You are given the number of rows n_rows
and number of columns n_cols
of a 2D binary matrix where all values are initially 0. Write a function flip
which chooses a 0 value uniformly at random, changes it to 1, and then returns the position [row.id, col.id]
of that value. Also, write a function reset
which sets all values back to 0. Try to minimize the number of calls to system's Math.random() and optimize the time and space complexity.
Note:
1 <= n_rows, n_cols <= 10000
0 <= row.id < n_rows
and0 <= col.id < n_cols
flip
will not be called when the matrix has no 0 values left.- the total number of calls to
flip
andreset
will not exceed 1000.
Example 1:
Input: ["Solution","flip","flip","flip","flip"] [[2,3],[],[],[],[]] Output: [null,[0,1],[1,2],[1,0],[1,1]]
Example 2:
Input: ["Solution","flip","flip","reset","flip"] [[1,2],[],[],[],[]] Output: [null,[0,0],[0,1],null,[0,0]]
Explanation of Input Syntax:
The input is two lists: the subroutines called and their arguments. Solution
's constructor has two arguments, n_rows
and n_cols
. flip
and reset
have no arguments. Arguments are always wrapped with a list, even if there aren't any.
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();
*/