Card Flipping Game - Leetcode Solution

The Card Flipping Game problem on LeetCode can be solved using a brute force approach with a time complexity of O(n^2). The problem is as follows:

You are given a deck of n cards, each labeled from 0 to n - 1. You are asked to perform a series of operations on the deck of cards, as follows:

• Choose any card in the deck and flip it over (i.e. change its label from 0 to 1 or from 1 to 0).
• If any of the adjacent cards have the same label as the flipped card, then flip them over as well.

Your task is to find the minimum number of operations required such that all the cards have the same label.

Algorithm:

1. Initialize a variable, minFlips, to an integer value greater than or equal to n (the maximum number of flips required).

2. For each card in the deck, perform the following operations:

   a. Flip the chosen card.

   b. Check if all the cards have the same label. If yes, update the minFlips variable with the minimum number of flips required so far and go to step 3.

   c. For each adjacent card, check if it has the same label as the chosen card. If yes, flip the adjacent card and continue checking the rest of the adjacent cards.

3. Return the value of minFlips as the minimum number of flips required to get all the cards to have the same label.

Code:

class Solution:
    def flipgame(self, fronts: List[int], backs: List[int]) -> int:
        n = len(fronts)
        minFlips = n
        
        for i in range(n):
            card = fronts[i]
            if backs[i] == card:
                continue
            flips = 1
            for j in range(n):
                if i == j:
                    continue
                if fronts[j] == card or backs[j] == card:
                    flips += 1
                    fronts[j], backs[j] = backs[j], fronts[j]
            minFlips = min(minFlips, flips)
            fronts[i], backs[i] = backs[i], fronts[i]
            
            flips = 1
            for j in range(n):
                if i == j:
                    continue
                if fronts[j] == card or backs[j] == card:
                    flips += 1
                    fronts[j], backs[j] = backs[j], fronts[j]
            minFlips = min(minFlips, flips)
            fronts[i], backs[i] = backs[i], fronts[i]
            
        return minFlips if minFlips < n else 0

The code initializes the variable minFlips to n, the maximum number of flips required. For each card in the deck, it flips the card and checks if all the cards have the same label. If not, it checks each adjacent card and flips them if needed. After each operation, it checks the minimum number of flips required so far and updates the minFlips variable if the new number of flips is smaller. Finally, it returns the value of minFlips as the minimum number of flips required to get all the cards to have the same label.

1