LeetCode-in-Java

519. Random Flip Matrix

Medium

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.

Solution

import java.util.HashSet;
import java.util.Random;
import java.util.Set;

@SuppressWarnings("java:S2245")
public 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();
    }
}

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