519. Random Flip Matrix
Explanation
To solve this problem efficiently, we can use a combination of techniques. We can map the 2D grid to a 1D array to optimize memory usage. Then, we can use a map or a set to keep track of the available indices that can be flipped. When flipping an index, we need to ensure that it is removed from the available indices set and update the mapping accordingly. Resetting the matrix simply involves clearing the set of available indices.
The time complexity for each flip operation is O(1), as we are choosing an available index randomly. The space complexity is O(m*n) for storing the matrix and O(k) for storing the set of available indices, where k is the number of available indices.
import java.util.*;
class Solution {
private int rows, cols, total;
private Random rand;
private Map<Integer, Integer> map;
public Solution(int m, int n) {
rows = m;
cols = n;
total = m * n;
rand = new Random();
map = new HashMap<>();
}
public int[] flip() {
int r = rand.nextInt(total--);
int x = map.getOrDefault(r, r);
map.put(r, map.getOrDefault(total, total));
return new int[]{x / cols, x % cols};
}
public void reset() {
map.clear();
total = rows * cols;
}
}Code Editor (Testing phase)
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