[LeetCode Sync] Runtime - 255 ms (84.21%), Memory - 17.6 MB (100.00%)

Author: NamVr

2061-painting-a-grid-with-three-different-colors/README.md

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+<p>You are given two integers <code>m</code> and <code>n</code>. Consider an <code>m x n</code> grid where each cell is initially white. You can paint each cell <strong>red</strong>, <strong>green</strong>, or <strong>blue</strong>. All cells <strong>must</strong> be painted.</p>
+
+<p>Return<em> the number of ways to color the grid with <strong>no two adjacent cells having the same color</strong></em>. Since the answer can be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/06/22/colorthegrid.png" style="width: 200px; height: 50px;" />
+<pre>
+<strong>Input:</strong> m = 1, n = 1
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> The three possible colorings are shown in the image above.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/06/22/copy-of-colorthegrid.png" style="width: 321px; height: 121px;" />
+<pre>
+<strong>Input:</strong> m = 1, n = 2
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> The six possible colorings are shown in the image above.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> m = 5, n = 5
+<strong>Output:</strong> 580986
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= m &lt;= 5</code></li>
+	<li><code>1 &lt;= n &lt;= 1000</code></li>
+</ul>

2061-painting-a-grid-with-three-different-colors/solution.py

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+class Solution:
+    def colorTheGrid(self, m: int, n: int) -> int:
+        mod = 10**9 + 7
+        # Hash mapping stores all valid coloration schemes for a single row that meet the requirements
+        # The key represents mask, and the value represents the ternary string of mask (stored as a list)
+        valid = dict()
+
+        # Enumerate masks that meet the requirements within the range [0, 3^m)
+        for mask in range(3**m):
+            color = list()
+            mm = mask
+            for i in range(m):
+                color.append(mm % 3)
+                mm //= 3
+            if any(color[i] == color[i + 1] for i in range(m - 1)):
+                continue
+            valid[mask] = color
+
+        # Preprocess all (mask1, mask2) binary tuples, satisfying mask1 and mask2 When adjacent rows, the colors of the two cells in the same column are different
+        adjacent = defaultdict(list)
+        for mask1, color1 in valid.items():
+            for mask2, color2 in valid.items():
+                if not any(x == y for x, y in zip(color1, color2)):
+                    adjacent[mask1].append(mask2)
+
+        f = [int(mask in valid) for mask in range(3**m)]
+        for i in range(1, n):
+            g = [0] * (3**m)
+            for mask2 in valid.keys():
+                for mask1 in adjacent[mask2]:
+                    g[mask2] += f[mask1]
+                    if g[mask2] >= mod:
+                        g[mask2] -= mod
+            f = g
+
+        return sum(f) % mod