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 <= 10000
There will be at least one free cell for each call to flip.
At most 1000 calls will be made to flip and reset.

Problem Summary #

You are given an m x n binary matrix matrix, and all values are initialized to 0. Design an algorithm to randomly select an index (i, j) such that matrix[i][j] == 0, and change its value to 1. All indices (i, j) that satisfy matrix[i][j] == 0 should have an equal probability of being selected.

Minimize the number of calls to the built-in random function as much as possible, and optimize time and space complexity.

Implement the Solution class:

Solution(int m, int n) Initializes the object using the size m and n of the binary matrix
int[] flip() Returns a random index [i, j] such that matrix[i][j] == 0, and changes the value in the corresponding cell to 1
void reset() Resets all values in the matrix to 0

Solution Approach #

Convert the two-dimensional matrix into one dimension using a hash table. Each time, randomly select any element in the one-dimensional representation, then swap it with the last element, and reduce the total number of one-dimensional elements by one
The default mapping in the hash table is x->x, and then store the special key-value pairs that do not satisfy this mapping in the hash table

Code #

package leetcode

import "math/rand"

type Solution struct {
	r     int
	c     int
	total int
	mp    map[int]int
}

func Constructor(m int, n int) Solution {
	return Solution{
		r:     m,
		c:     n,
		total: m * n,
		mp:    map[int]int{},
	}
}

func (this *Solution) Flip() []int {
	k := rand.Intn(this.total)
	val := k
	if v, ok := this.mp[k]; ok {
		val = v
	}
	if _, ok := this.mp[this.total-1]; ok {
		this.mp[k] = this.mp[this.total-1]
	} else {
		this.mp[k] = this.total - 1
	}
	delete(this.mp, this.total - 1)
	this.total--
	newR, newC := val/this.c, val%this.c
	return []int{newR, newC}
}

func (this *Solution) Reset() {
	this.total = this.r * this.c
	this.mp = map[int]int{}
}