Formatted question description: https://leetcode.ca/all/519.html

# 519. Random Flip Matrix (Medium)

You are given the number of rows n_rows and number of columns n_cols of a 2D binary matrix where all values are initially 0. Write a function flip which chooses a 0 value uniformly at random, changes it to 1, and then returns the position [row.id, col.id] of that value. Also, write a function reset which sets all values back to 0. Try to minimize the number of calls to system's Math.random() and optimize the time and space complexity.

Note:

1. 1 <= n_rows, n_cols <= 10000
2. 0 <= row.id < n_rows and 0 <= col.id < n_cols
3. flip will not be called when the matrix has no 0 values left.
4. the total number of calls to flip and reset will not exceed 1000.

Example 1:

Input:
["Solution","flip","flip","flip","flip"]
[[2,3],[],[],[],[]]
Output: [null,[0,1],[1,2],[1,0],[1,1]]


Example 2:

Input:
["Solution","flip","flip","reset","flip"]
[[1,2],[],[],[],[]]
Output: [null,[0,0],[0,1],null,[0,0]]

Explanation of Input Syntax:

The input is two lists: the subroutines called and their arguments. Solution's constructor has two arguments, n_rows and n_colsflip and reset have no arguments. Arguments are always wrapped with a list, even if there aren't any.

Companies:

Related Topics:
Random

## Solution 1.

// OJ: https://leetcode.com/problems/random-flip-matrix/

// Time:
//   Solution: O(MN)
//   flip: O(1)
//   reset: O(1)
// Space: O(MN)
class Solution {
private:
vector<int> v;
int size, M, N;
public:
Solution(int M, int N): M(M), N(N), size(M * N), v(M * N) {
for (int i = 0; i < size; ++i) v[i] = i;
srand(time(NULL));
}

vector<int> flip() {
swap(v[rand() % size], v[size - 1]);
--size;
return { v[size] / N, v[size] % N };
}

void reset() {
size = M * N;
}
};


## Solution 2.

Same idea as Solution 1, but use unordered_map to save space, and time if we only flip a small set of the points.

// OJ: https://leetcode.com/problems/random-flip-matrix/

// Time:
//   Solution: O(1)
//   flip: O(1)
//   reset: O(MN) in worst case
// Space: O(MN) in worst case
// Ref: https://leetcode.com/problems/random-flip-matrix/solution/
class Solution {
private:
unordered_map<int, int> m;
int size, M, N;
//c++11 random integer generation
mt19937 rng{random_device{}()};
//uniform random integer in [0, bound]
int randint(int bound) {
uniform_int_distribution<int> uni(0, bound);
return uni(rng);
}
public:
Solution(int M, int N): M(M), N(N), size(M * N) {}

vector<int> flip() {
int r = randint(--size);
int x = m.count(r) ? m[r] : r;
m[r] = m.count(size) ? m[size] : size;
return { x / N, x % N };
}

void reset() {
m.clear();
size = M * N;
}
};


Java java class Solution { int n_rows; int n_cols; int remaining; Map<Integer, Integer> map; Random random;

public Solution(int n_rows, int n_cols) {
this.n_rows = n_rows;
this.n_cols = n_cols;
remaining = n_rows * n_cols;
map = new HashMap<Integer, Integer>();
random = new Random();
}

public int[] flip() {
int randNum = random.nextInt(remaining);
remaining--;
int index = map.getOrDefault(randNum, randNum);
int value = map.getOrDefault(remaining, remaining);
map.put(randNum, value);
int[] rowColumn = {index / n_cols, index % n_cols};
return rowColumn;
}

public void reset() {
map.clear();
remaining = n_rows * n_cols;
} }


/**

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