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
.
class Solution:
def __init__(self, m: int, n: int):
self.m = m
self.n = n
self.total = m * n
self.mp = {}
def flip(self) -> List[int]:
self.total -= 1
x = random.randint(0, self.total)
idx = self.mp.get(x, x)
self.mp[x] = self.mp.get(self.total, self.total)
return [idx // self.n, idx % self.n]
def reset(self) -> None:
self.total = self.m * self.n
self.mp.clear()
# Your Solution object will be instantiated and called as such:
# obj = Solution(m, n)
# param_1 = obj.flip()
# obj.reset()
class Solution {
private int m;
private int n;
private int total;
private Random rand = new Random();
private Map<Integer, Integer> mp = new HashMap<>();
public Solution(int m, int n) {
this.m = m;
this.n = n;
this.total = m * n;
}
public int[] flip() {
int x = rand.nextInt(total--);
int idx = mp.getOrDefault(x, x);
mp.put(x, mp.getOrDefault(total, total));
return new int[] {idx / n, idx % n};
}
public void reset() {
total = m * n;
mp.clear();
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(m, n);
* int[] param_1 = obj.flip();
* obj.reset();
*/