<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>

class Solution {
public:
    int orderOfLargestPlusSign(int n, std::vector<std::vector<int>>& mines) {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cout.tie(nullptr);
        vector<vector<int>> grid(n, vector<int>(n, 1));
        for (auto& mine : mines) {
            grid[mine[0]][mine[1]] = 0;
        }
        vector<vector<int>> left(n, vector<int>(n, 0));
        vector<vector<int>> right(n, vector<int>(n, 0));
        vector<vector<int>> top(n, vector<int>(n, 0));
        vector<vector<int>> bottom(n, vector<int>(n, 0));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (j == 0) {
                    left[i][j] = grid[i][j];
                }
                else if (grid[i][j]) {
                    left[i][j] = left[i][j - 1] + 1;
                }
            }
            for (int j = n - 1; j >= 0; j--) {
                if (j == n - 1) {
                    right[i][j] = grid[i][j];
                }
                else if (grid[i][j]) {
                    right[i][j] = right[i][j + 1] + 1;
                }
            }
        }
        for (int j = 0; j < n; j++) {
            for (int i = 0; i < n; i++) {
                if (i == 0) {
                    top[i][j] = grid[i][j];
                }
                else if (grid[i][j]) {
                    top[i][j] = top[i - 1][j] + 1;
                }
            }
            for (int i = n - 1; i >= 0; i--) {
                if (i == n - 1) {
                    bottom[i][j] = grid[i][j];
                }
                else if (grid[i][j]) {
                    bottom[i][j] = bottom[i + 1][j] + 1;
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                ans = max(ans, min({left[i][j], right[i][j], top[i][j], bottom[i][j]}));
            }
        }
        return ans;
    }
};
93668