LeetCode 822 Card Flipping Game - Medium

822. Card Flipping Game — Medium

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Problem

822. Card Flipping Game -- Medium

On a table are N cards, with a positive integer printed on the front and back of each card (possibly different).

We flip any number of cards, and after we choose one card.

If the number X on the back of the chosen card is not on the front of any card, then this number X is good.

What is the smallest number that is good? If no number is good, output 0.

Here, fronts[i] and backs[i] represent the number on the front and back of card i.

A flip swaps the front and back numbers, so the value on the front is now on the back and vice versa.

Example:

Input: fronts = [1,2,4,4,7], backs = [1,3,4,1,3] Output: 2 Explanation: If we flip the second card, the fronts are [1,3,4,4,7] and the backs are [1,2,4,1,3]. We choose the second card, which has number 2 on the back, and it isn't on the front of any card, so 2 is good.

Note:

1 <= fronts.length == backs.length <= 1000. 1 <= fronts[i] <= 2000. 1 <= backs[i] <= 2000.

Solution

class Solution:
    def flipgame(self, fronts: List[int], backs: List[int]) -> int:
        cantPick = set()
        for i in range(len(fronts)):
            if fronts[i] == backs[i]:
                cantPick.add(fronts[i])
                
        res = float('inf')
        for number in (fronts + backs):
            if number not in cantPick:
                res = min(res, number)
        
        return res if res != float('inf') else 0