LeetCode 519: Random Flip Matrix Solution in Python – A Step-by-Step Guide

Imagine you’re playing a game with a grid of zeros—like a giant tic-tac-toe board—and your task is to flip random cells to ones, keeping track of which ones you’ve flipped, all while ensuring each flip is unique. That’s the intriguing challenge of LeetCode 519: Random Flip Matrix, a medium-level problem that’s a fantastic way to practice randomization and data structures in Python. We’ll explore two solutions: the Best Solution, a hash map with index mapping that’s efficient and clever, and an Alternative Solution, an array-based tracking approach that’s simpler but less space-efficient. With detailed examples, clear code, and a friendly tone—especially for the hash map trick—this guide will help you flip that matrix, whether you’re new to coding or leveling up. Let’s roll the dice and start flipping!

What Is LeetCode 519: Random Flip Matrix?

In LeetCode 519: Random Flip Matrix, you’re given a binary matrix of size m x n initialized with all zeros. You need to implement a class with two methods: flip(), which randomly selects a zero cell, flips it to one, and returns its coordinates, and reset(), which sets all cells back to zero. The catch? Each call to flip() must pick a unique unflipped cell, and you must do this efficiently. For example, in a 2x3 matrix, you might flip (0,1), then (1,2), ensuring no repeats. This problem tests randomization and state management, related to LeetCode 384: Shuffle an Array.

Problem Statement

Input:

Constructor: m (rows), n (cols)—matrix dimensions.
flip(): No input.
reset(): No input.

Output:

Constructor: None.
flip(): List[int]—[row, col] of flipped cell.
reset(): None.

Rules: Randomly flip a 0 to 1; no repeats; reset to all 0s.

Constraints

1 <= m, n <= 10⁴
m * n <= 10⁵
At most 1000 calls to flip() and reset() combined.

Examples

Input:

Output:
Explanation: 2x3 matrix, flip random zeros, reset, flip again.

Understanding the Problem: Flipping Randomly

To solve LeetCode 519: Random Flip Matrix in Python, we need a class that efficiently picks random unflipped cells in an m x n matrix and flips them, ensuring uniqueness, then resets the state. A naive approach might use a 2D array and scan for zeros each time, but with up to 10⁵ cells, that’s too slow. We’ll explore:

Best Solution (Hash Map with Index Mapping): O(1) time per call, O(k) space (k = flipped cells)—fast and smart.
Alternative Solution (Array-Based Tracking): O(1) time per call, O(m * n) space—simple but memory-heavy.

Let’s dive into the hash map solution with a friendly breakdown!

Best Solution: Hash Map with Index Mapping

Why Hash Map Wins

The hash map with index mapping is the best for LeetCode 519 because it achieves O(1) time complexity for both flip() and reset() while using O(k) space (k = number of flips), which is minimal since k ≤ 1000. It avoids storing the full matrix by mapping flipped indices to unflipped ones, using randomization efficiently. It’s like playing a memory game with a clever shortcut!

How It Works

Think of this as a number-shuffling trick:

Step 1: Setup:

Total cells = m * n.
Use a hash map to track flipped indices.
Track remaining unflipped cells (total).

Step 2: Flip:

Pick a random index from 0 to total-1.
Map to row, col: row = idx // n, col = idx % n.
If index was flipped, use mapped value; else, use original.
Map this index to last unflipped index, decrement total.
Return coordinates.

Step 3: Reset:

Clear map, reset total.

Why It Works:

Mapping ensures unique flips.
Hash map avoids full matrix storage.

It’s like a random flip magician!

Step-by-Step Example

Example: m = 2, n = 3

Init:

total = 6, flipped = {}.
Matrix (virtual): [[0,0,0], [0,0,0]].

Flip 1:

randint(0, 5) = 2.
row = 2 // 3 = 0, col = 2 % 3 = 2.
flipped[2] = 5, total = 5.
Return [0, 2].

Flip 2:

randint(0, 4) = 1.
row = 1 // 3 = 0, col = 1 % 3 = 1.
flipped[1] = 4, total = 4.
Return [0, 1].

Reset:

flipped = {}, total = 6.

Flip 3:

randint(0, 5) = 0.
row = 0, col = 0.
flipped[0] = 5, total = 5.
Return [0, 0].

