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1905. Count Sub Islands 👍

  • Time: $O(mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  int countSubIslands(vector<vector<int>>& grid1, vector<vector<int>>& grid2) {
    int ans = 0;

    for (int i = 0; i < grid2.size(); ++i)
      for (int j = 0; j < grid2[0].size(); ++j)
        if (grid2[i][j] == 1)
          ans += dfs(grid1, grid2, i, j);

    return ans;
  }

 private:
  int dfs(const vector<vector<int>>& grid1, vector<vector<int>>& grid2, int i,
          int j) {
    if (i < 0 || i == grid1.size() || j < 0 || j == grid2[0].size())
      return 1;
    if (grid2[i][j] != 1)
      return 1;

    grid2[i][j] = 2;  // Mark 2 as visited.

    return dfs(grid1, grid2, i + 1, j) & dfs(grid1, grid2, i - 1, j) &
           dfs(grid1, grid2, i, j + 1) & dfs(grid1, grid2, i, j - 1) &
           grid1[i][j];
  }
};

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class Solution {
  public int countSubIslands(int[][] grid1, int[][] grid2) {
    int ans = 0;

    for (int i = 0; i < grid2.length; ++i)
      for (int j = 0; j < grid2[0].length; ++j)
        if (grid2[i][j] == 1)
          ans += dfs(grid1, grid2, i, j);

    return ans;
  }

  private int dfs(int[][] grid1, int[][] grid2, int i, int j) {
    if (i < 0 || i == grid1.length || j < 0 || j == grid2[0].length)
      return 1;
    if (grid2[i][j] != 1)
      return 1;

    grid2[i][j] = 2; // Mark 2 as visited.

    return                                                          //
        dfs(grid1, grid2, i + 1, j) & dfs(grid1, grid2, i - 1, j) & //
        dfs(grid1, grid2, i, j + 1) & dfs(grid1, grid2, i, j - 1) & grid1[i][j];
  }
}

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class Solution:
  def countSubIslands(
      self,
      grid1: list[list[int]],
      grid2: list[list[int]],
  ) -> int:
    m = len(grid2)
    n = len(grid2[0])

    def dfs(i: int, j: int) -> int:
      if i < 0 or i == m or j < 0 or j == n:
        return 1
      if grid2[i][j] != 1:
        return 1

      grid2[i][j] = 2  # Mark 2 as visited.

      return (dfs(i + 1, j) & dfs(i - 1, j) &
              dfs(i, j + 1) & dfs(i, j - 1) & grid1[i][j])

    ans = 0

    for i in range(m):
      for j in range(n):
        if grid2[i][j] == 1:
          ans += dfs(i, j)

    return ans