, boolean[][] seen) {
    seen[i][j] = true;
    for (int k = 0; k < 4; ++k) {
      int x = i + dirs[k];
      int y = j + dirs[k + 1];
      if (x < 0 || x == grid.length || y < 0 || y == grid[0].length)
        continue;
      if (grid[x][y] == 0 || seen[x][y])
        continue;
      dfs(grid, x, y, seen);
    }
  }
}

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class Solution:
  def minDays(self, grid: List[List[int]]) -> int:
    m = len(grid)
    n = len(grid[0])
    dirs = [0, 1, 0, -1, 0]

    def dfs(grid: List[List[int]], i: int, j: int, seen: Set[Tuple[int, int]]):
      seen.add((i, j))
      for k in range(4):
        x = i + dirs[k]
        y = j + dirs[k + 1]
        if x < 0 or x == m or y < 0 or y == n:
          continue
        if grid[x][y] == 0 or (x, y) in seen:
          continue
        dfs(grid, x, y, seen)

    def disconnected(grid: List[List[int]]) -> bool:
      islandsCount = 0
      seen = set()
      for i in range(m):
        for j in range(n):
          if grid[i][j] == 0 or (i, j) in seen:
            continue
          if islandsCount > 1:
            return True
          islandsCount += 1
          dfs(grid, i, j, seen)
      return islandsCount != 1

    if disconnected(grid):
      return 0

    # Try to remove 1 land.
    for i in range(m):
      for j in range(n):
        if grid[i][j] == 1:
          grid[i][j] = 0
          if disconnected(grid):
            return 1
          grid[i][j] = 1

    # Remove 2 lands.
    return 2