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

import random

class Solution:
    def __init__(self, m: int, n: int):
        # Step 1: Initialize dimensions and state
        self.m = m
        self.n = n
        self.total = m * n  # Total unflipped cells
        self.flipped = {}   # Map flipped idx to new idx

    def flip(self) -> List[int]:
        # Step 2: Pick random unflipped index
        idx = random.randint(0, self.total - 1)
        # Get actual index (mapped or original)
        actual = self.flipped.get(idx, idx)

        # Map this index to last unflipped
        self.flipped[idx] = self.flipped.get(self.total - 1, self.total - 1)
        self.total -= 1

        # Convert to row, col
        row = actual // self.n
        col = actual % self.n
        return [row, col]

    def reset(self) -> None:
        # Step 3: Reset state
        self.flipped = {}
        self.total = self.m * self.n

Lines 5-8: Constructor:

Store m, n, compute total cells, init empty map.

Lines 11-20: flip():

Random index from remaining.
Get mapped or original index.
Map to last unflipped, decrement total.
Convert to coordinates.

Lines 23-25: reset():

Clear map, restore total.

Time Complexity: O(1) per call—random and hash ops.
Space Complexity: O(k)—map stores flipped cells.

It’s like a flip-tracking maestro!

Alternative Solution: Array-Based Tracking

Why an Alternative Approach?

The array-based solution uses a 2D list to explicitly track the matrix state, flipping zeros as needed. It’s O(1) time per call but O(m * n) space—simple but memory-intensive. Great for visualizing the process!

How It Works

Picture this as a matrix game board:

Step 1: Init m x n matrix of zeros.
Step 2: flip()—Pick random zero, flip to 1, track count.
Step 3: reset()—Set all back to 0, reset count.

It’s like a physical flip board!

Step-by-Step Example

Example: m = 2, n = 2

Init:

matrix = [[0,0], [0,0]], flipped = 0.

Flip 1:

Random (1,0), matrix[1][0] = 1, flipped = 1.
Return [1, 0].

Flip 2:

Random (0,1), matrix[0][1] = 1, flipped = 2.
Return [0, 1].

Reset:

matrix = [[0,0], [0,0]], flipped = 0.

Code for Array Approach

import random

class Solution:
    def __init__(self, m: int, n: int):
        self.m = m
        self.n = n
        self.matrix = [[0] * n for _ in range(m)]
        self.flipped = 0

    def flip(self) -> List[int]:
        while True:
            row = random.randint(0, self.m - 1)
            col = random.randint(0, self.n - 1)
            if self.matrix[row][col] == 0 and self.flipped < self.m * self.n:
                self.matrix[row][col] = 1
                self.flipped += 1
                return [row, col]

    def reset(self) -> None:
        self.matrix = [[0] * self.n for _ in range(self.m)]
        self.flipped = 0

Time Complexity: O(1) amortized—may loop to find zero.
Space Complexity: O(m * n)—full matrix.

It’s a matrix-flipping tracker!

Comparing the Two Solutions

Hash Map (Best):

Pros: O(1), O(k) space, efficient.
Cons: Mapping logic.

Array (Alternative):

Pros: O(1), simple.
Cons: O(m * n) space, slower flips.

Hash map wins for efficiency!

Additional Examples and Edge Cases

1x1: Flip [0,0], reset.
2x1: Flip [0,0], [1,0].
1x3: Flip [0,1], [0,0].

Hash map handles them all!

Complexity Recap

Hash Map: Time O(1), Space O(k).
Array: Time O(1) amortized, Space O(m * n).

Hash map’s the space-saving star!

Key Takeaways

Hash Map: Clever mapping—learn at Python Basics!
Array: Simple but costly.
Randomization: Fun with indices.
Python: Dicts or lists, your choice!

Final Thoughts: Flip That Matrix!

LeetCode 519: Random Flip Matrix in Python is a clever randomization challenge. The hash map solution is your fast track, while the array approach keeps it basic. Want more? Try LeetCode 384: Shuffle an Array or LeetCode 398: Random Pick Index. Ready to flip? Head to Solve LeetCode 519 on LeetCode and randomize that matrix today!