# 519. Random Flip Matrix

## 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

• 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();
*/

• 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()