How do you solve Card Flipping Game on LeetCode?

Card Flipping Game (LeetCode #822) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to track numbers efficiently using sets can greatly reduce complexity.

What is the brute force solution for Card Flipping Game?

The brute force approach for Card Flipping Game is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try flipping each card in every possible combination and check which integers are facing down but not up. This approach is straightforward but inefficient as it checks all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Card Flipping Game?

Card Flipping Game uses the following concepts: Array, Hash Table. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Card Flipping Game?

Always clarify the problem statement and constraints before diving into coding. Think about the data structures that can help you solve the problem efficiently. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving Card Flipping Game?

Not considering edge cases where all numbers might be facing up. Overcomplicating the problem by trying to generate all combinations instead of using sets.

#822

Card Flipping Game

Medium

ArrayHash TableHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

Instead of checking all combinations, we can use sets to track numbers that are facing up and down, allowing us to find the minimum good integer efficiently.

⚙️

Algorithm

3 steps

1Step 1: Create a set for numbers facing up and another for numbers facing down.
2Step 2: Iterate through the cards, adding the front numbers to the up set and the back numbers to the down set.
3Step 3: For each number in the down set, check if it is not in the up set and track the minimum.

solution.py11 lines

1# Full working Python code
2
3def minGoodInteger(fronts, backs):
4    up = set(fronts)
5    down = set(backs)
6    for num in fronts:
7        down.add(num)
8    for num in backs:
9        up.add(num)
10    good_integers = [num for num in down if num not in up]
11    return min(good_integers) if good_integers else 0

ℹ

Complexity note: We only traverse the arrays a couple of times, leading to linear time complexity. The space complexity is due to the sets storing up and down numbers.

1Understanding how to track numbers efficiently using sets can greatly reduce complexity.
2Recognizing that we only need to check down numbers against up numbers helps streamline the solution.

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