519. Random Flip Matrix

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 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 <= 10^4

• There will be at least one free cell for each call to flip.

• At most 1000 calls will be made to flip and reset.

Solution (Java)

class Solution {
    private final int cols;
    private final int total;
    private final Random rand;
    private final Set<Integer> available;

    public Solution(int nRows, int nCols) {
        this.cols = nCols;
        this.rand = new Random();
        this.available = new HashSet<>();
        this.total = nRows * this.cols;
    }

    public int[] flip() {
        int x = rand.nextInt(this.total);
        while (available.contains(x)) {
            x = rand.nextInt(this.total);
        }

        this.available.add(x);
        return new int[] {x / this.cols, x % this.cols};
    }

    public void reset() {
        this.available.clear();
    }
}

/**
 * Your Solution object will be instantiated and called as such:
 * Solution obj = new Solution(m, n);
 * int[] param_1 = obj.flip();
 * obj.reset();
 */

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).