<p>Given an <code>m x n</code> binary <code>matrix</code> filled with <code>0</code>&#39;s and <code>1</code>&#39;s, <em>find the largest square containing only</em> <code>1</code>&#39;s <em>and return its area</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/max1grid.jpg" style="width: 400px; height: 319px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 4
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/26/max2grid.jpg" style="width: 165px; height: 165px;" />
<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;,&quot;1&quot;],[&quot;1&quot;,&quot;0&quot;]]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> matrix = [[&quot;0&quot;]]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>matrix[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>

class Solution {
public:
    int maximalSquare(vector<vector<char>>& matrix) {
        struct Data
        {
            int numberOfConsecutiveSquaresLeft = 0;
            int numberOfConsecutiveSquaresUp = 0;
            int sizeOfSquare = 0;
        };

        auto data = vector<vector<Data>>(matrix.size() + 1, vector<Data>(matrix[0].size() + 1));
        auto maximumSizeOfSquare = 0;

        for (auto y = 1; y <= matrix.size(); ++y)
        {
            for (auto x = 1; x <= matrix[0].size(); ++x)
            {
                if (matrix[y - 1][x - 1] == '1')
                {
                    data[y][x].numberOfConsecutiveSquaresLeft = data[y][x - 1].numberOfConsecutiveSquaresLeft + 1;
                    data[y][x].numberOfConsecutiveSquaresUp = data[y - 1][x].numberOfConsecutiveSquaresUp + 1;
                    // This is a square if numberOfConsecutiveSquaresLeft and numberOfConsecutiveSquaresUp are both
                    // at least one larger than the sizeOfSquare in the top left. The size of the new square is the
                    // size of the top left square + 1.
                    auto shortestDirection = std::min(data[y][x].numberOfConsecutiveSquaresLeft, data[y][x].numberOfConsecutiveSquaresUp);
                    if (shortestDirection >= data[y - 1][x - 1].sizeOfSquare + 1)
                    {
                        data[y][x].sizeOfSquare = data[y - 1][x - 1].sizeOfSquare + 1;
                    }
                    else
                    {
                        data[y][x].sizeOfSquare = std::min(shortestDirection, data[y - 1][x - 1].sizeOfSquare);
                    }

                    maximumSizeOfSquare = std::max(maximumSizeOfSquare, data[y][x].sizeOfSquare);

                    // If we can't continue to expand the square, we can at least make a smaller square up to the
                    // limits of std::min(numberOfConsecutiveSquaresLeft, numberOfConsecutiveSquaresUp)

                    // [["0","0","0","1"],
                    //  ["1","1","0","1"],
                    //  ["1","1","1","1"],
                    //  ["0","1","1","1"],
                    //  ["0","1","1","1"]]
                }
            }
        }

        return maximumSizeOfSquare * maximumSizeOfSquare;
    }
};
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