<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:
    """
    0 0 0 1
    1 1 0 1
    1 1 1 1 
    0 1 1 1
    0 1 1 1
    """
    def maximalSquare(self, matrix: List[List[str]]) -> int:
        dp = [[(float('inf'), float('inf')) for i in range(len(matrix[0]))] for j in range(len(matrix))] #Record the top left corner of the square
        
        out = 0
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                if matrix[i][j] == "0":
                    continue

                dp[i][j] = (i, j) #2 3
                out = max(out, 1)
                if i == 0 or j == 0:
                    continue

                new_top_left = dp[i - 1][j - 1] #1, 2
                if new_top_left != (float('inf'), float('inf')):
                    left_res_y, left_res_x = dp[i][j - 1] # 2 2
                    up_res_y, up_res_x = dp[i - 1][j] # 1 3

                    new_coords = (max(up_res_y, new_top_left[0]), max(left_res_x, new_top_left[1]))
                    diagonal = min(i - new_coords[0], j - new_coords[1])

                    if new_coords[0] == float('inf') or new_coords[1] == float('inf'):
                        continue
                    dp[i][j] = (i - diagonal, j - diagonal)
                out = max(out, (i - dp[i][j][0] + 1) * (j - dp[i][j][1] + 1))
        
        return out
                




                


30025