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.

SOLUTION:

from random import randint

class Solution:

    def __init__(self, m: int, n: int):
        self.m = m
        self.n = n
        self.ones = set()

    def flip(self) -> List[int]:
        i, j = randint(0, self.m - 1), randint(0, self.n - 1)
        while (i, j) in self.ones:
            i, j = randint(0, self.m - 1), randint(0, self.n - 1)
        self.ones.add((i, j))
        return [i, j]

    def reset(self) -> None:
        self.ones.clear()

# Your Solution object will be instantiated and called as such:
# obj = Solution(m, n)
# param_1 = obj.flip()
# obj.reset()