Sync LeetCode submission Runtime - 71 ms (28.37%), Memory - 21.6 MB (16.62%)

Author: jimit105

0302-smallest-rectangle-enclosing-black-pixels/README.md

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+<p>You are given an <code>m x n</code> binary matrix <code>image</code> where <code>0</code> represents a white pixel and <code>1</code> represents a black pixel.</p>
+
+<p>The black pixels are connected (i.e., there is only one black region). Pixels are connected horizontally and vertically.</p>
+
+<p>Given two integers <code>x</code> and <code>y</code> that represents the location of one of the black pixels, return <em>the area of the smallest (axis-aligned) rectangle that encloses all black pixels</em>.</p>
+
+<p>You must write an algorithm with less than <code>O(mn)</code> runtime complexity</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/pixel-grid.jpg" style="width: 333px; height: 253px;" />
+<pre>
+<strong>Input:</strong> image = [[&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;],[&quot;0&quot;,&quot;1&quot;,&quot;1&quot;,&quot;0&quot;],[&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;]], x = 0, y = 2
+<strong>Output:</strong> 6
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> image = [[&quot;1&quot;]], x = 0, y = 0
+<strong>Output:</strong> 1
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>m == image.length</code></li>
+	<li><code>n == image[i].length</code></li>
+	<li><code>1 &lt;= m, n &lt;= 100</code></li>
+	<li><code>image[i][j]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
+	<li><code>0 &lt;= x &lt; m</code></li>
+	<li><code>0 &lt;= y &lt; n</code></li>
+	<li><code>image[x][y] == &#39;1&#39;.</code></li>
+	<li>The black pixels in the <code>image</code> only form <strong>one component</strong>.</li>
+</ul>

0302-smallest-rectangle-enclosing-black-pixels/solution.py

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+# Approach 2: DFS
+
+# Time: O(mn)
+# Space: O(mn)
+
+class Solution:
+    def minArea(self, image: List[List[str]], x: int, y: int) -> int:
+        if not image or not image[0]:
+            return 0
+
+        self.top = self.bottom = x
+        self.left = self.right = y
+
+        def dfs(x, y):
+            if (x < 0 or y < 0 or
+                x >= len(image) or y >= len(image[0]) or
+                image[x][y] == '0'):
+                return
+
+            image[x][y] = '0' # Mark visited black pixel as white
+
+            self.top = min(self.top, x)
+            self.bottom = max(self.bottom, x + 1)
+            self.left = min(self.left, y)
+            self.right = max(self.right, y + 1)
+
+            dfs(x + 1, y)
+            dfs(x - 1, y)
+            dfs(x, y - 1)
+            dfs(x, y + 1)
+
+        dfs(x, y)
+        return (self.right - self.left) * (self.bottom - self.top)
+