Introduction

In this blog post, we will explore how to solve the LeetCode problem "519" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's the solution to the problem, focusing on efficiency and optimization:

• Key Idea: Instead of using a full matrix, we maintain a mapping (dictionary) that represents the available indices and their corresponding 'true' values. This avoids the space overhead of storing a large matrix. We treat the available indices as a contiguous range [0, total_cells), and when a cell is flipped, we swap it with the last available cell in the range and shrink the range. This ensures uniform random sampling.
• Optimization: We use a dictionary to store the mapping between the random index generated and the actual index. This approach avoids repeated calls to random number generator and reduces the time complexity.
• Complexity:
  • Runtime: O(1) for flip and reset operations.
  • Storage: O(K), where K is the number of flip operations performed (at most 1000).

Code

    import random

class Solution:
    def __init__(self, m: int, n: int):
        self.m = m
        self.n = n
        self.total_cells = m * n
        self.mapping = {}  # Maps index to its actual value after flips

    def flip(self) -> list[int]:
        # Generate a random index within the current range
        rand_index = random.randrange(self.total_cells)

        # Get the actual value at the random index. If it has been remapped,
        # use the remapped value; otherwise, it's just the index itself.
        actual_index = self.mapping.get(rand_index, rand_index)

        # Get the actual value of the last element in the range
        last_index = self.total_cells - 1
        last_actual_index = self.mapping.get(last_index, last_index)

        # Swap the random index with the last index. This means the random
        # index now points to what the last index used to point to.
        self.mapping[rand_index] = last_actual_index

        # Decrement the total number of cells, effectively removing the last
        # element from the range.
        self.total_cells -= 1

        # Convert the actual index to row and column
        row = actual_index // self.n
        col = actual_index % self.n

        return [row, col]

    def reset(self) -> None:
        self.total_cells = self.m * self.n
        self.mapping = {}  # Clear the mapping