Question

1139. Largest 1-Bordered Square

中文文档

Description

Given a 2D grid of 0s and 1s, return the number of elements in the largest square subgrid that has all 1s on its border, or 0 if such a subgrid doesn't exist in the grid.

 

Example 1:

Input: grid = [[1,1,1],[1,0,1],[1,1,1]]

Output: 9

Example 2:

Input: grid = [[1,1,0,0]]

Output: 1

 

Constraints:

• 1 <= grid.length <= 100
• 1 <= grid[0].length <= 100
• grid[i][j] is 0 or 1

Solutions

Solution 1: Prefix Sum + Enumeration

We can use the prefix sum method to preprocess the number of consecutive 1s down and to the right of each position, denoted as down[i][j] and right[i][j].

Then we enumerate the side length $k$ of the square, starting from the largest side length. Then we enumerate the upper left corner position $(i, j)$ of the square. If it meets the condition, we can return $k^2$.

The time complexity is $O(m \times n \times \min(m, n))$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the grid, respectively.

class Solution:
  def largest1BorderedSquare(self, grid: List[List[int]]) -> int:
    m, n = len(grid), len(grid[0])
    down = [[0] * n for _ in range(m)]
    right = [[0] * n for _ in range(m)]
    for i in range(m - 1, -1, -1):
      for j in range(n - 1, -1, -1):
        if grid[i][j]:
          down[i][j] = down[i + 1][j] + 1 if i + 1 < m else 1
          right[i][j] = right[i][j + 1] + 1 if j + 1 < n else 1
    for k in range(min(m, n), 0, -1):
      for i in range(m - k + 1):
        for j in range(n - k + 1):
          if (
            down[i][j] >= k
            and right[i][j] >= k
            and right[i + k - 1][j] >= k
            and down[i][j + k - 1] >= k
          ):
            return k * k
    return 0

class Solution {
  public int largest1BorderedSquare(int[][] grid) {
    int m = grid.length, n = grid[0].length;
    int[][] down = new int[m][n];
    int[][] right = new int[m][n];
    for (int i = m - 1; i >= 0; --i) {
      for (int j = n - 1; j >= 0; --j) {
        if (grid[i][j] == 1) {
          down[i][j] = i + 1 < m ? down[i + 1][j] + 1 : 1;
          right[i][j] = j + 1 < n ? right[i][j + 1] + 1 : 1;
        }
      }
    }
    for (int k = Math.min(m, n); k > 0; --k) {
      for (int i = 0; i <= m - k; ++i) {
        for (int j = 0; j <= n - k; ++j) {
          if (down[i][j] >= k && right[i][j] >= k && right[i + k - 1][j] >= k
            && down[i][j + k - 1] >= k) {
            return k * k;
          }
        }
      }
    }
    return 0;
  }
}

class Solution {
public:
  int largest1BorderedSquare(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    int down[m][n];
    int right[m][n];
    memset(down, 0, sizeof down);
    memset(right, 0, sizeof right);
    for (int i = m - 1; i >= 0; --i) {
      for (int j = n - 1; j >= 0; --j) {
        if (grid[i][j] == 1) {
          down[i][j] = i + 1 < m ? down[i + 1][j] + 1 : 1;
          right[i][j] = j + 1 < n ? right[i][j + 1] + 1 : 1;
        }
      }
    }
    for (int k = min(m, n); k > 0; --k) {
      for (int i = 0; i <= m - k; ++i) {
        for (int j = 0; j <= n - k; ++j) {
          if (down[i][j] >= k && right[i][j] >= k && right[i + k - 1][j] >= k
            && down[i][j + k - 1] >= k) {
            return k * k;
          }
        }
      }
    }
    return 0;
  }
};

func largest1BorderedSquare(grid [][]int) int {
  m, n := len(grid), len(grid[0])
  down := make([][]int, m)
  right := make([][]int, m)
  for i := range down {
    down[i] = make([]int, n)
    right[i] = make([]int, n)
  }
  for i := m - 1; i >= 0; i-- {
    for j := n - 1; j >= 0; j-- {
      if grid[i][j] == 1 {
        down[i][j], right[i][j] = 1, 1
        if i+1 < m {
          down[i][j] += down[i+1][j]
        }
        if j+1 < n {
          right[i][j] += right[i][j+1]
        }
      }
    }
  }
  for k := min(m, n); k > 0; k-- {
    for i := 0; i <= m-k; i++ {
      for j := 0; j <= n-k; j++ {
        if down[i][j] >= k && right[i][j] >= k && right[i+k-1][j] >= k && down[i][j+k-1] >= k {
          return k * k
        }
      }
    }
  }
  return 0
}