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519\. Random Flip Matrix
Medium
There is an `m x n` binary grid `matrix` with all the values set `0` initially. Design an algorithm to randomly pick an index `(i, j)` where `matrix[i][j] == 0` and flips it to `1`. All the indices `(i, j)` where `matrix[i][j] == 0` 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 `Solution` class:
* `Solution(int m, int n)` Initializes the object with the size of the binary matrix `m` and `n`.
* `int[] flip()` Returns a random index `[i, j]` of the matrix where `matrix[i][j] == 0` and flips it to `1`.
* `void reset()` Resets all the values of the matrix to be `0`.
**Example 1:**
**Input** ["Solution", "flip", "flip", "flip", "reset", "flip"] [[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.
**Constraints:**
* 1 <= m, n <= 104
* There will be at least one free cell for each call to `flip`.
* At most `1000` calls will be made to `flip` and `reset`.
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