random-flip-matrix

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Algorithms

There is an <code>m x n</code> binary grid <code>matrix</code> with all the values set <code>0</code> initially. Design an algorithm to randomly pick an index <code>(i, j)</code> where <code>matrix[i][j] == 0</code> and flips it to <code>1</code>. All the indices <code>(i, j)</code> where <code>matrix[i][j] == 0</code> should be equally likely to be returned.

Optimize your algorithm to minimize the number of calls made to the built-in random function of your language and optimize the time and space complexity.

Implement the <code>Solution</code> class:

<ul>
	<li><code>Solution(int m, int n)</code> Initializes the object with the size of the binary matrix <code>m</code> and <code>n</code>.</li>
	<li><code>int[] flip()</code> Returns a random index <code>[i, j]</code> of the matrix where <code>matrix[i][j] == 0</code> and flips it to <code>1</code>.</li>
	<li><code>void reset()</code> Resets all the values of the matrix to be <code>0</code>.</li>
</ul>

&nbsp;
Example 1:

<pre>
Input
[&quot;Solution&quot;, &quot;flip&quot;, &quot;flip&quot;, &quot;flip&quot;, &quot;reset&quot;, &quot;flip&quot;]
[[3, 1], [], [], [], [], []]
Output
[null, [1, 0], [2, 0], [0, 0], null, [2, 0]]

Explanation
Solution solution = new Solution(3, 1);
solution.flip(); // return [1, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.
solution.flip(); // return [2, 0], Since [1,0] was returned, [2,0] and [0,0]
solution.flip(); // return [0, 0], Based on the previously returned indices, only [0,0] can be returned.
solution.reset(); // All the values are reset to 0 and can be returned.
solution.flip(); // return [2, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= m, n &lt;= 104</code></li>
	<li>There will be at least one free cell for each call to <code>flip</code>.</li>
	<li>At most <code>1000</code> calls will be made to <code>flip</code> and <code>reset</code>.</li>
</ul>

Random Flip Matrix

Reservoir Sampling

Hash Table

Math

Randomized

给你一个 <code>m x n</code> 的二元矩阵 <code>matrix</code> ，且所有值被初始化为 <code>0</code> 。请你设计一个算法，随机选取一个满足&nbsp;<code>matrix[i][j] == 0</code> 的下标&nbsp;<code>(i, j)</code> ，并将它的值变为 <code>1</code> 。所有满足 <code>matrix[i][j] == 0</code> 的下标 <code>(i, j)</code> 被选取的概率应当均等。

尽量最少调用内置的随机函数，并且优化时间和空间复杂度。

实现 <code>Solution</code> 类：

<ul>
	<li><code>Solution(int m, int n)</code> 使用二元矩阵的大小 <code>m</code> 和 <code>n</code> 初始化该对象</li>
	<li><code>int[] flip()</code> 返回一个满足&nbsp;<code>matrix[i][j] == 0</code> 的随机下标 <code>[i, j]</code> ，并将其对应格子中的值变为 <code>1</code></li>
	<li><code>void reset()</code> 将矩阵中所有的值重置为 <code>0</code></li>
</ul>

&nbsp;

示例：

<pre>
输入
["Solution", "flip", "flip", "flip", "reset", "flip"]
[[3, 1], [], [], [], [], []]
输出
[null, [1, 0], [2, 0], [0, 0], null, [2, 0]]

解释
Solution solution = new Solution(3, 1);
solution.flip(); // 返回 [1, 0]，此时返回 [0,0]、[1,0] 和 [2,0] 的概率应当相同
solution.flip(); // 返回 [2, 0]，因为 [1,0] 已经返回过了，此时返回 [2,0] 和 [0,0] 的概率应当相同
solution.flip(); // 返回 [0, 0]，根据前面已经返回过的下标，此时只能返回 [0,0]
solution.reset(); // 所有值都重置为 0 ，并可以再次选择下标返回
solution.flip(); // 返回 [2, 0]，此时返回 [0,0]、[1,0] 和 [2,0] 的概率应当相同</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= m, n &lt;= 104</code></li>
	<li>每次调用<code>flip</code> 时，矩阵中至少存在一个值为 0 的格子。</li>
	<li>最多调用 <code>1000</code> 次 <code>flip</code> 和 <code>reset</code> 方法。</li>
</ul>