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LeetCode Problem 822: Card Flipping Game

On each card of a deck, written a unique integer card[i] is written on its front and a unique integer on its back. You can flip the card any number of times you want, but you cannot flip any card that would result in both sides showing the same integer.

A card is flipped if it shows the integer b on its front but not on its back, and b is less than any integer that would appear on both the front and the back of any card.

Write a solution that returns the minimum such b. If no such number exists, return 0.

Clarifying Questions

  1. What are the constraints on the number of cards and the integers on the cards?
    • We assume constraints consistent with typical competitive programming problems, such as 1 <= len(front), len(back) <= 1000 and 1 <= card[i] <= 2000.
  2. Can the integers on the front and back of a single card be the same?
    • Yes, they can be, but we will use the constraint from the problem.
  3. Should the result be for the minimum possible integer on the front that can be flipped to show an integer not present on the back of the same card?
    • Yes, that is the essence of the problem: find the minimum such integer b.

Strategy

  1. Identify and Eliminate Duplicates:
    • Determine which integers are present on both the front and the back of the same card; these cannot create a suitable b.
  2. Candidate Identification:
    • Collect all integers from the front of the cards which do not have the same integer on the back of the same card.
  3. Finding Minimum:
    • From the collected candidates, find the smallest integer.

Code

Here’s a Python function that implements the described strategy:

def flipgame(fronts, backs):
    # Identify numbers that appear on both sides of the same card
    same_numbers = {fronts[i] for i in range(len(fronts)) if fronts[i] == backs[i]}
    
    # Initialize result as infinity for comparison purposes
    result = float('inf')
    
    # Iterate through fronts and backs to find the minimum candidate
    for num in fronts + backs:
        if num not in same_numbers:
            result = min(result, num)
    
    # If result was updated, return it, otherwise return 0
    return result if result != float('inf') else 0

# Example usage
fronts = [1, 2, 4, 4, 7]
backs = [1, 3, 4, 1, 3]
print(flipgame(fronts, backs))  # Output: 2

Time Complexity

Overall, the time complexity is O(n) for this approach, which is efficient given the problem constraints.

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