<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, vector<vector<int>>& mines) {
        using TData = std::pair<int, int>;
        // Hash function
        struct hashFunction
        {
            size_t operator()(const TData& x) const { return x.first ^ x.second; }
        };
        std::unordered_set<TData, hashFunction> us;

        for (const auto& mine : mines)
        {
            us.insert(std::make_pair(mine[0], mine[1]));
        }

        auto dp = std::vector<std::vector<TData>>(n, std::vector<TData>(n));

        int result = 0;
        if (n > 0)
        {
            int w = 0;
            for (int j = 0; j < n; ++j)
            {
                if (us.find(std::make_pair(0, j)) == us.end())
                {
                    ++w;
                    dp[0][j] = std::make_pair(1, w);
                    result = 1;
                }
                else
                {
                    w = 0;
                    dp[0][j] = std::make_pair(0, w);
                }
            }
        }

        auto funcNextCheckN = [](int n) {
            return (n * 2) + 1;
        };

        int checkN = funcNextCheckN(result);
        std::vector<TData>* last = &(dp[0]);
        for (int i = 1; i < n; ++i)
        {
            int w = 0;
            for (int j = 0; j < n; ++j)
            {
                if (us.find(std::make_pair(i, j)) == us.end())
                {
                    ++w;
                    int nw = (*last)[j].first + 1;
                    dp[i][j] = std::make_pair(nw, w);
                    if (nw >= checkN && (j + result) < n && (i - result < n) &&
                        dp[i - result][j+result].second >= checkN)
                    {
                        result++;
                        checkN = funcNextCheckN(result);
                        if (checkN > n)
                        {
                            break;
                        }
                    }
                }
                else
                {
                    w = 0;
                    dp[i][j] = std::make_pair(0, w);
                }
            }
            if (checkN > n)
            {
                break;
            }
            last = &(dp[i]);
        }

        return result;
    }
};
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