Card Flipping Game

Medium

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Topics:

Arrays

You are given two 0-indexed integer arrays fronts and backs of length n, where the i^th card has the positive integer fronts[i] printed on the front and backs[i] printed on the back. Initially, each card is placed on a table such that the front number is facing up and the other is facing down. You may flip over any number of cards (possibly zero).

After flipping the cards, an integer is considered good if it is facing down on some card and not facing up on any card.

Return the minimum possible good integer after flipping the cards. If there are no good integers, return 0.

Example 1:

Input: fronts = [1,2,4,4,7], backs = [1,3,4,1,3]
Output: 2
Explanation:
If we flip the second card, the face up numbers are [1,3,4,4,7] and the face down are [1,2,4,1,3].
2 is the minimum good integer as it appears facing down but not facing up.
It can be shown that 2 is the minimum possible good integer obtainable after flipping some cards.

Example 2:

Input: fronts = [1], backs = [1]
Output: 0
Explanation:
There are no good integers no matter how we flip the cards, so we return 0.

Constraints:

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

Try coding it in LeetCode

Card Flipping Game

Medium

Asked by:

5 views

Topics:

Arrays

You are given two 0-indexed integer arrays fronts and backs of length n, where the i^th card has the positive integer fronts[i] printed on the front and backs[i] printed on the back. Initially, each card is placed on a table such that the front number is facing up and the other is facing down. You may flip over any number of cards (possibly zero).

After flipping the cards, an integer is considered good if it is facing down on some card and not facing up on any card.

Return the minimum possible good integer after flipping the cards. If there are no good integers, return 0.

Example 1:

Input: fronts = [1,2,4,4,7], backs = [1,3,4,1,3]
Output: 2
Explanation:
If we flip the second card, the face up numbers are [1,3,4,4,7] and the face down are [1,2,4,1,3].
2 is the minimum good integer as it appears facing down but not facing up.
It can be shown that 2 is the minimum possible good integer obtainable after flipping some cards.

Example 2:

Input: fronts = [1], backs = [1]
Output: 0
Explanation:
There are no good integers no matter how we flip the cards, so we return 0.

Constraints:

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values on the cards, and can there be negative or zero values?
Can the array of cards be empty or contain only one card?
If there are multiple pairs of cards that satisfy the condition, should I return any one of them, or is there a specific pair I should prioritize?
What should I return if no such card flipping is possible to achieve the target?
Are there duplicate values on the cards, and if so, how should I handle them in determining the outcome?

Brute Force Solution

Approach

Imagine you have a deck of cards, some face up and some face down. We want to flip some cards so that all the cards show the same side. The brute force approach is to try flipping every possible combination of cards.

Here's how the algorithm would work step-by-step:

First, consider flipping no cards at all, and check if all cards are showing the same side.
Next, consider flipping only the first card, and check if all cards are showing the same side.
Then, consider flipping only the second card, and check if all cards are showing the same side.
Continue this pattern, checking the result of flipping each card individually.
Now, consider flipping the first and second cards together, and check if all cards are showing the same side.
Repeat this process for every possible pair of cards, then every possible group of three cards, then four, and so on, until we consider flipping all the cards.
For each of these combinations, check if the cards all show the same side. If we find a combination that works, we note the number of cards we flipped.
Finally, from all the combinations that work, we choose the one where we flipped the fewest number of cards.

Code Implementation

def card_flipping_game_brute_force(cards):
    number_of_cards = len(cards)
    minimum_flips = number_of_cards + 1

    for i in range(2**number_of_cards):
        # Generate all possible combinations of flips using binary representation
        binary_representation = bin(i)[2:].zfill(number_of_cards)
        flipped_cards = list(cards)
        number_of_flips_for_combination = 0

        for card_index in range(number_of_cards):
            if binary_representation[card_index] == '1':
                # If the bit is 1, flip the card
                flipped_cards[card_index] = not flipped_cards[card_index]
                number_of_flips_for_combination += 1

        # Check if all cards are the same after flipping
        all_same = True
        for card_index in range(1, number_of_cards):
            if flipped_cards[card_index] != flipped_cards[0]:
                all_same = False
                break

        if all_same:
            # Update minimum flips if this combination is better
            minimum_flips = min(minimum_flips, number_of_flips_for_combination)

    if minimum_flips == number_of_cards + 1:
        return -1
    else:
        return minimum_flips

Big(O) Analysis

Time Complexity

O(2^n * n) – The algorithm iterates through all possible subsets of the input cards. For a deck of n cards, there are 2^n possible subsets (combinations of cards to flip). For each subset, it checks if all cards show the same side, which takes O(n) time. Therefore, the overall time complexity is O(2^n * n), representing the time to generate and evaluate all possible card flipping combinations.

