Sync LeetCode submission Runtime - 1458 ms (23.23%), Memory - 30.6 MB (41.88%)

Author: jimit105

0854-making-a-large-island/README.md

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+<p>You are given an <code>n x n</code> binary matrix <code>grid</code>. You are allowed to change <strong>at most one</strong> <code>0</code> to be <code>1</code>.</p>
+
+<p>Return <em>the size of the largest <strong>island</strong> in</em> <code>grid</code> <em>after applying this operation</em>.</p>
+
+<p>An <strong>island</strong> is a 4-directionally connected group of <code>1</code>s.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> grid = [[1,0],[0,1]]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> Change one 0 to 1 and connect two 1s, then we get an island with area = 3.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> grid = [[1,1],[1,0]]
+<strong>Output:</strong> 4
+<strong>Explanation: </strong>Change the 0 to 1 and make the island bigger, only one island with area = 4.</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> grid = [[1,1],[1,1]]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> Can&#39;t change any 0 to 1, only one island with area = 4.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>n == grid.length</code></li>
+	<li><code>n == grid[i].length</code></li>
+	<li><code>1 &lt;= n &lt;= 500</code></li>
+	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
+</ul>

0854-making-a-large-island/solution.py

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+# Approach 2: Component Sizes
+
+# Time: O(n ^ 2)
+# Space: O(n ^ 2)
+
+class Solution:
+    def largestIsland(self, grid: List[List[int]]) -> int:
+        n = len(grid)
+
+        def get_neighbors(r, c):
+            for nr, nc in ((r - 1, c), (r + 1, c), (r, c - 1), (r, c + 1)):
+                if 0 <= nr < n and 0 <= nc < n:
+                    yield nr, nc
+
+        def dfs(r, c, index):
+            ans = 1
+            grid[r][c] = index
+            for nr, nc in get_neighbors(r, c):
+                if grid[nr][nc] == 1:
+                    ans += dfs(nr, nc, index)
+            return ans
+
+        area = {}
+        index = 2
+        for r in range(n):
+            for c in range(n):
+                if grid[r][c] == 1:
+                    area[index] = dfs(r, c, index)
+                    index += 1
+
+        ans = max(area.values() or [0])
+        for r in range(n):
+            for c in range(n):
+                if grid[r][c] == 0:
+                    seen = {grid[nr][nc] for nr, nc in get_neighbors(r, c) if grid[nr][nc] > 1}
+                    ans = max(ans, 1 + sum(area[i] for i in seen))
+
+        return ans
+