How do you solve Random Flip Matrix on LeetCode?

Random Flip Matrix (LeetCode #519) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(n) space complexity. Key insight: Using a HashMap allows for efficient tracking of flipped indices.

What is the brute force solution for Random Flip Matrix?

The brute force approach for Random Flip Matrix is Brute Force, with O(n²) time complexity and O(n²) space complexity. In this approach, we will iterate through the entire matrix to find all the indices where the value is 0. Then, we will randomly select one of these indices to flip it to 1. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Random Flip Matrix?

Random Flip Matrix uses the following concepts: Hash Table, Math, Reservoir Sampling, Randomized. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Random Flip Matrix?

Explain your thought process clearly while coding. Discuss trade-offs between different approaches. Practice coding on a whiteboard or online platform to simulate interview conditions.

What are common mistakes when solving Random Flip Matrix?

Not resetting the matrix properly after flips. Failing to account for the mapping of flipped indices.

#519

Random Flip Matrix

Medium

Hash TableMathReservoir SamplingRandomizedHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(n²)	O(n)

💡

Intuition

Time O(1)Space O(n)

This approach uses a mapping technique to efficiently track the available indices without needing to maintain a separate list. We can use a HashMap to map the flipped indices to the last available index, allowing us to select random indices without iterating through the entire matrix.

⚙️

Algorithm

4 steps

1Step 1: Maintain a count of total available cells and a mapping of flipped indices to available indices.
2Step 2: When flipping, randomly select an index and swap it with the last available index in the mapping.
3Step 3: Update the mapping and decrement the available count.
4Step 4: On reset, clear the mapping and reset the count.

solution.py20 lines

1import random
2
3class Solution:
4    def __init__(self, m: int, n: int):
5        self.m = m
6        self.n = n
7        self.total = m * n
8        self.flipped = {}
9
10    def flip(self):
11        idx = random.randint(0, self.total - 1)
12        if idx in self.flipped:
13            idx = self.flipped[idx]
14        self.flipped[idx] = self.flipped.get(self.total - 1, self.total - 1)
15        self.total -= 1
16        return [idx // self.n, idx % self.n]
17
18    def reset(self):
19        self.flipped.clear()
20        self.total = self.m * self.n

ℹ

Complexity note: The time complexity is O(1) for the flip operation because we are only performing a constant amount of work regardless of the size of the matrix. The space complexity is O(n) for storing the flipped indices in the HashMap.

1Using a HashMap allows for efficient tracking of flipped indices.
2Random selection without needing to iterate through the entire matrix reduces time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.