- Difficulty: Medium
- Tags: LeetCode, Medium, Reservoir Sampling, Hash Table, Math, Randomized, leetcode-519, O(n)
Problem
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
.