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Kotlin-based LeetCode algorithm problem solutions, regularly updated

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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`.