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 matrixm
andn
. -
int[] flip()
Returns a random index[i, j]
of the matrix wherematrix[i][j] == 0
and flips it to1
. -
void reset()
Resets all the values of the matrix to be0
.
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 toflip
andreset
.
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()
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