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You are given the number of rows m and the number of columns n of a 2D binary matrix where all values are initially 0. Implement the FlipMatrix class:

• FlipMatrix(m: int, n: int): Initializes the object with the size of the binary matrix m x n.
• flip() -> List[int]: Returns a random position [row, col] of a 0 in the matrix. The 0 at this position will now be set to 1. This function should ensure that each position is flipped at most once.
• reset() -> None: Resets all values of the matrix back to 0.

You need to implement these functions such that each flip is done in O(1) time on average, and the memory used is O(m * n).

Clarifying Questions

1. Are the arguments always within the expected range?
   • Yes, you can assume that m and n are positive integers within reasonable limits.
2. How do we ensure randomness?
   • Randomness should be uniform across the matrix positions that haven’t been flipped yet.
3. Can we use extra space for keeping track of the flipped positions?
   • Yes, but the solution should still be efficient in terms of both time and space.

Strategy

1. Initialization:
   • We’ll treat the matrix as a flattened list of size m * n. This helps us to simplify the problem to a 1D array.
   • We will use a dictionary to map flipped positions to ensure constant time operations.
2. Flipping:
   • We’ll randomly pick an index from the available positions in the 1D array.
   • We keep track of which indices have been flipped by using a dictionary that maps from the picked position to the last available position in the array.
3. Resetting:
   • Simply reset the tracking dictionary and counter.

By storing only the necessary information to remap positions, we efficiently manage and maintain matrix flips.

Code

import random

class FlipMatrix:
    def __init__(self, m: int, n: int):
        self.m = m
        self.n = n
        self.total = m * n
        self.flips_map = {}
        self.available = self.total

    def flip(self) -> List[int]:
        rand_index = random.randint(0, self.available - 1)
        self.available -= 1

        # Get the mapped position or default to rand_index
        flip_pos = self.flips_map.get(rand_index, rand_index)

        # Update the map to reflect this position has been flipped
        self.flips_map[rand_index] = self.flips_map.get(self.available, self.available)

        return [flip_pos // self.n, flip_pos % self.n]

    def reset(self) -> None:
        self.flips_map = {}
        self.available = self.total

# Example usage:
# obj = FlipMatrix(m, n)
# param_1 = obj.flip()
# obj.reset()

Time Complexity

• Initialization: O(1), setting up initial variables.
• flip(): O(1) on average, due to constant-time updates and dictionary access.
• reset(): O(1), resetting the dictionary and counter.

This ensures that the operations are efficient and scalable with respect to the matrix size.

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