# 519. Random Flip Matrix LeetCode Solution | Easy Approach Share:

Random Flip Matrix 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`.

Time: O(n)
Space: O(n)

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