leetcode/solution/0500-0599/0519.Random Flip Matrix

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0500-0599/0519.Random%20Flip%20Matrix/README_EN.md	Reservoir Sampling Hash Table Math Randomized

519. Random Flip Matrix

中文文档

Description

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 matrix m and n.
• int[] flip() Returns a random index [i, j] of the matrix where matrix[i][j] == 0 and flips it to 1.
• void reset() Resets all the values of the matrix to be 0.

 

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 to flip and reset.

Solutions

Solution 1

Python3

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()

Java

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();
 */