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Copy file name to clipboardexpand all lines: solution/0700-0799/0735.Asteroid Collision/README_EN.md
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<p>We are given an array <code>asteroids</code> of integers representing asteroids in a row.</p>
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<p>We are given an array <code>asteroids</code> of integers representing asteroids in a row. The indices of the asteriod in the array represent their relative position in space.</p>
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<p>For each asteroid, the absolute value represents its size, and the sign represents its direction (positive meaning right, negative meaning left). Each asteroid moves at the same speed.</p>
# [3391. Design a 3D Binary Matrix with Efficient Layer Tracking 🔒](https://leetcode.cn/problems/design-a-3d-binary-matrix-with-efficient-layer-tracking)
<p>You are given a <code>n x n x n</code> <strong>binary</strong> 3D array <code>matrix</code>.</p>
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<p>给定一个 <code>n x n x n</code> 的 <strong>二进制 </strong>三维数组 <code>matrix</code>。</p>
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<p>Implement the <code>matrix3D</code> class:</p>
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<p>实现 <code>Matrix3D</code> 类：</p>
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<ul>
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<li><code>matrix3D(int n)</code> Initializes the object with the 3D binary array <code>matrix</code>, where <strong>all</strong> elements are initially set to 0.</li>
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<li><code>void setCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 1.</li>
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<li><code>void unsetCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 0.</li>
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<li><code>int largestMatrix()</code> Returns the index <code>x</code> where <code>matrix[x]</code> contains the most number of 1's. If there are multiple such indices, return the <strong>largest</strong><code>x</code>.</li>
matrix3D matrix3D = new matrix3D(3); // Initializes a <code>3 x 3 x 3</code> 3D array <code>matrix</code>, filled with all 0's.<br />
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matrix3D.setCell(0, 0, 0); // Sets <code>matrix[0][0][0]</code> to 1.<br />
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matrix3D.largestMatrix(); // Returns 0. <code>matrix[0]</code> has the most number of 1's.<br />
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matrix3D.setCell(1, 1, 2); // Sets <code>matrix[1][1][2]</code> to 1.<br />
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matrix3D.largestMatrix(); // Returns 1. <code>matrix[0]</code> and <code>matrix[1]</code> tie with the most number of 1's, but index 1 is bigger.<br />
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matrix3D.setCell(0, 0, 1); // Sets <code>matrix[0][0][1]</code> to 1.<br />
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matrix3D.largestMatrix(); // Returns 0. <code>matrix[0]</code> has the most number of 1's.</div>
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<p><strong>解释：</strong></p>
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Matrix3D matrix3D = new Matrix3D(3); // 初始化一个 <code>3 x 3 x 3</code> 的三维数组 <code>matrix</code>，用全 0 填充。<br />
<p>You are given a <code>n x n x n</code> <strong>binary</strong> 3D array <code>matrix</code>.</p>
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<p>Implement the <code>matrix3D</code> class:</p>
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<p>Implement the <code>Matrix3D</code> class:</p>
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<ul>
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<li><code>matrix3D(int n)</code> Initializes the object with the 3D binary array <code>matrix</code>, where <strong>all</strong> elements are initially set to 0.</li>
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<li><code>Matrix3D(int n)</code> Initializes the object with the 3D binary array <code>matrix</code>, where <strong>all</strong> elements are initially set to 0.</li>
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<li><code>void setCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 1.</li>
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<li><code>void unsetCell(int x, int y, int z)</code> Sets the value at <code>matrix[x][y][z]</code> to 0.</li>
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<li><code>int largestMatrix()</code> Returns the index <code>x</code> where <code>matrix[x]</code> contains the most number of 1's. If there are multiple such indices, return the <strong>largest</strong> <code>x</code>.</li>
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