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1139. Largest 1-Bordered Square

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.

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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

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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;
    }
}

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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;
    }
};

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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
}

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