<p>You are given an integer <code>n</code>. You have an <code>n x n</code> binary grid <code>grid</code> with all values initially <code>1</code>&#39;s except for some indices given in the array <code>mines</code>. The <code>i<sup>th</sup></code> element of the array <code>mines</code> is defined as <code>mines[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> where <code>grid[x<sub>i</sub>][y<sub>i</sub>] == 0</code>.</p>

<p>Return <em>the order of the largest <strong>axis-aligned</strong> plus sign of </em>1<em>&#39;s contained in </em><code>grid</code>. If there is none, return <code>0</code>.</p>

<p>An <strong>axis-aligned plus sign</strong> of <code>1</code>&#39;s of order <code>k</code> has some center <code>grid[r][c] == 1</code> along with four arms of length <code>k - 1</code> going up, down, left, and right, and made of <code>1</code>&#39;s. Note that there could be <code>0</code>&#39;s or <code>1</code>&#39;s beyond the arms of the plus sign, only the relevant area of the plus sign is checked for <code>1</code>&#39;s.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/13/plus1-grid.jpg" style="width: 404px; height: 405px;" />
<pre>
<strong>Input:</strong> n = 5, mines = [[4,2]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> In the above grid, the largest plus sign can only be of order 2. One of them is shown.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/13/plus2-grid.jpg" style="width: 84px; height: 85px;" />
<pre>
<strong>Input:</strong> n = 1, mines = [[0,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> There is no plus sign, so return 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 500</code></li>
	<li><code>1 &lt;= mines.length &lt;= 5000</code></li>
	<li><code>0 &lt;= x<sub>i</sub>, y<sub>i</sub> &lt; n</code></li>
	<li>All the pairs <code>(x<sub>i</sub>, y<sub>i</sub>)</code> are <strong>unique</strong>.</li>
</ul>

from collections import deque
class Solution:
    def orderOfLargestPlusSign(self, n: int, mines: List[List[int]]) -> int:
        p = print

        mat = [[1] * n for _ in range(n)]
        for i, j in mines:
            mat[i][j] = 0

        dp_top_to_bottom = [[0] * n for _ in range(n)]
        for i in range(n):
            for j in range(n):
                if i == 0 and mat[0][j] == 1:
                    dp_top_to_bottom[0][j] = 1
                elif i == 0:
                    dp_top_to_bottom[0][j] = 0
                elif mat[i][j] == 1:
                    dp_top_to_bottom[i][j] = dp_top_to_bottom[i-1][j] + 1

        dp_bottom_to_top = [[0] * n for _ in range(n)]
        for i in range(n-1, -1, -1):
            for j in range(n):
                if i == n-1 and mat[n-1][j] == 1:
                    dp_bottom_to_top[n-1][j] = 1
                elif i == n-1:
                    dp_bottom_to_top[n-1][j] = 0
                elif mat[i][j] == 1:
                    dp_bottom_to_top[i][j] = dp_bottom_to_top[i+1][j] + 1

        dp_left_to_right = [[0] * n for _ in range(n)]
        for i in range(n):
            for j in range(n):
                if j == 0 and mat[i][j] == 1:
                    dp_left_to_right[i][0] = 1
                elif j == 0:
                    dp_left_to_right[i][0] = 0
                elif mat[i][j] == 1:
                    dp_left_to_right[i][j] = dp_left_to_right[i][j-1] + 1        

        dp_right_to_left = [[0] * n for _ in range(n)]
        for i in range(n):
            for j in range(n-1, -1, -1):
                if j == n-1 and mat[i][n-1] == 1:
                    dp_right_to_left[i][n-1] = 1
                elif j == n-1:
                    dp_right_to_left[i][n-1] = 0
                elif mat[i][j] == 1:
                    dp_right_to_left[i][j] = dp_right_to_left[i][j+1] + 1               

        # p(dp_top_to_bottom)
        # p(dp_bottom_to_top)
        # p(dp_left_to_right)
        # p(dp_right_to_left)
        dp = [[0] * n for _ in range(n)]
        max_len = 0
        for i in range(n):
            for j in range(n):
                # dp[i][j] = dp_top_to_bottom[i][j] + dp_bottom_to_top[i][j] + dp_left_to_right[i][j] + dp_right_to_left[i][j]
                dp[i][j] = min(dp_top_to_bottom[i][j] , dp_bottom_to_top[i][j] , dp_left_to_right[i][j] , dp_right_to_left[i][j] )
                max_len = max(max_len, dp[i][j])
        
        p(dp)
        p(max_len)

        return max_len


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