Space Complexity

O(1) – The provided solution explores all possible combinations of card flips, but it does not explicitly store these combinations. It only needs to keep track of the current combination being tested and the minimum number of flips found so far. The algorithm uses a constant amount of extra space for variables to store the minimum flips and potentially a counter, which is independent of the input size N (the number of cards). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The core idea is to identify which cards, when flipped, will maximize the number of cards showing the same side. We want to count how many cards initially show a certain side and determine if flipping a specific card will result in more cards showing that same side.

Here's how the algorithm would work step-by-step:

Count how many cards initially have the same number showing.
Go through each card and imagine flipping it.
Figure out how flipping that card would change the counts of the numbers shown. If flipping it increases the number of cards showing the dominant side, remember this card.
Find the card which, when flipped, gives the biggest increase in the count of the more frequent side.
Flip that card.
Repeat the process until no more flips improve the count of the more frequent side.

Code Implementation

def card_flipping_game(cards):
    cards = list(cards)
    while True:
        counts = {}
        for card in cards:
            counts[card] = counts.get(card, 0) + 1
        
        max_count = 0
        dominant_side = None
        for side, count in counts.items():
            if count > max_count:
                max_count = count
                dominant_side = side
        
        best_flip_index = -1
        best_flip_increase = 0

        # Iterate through cards to find best flip
        for index in range(len(cards)): 
            flipped_card = 1 - cards[index]
            
            original_count = counts.get(cards[index], 0)
            flipped_count = counts.get(flipped_card, 0)
            
            increase = (flipped_count + 1) - original_count
            if flipped_card == dominant_side:
                increase = increase - (counts.get(dominant_side, 0) -1) + counts.get(dominant_side, 0)
            if increase > best_flip_increase:
                best_flip_increase = increase
                best_flip_index = index

        # Terminate if no flip improves the count
        if best_flip_index == -1:
            break
        
        # Flip the card
        cards[best_flip_index] = 1 - cards[best_flip_index]
    
    return cards

Big(O) Analysis

Time Complexity

O(n^2) – The algorithm iterates until no more flips improve the count. Inside the outer loop (repeated flips), we iterate through each of the n cards. For each card, we potentially need to recount the frequency of each number which requires iterating through all n cards again to determine the new dominant side count. Therefore, we have a nested loop structure where the outer loop's iterations are data dependent, and in the worst case, it runs a fixed small number of times, while the inner loop contains an O(n) recount operation. The dominant cost is iterating through all n cards to count the numbers and assess the impact of flipping each one in the inner loop, making the overall time complexity O(n*n), which simplifies to O(n^2).

Space Complexity

O(N) – The algorithm involves counting the occurrences of each number initially, which can be done using a hash map or an array. In the worst case, if all cards have different numbers, this data structure would store N key-value pairs, where N is the number of cards. Additionally, to track the potential increase in the dominant side's count, we might need to store intermediate results related to each card, potentially requiring extra space proportional to N. Therefore, the auxiliary space used is O(N).

Edge Cases

Case	How to Handle
Empty input arrays for front and back	Return 0, as no cards can be flipped.
front and back arrays of different lengths	Return 0 as it's impossible to compare them
Arrays with only one card each (front[0] == back[0])	If front[0] == back[0], no good number exists, return 0, else return front[0].
All cards have the same front and back values (e.g., all 1s)	Return 0, as there is no good number.
Large input arrays causing performance issues.	Optimize by early exiting when a good number is found and is less than all other potential candidates.
Integer overflow when calculating sums or comparing large numbers	Use a long data type to store sums and perform comparisons to prevent potential overflow.
All front values are distinct, and all back values are distinct, and there is no solution	Return 0 after checking all possibilities if no good number is found.
The smallest good number is a very large number	Initialize the smallest good number to a maximum possible value (e.g., Integer.MAX_VALUE) and update it when a smaller good number is found.