519. Random Flip Matrix

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.


  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:

Output: [null,[0,1],[1,2],[1,0],[1,1]]

Example 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.

Rust Solution

use rand::prelude::*;
use std::collections::HashMap;

struct Solution {
    size: usize,
    indexes: HashMap<usize, usize>,
    rng: ThreadRng,
    rows: usize,
    cols: usize,

impl Solution {
    fn new(n_rows: i32, n_cols: i32) -> Self {
        let rows = n_rows as usize;
        let cols = n_cols as usize;
        let size = rows * cols;
        let indexes = HashMap::new();
        let rng = thread_rng();
        Solution {

    fn flip(&mut self) -> Vec<i32> {
        let r = self.rng.gen_range(0, self.size);
        let x = *self.indexes.entry(r).or_insert(r);
        self.size -= 1;
        let y = *self.indexes.entry(self.size).or_insert(self.size);
        *self.indexes.entry(r).or_default() = y;
        let col = x % self.cols;
        let row = x / self.cols;
        vec![row as i32, col as i32]

    fn reset(&mut self) {
        self.size = self.rows * self.cols;
        self.indexes = HashMap::new();

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