Problem Description

You have n cards, each with two sides - a front and a back. The i-th card has a positive integer fronts[i] on the front side and backs[i] on the back side.

Initially, all cards are placed on a table with their front sides facing up (visible) and back sides facing down (hidden). You can flip any number of cards (including zero cards) to swap which side is facing up.

After flipping some cards, an integer is considered good if:

• It appears on at least one card's face-down side (hidden side)
• It does NOT appear on any card's face-up side (visible side)

Your task is to find the minimum possible good integer after optimally flipping the cards. If no good integer exists, return 0.

Example walkthrough:

• If a card has the same number on both front and back (like fronts[i] = backs[i] = 5), then 5 can never be a good integer. No matter how you flip this card, 5 will always be visible on one side.
• For cards with different numbers on front and back, you can choose which number to hide by flipping the card appropriately.

The solution identifies all numbers that appear on both sides of the same card (these can never be good), then finds the minimum number from all card values that isn't in this "impossible" set.

Intuition

Let's think about when a number can be "good". For a number to be good, we need to be able to hide it from all face-up positions while keeping it on at least one face-down position.

The key insight comes from considering individual cards:

1. Cards with different numbers on front and back: These cards give us flexibility. If we have a card with fronts[i] = 3 and backs[i] = 5, we can choose to show either 3 or 5 by flipping the card. This means we can potentially hide one of these numbers while showing the other.

2. Cards with the same number on both sides: This is the critical case. If fronts[i] = backs[i] = 7, then no matter how we flip this card, the number 7 will always be visible. There's no way to hide it from the face-up side. Therefore, 7 can never be a good number.

This observation leads us to the solution strategy:

• First, identify all "impossible" numbers - those that appear on both sides of at least one card. These numbers will always be visible no matter how we arrange the cards.
• Then, from all the numbers that appear on any card (front or back), find the smallest one that isn't "impossible".

Why does this work? Because for any number that isn't in the "impossible" set, we know it only appears on cards where the front and back are different. For each such card, we can flip it to hide this number on the face-down side while showing a different number on the face-up side.

The elegance of this solution is that we don't need to track which specific cards to flip - we just need to know which numbers are possible candidates for being "good", and then pick the minimum among them.

Solution Approach

The implementation follows directly from our intuition using a hash table approach:

Step 1: Identify impossible numbers

We create a hash set s to store all numbers that can never be "good". We iterate through all positions and check if fronts[i] == backs[i]. If they're equal, we add that number to our set:

s = {a for a, b in zip(fronts, backs) if a == b}

This set comprehension zips the fronts and backs arrays together and adds any number a where the front and back values are identical.

Step 2: Find the minimum valid number

We need to check all numbers that appear on any card (both fronts and backs). Using chain(fronts, backs), we create an iterator that goes through all values from both arrays.

For each number x in this combined list:

• If x is not in set s, it's a candidate for being "good"
• We want the minimum among all such candidates

return min((x for x in chain(fronts, backs) if x not in s), default=0)

The generator expression (x for x in chain(fronts, backs) if x not in s) produces all numbers that aren't in the "impossible" set. The min() function finds the smallest such number.

Edge case handling:

If no valid number exists (all numbers appear on both sides of at least one card), the generator will be empty. The default=0 parameter handles this case by returning 0 when there are no good integers.

Time Complexity: O(n) where n is the number of cards, as we iterate through the arrays a constant number of times.

Space Complexity: O(n) in the worst case for the hash set if all cards have the same number on both sides.

Example Walkthrough

Let's walk through a concrete example with 3 cards:

• Card 1: front = 1, back = 2
• Card 2: front = 3, back = 3
• Card 3: front = 2, back = 4

Step 1: Identify impossible numbers

We check each card to find numbers that appear on both sides:

• Card 1: front (1) ≠ back (2) → No impossible numbers
• Card 2: front (3) = back (3) → 3 is impossible (can never be hidden)
• Card 3: front (2) ≠ back (4) → No impossible numbers

Impossible set = {3}

Step 2: Find minimum valid number

All numbers appearing on cards: [1, 2, 3, 3, 2, 4] = {1, 2, 3, 4}

Check each number:

• 1: Not in impossible set → Valid candidate ✓
• 2: Not in impossible set → Valid candidate ✓
• 3: In impossible set → Cannot be good ✗
• 4: Not in impossible set → Valid candidate ✓

Minimum valid number = 1

Verification: We can achieve this by:

• Keep Card 1 as is (showing front = 1, hiding back = 2)
• Keep Card 2 as is (doesn't matter since 3 is always visible)
• Flip Card 3 (showing back = 4, hiding front = 2)

Result: Face-up shows [1, 3, 4], Face-down has [2, 3, 2]

• Number 2 is hidden on face-down sides but not visible on any face-up side → 2 is good
• Number 1 is visible on face-up side → 1 is not good
• Actually, we need to flip Card 1 to hide 1!

Let me reconsider:

• Flip Card 1 (showing back = 2, hiding front = 1)
• Keep Card 2 (showing 3)
• Keep Card 3 (showing front = 2, hiding back = 4)

Result: Face-up shows [2, 3, 2], Face-down has [1, 3, 4]

• Number 1 appears on face-down side only → 1 is good ✓
• Number 4 appears on face-down side only → 4 is good ✓

The minimum good integer is 1.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm performs the following operations:

• Creating set s by iterating through fronts and backs with zip(): O(n) time
• The set comprehension checks condition a == b for each pair: O(1) per element
• chain(fronts, backs) creates an iterator over 2n elements
• The generator expression iterates through at most 2n elements
• For each element, checking membership in set s takes O(1) average time
• Finding the minimum of the filtered elements takes O(n) in total

Overall time complexity: O(n) + O(2n) = O(n)

Space Complexity: O(n)

The space usage includes:

• Set s stores at most n unique values (worst case when all pairs have matching front and back): O(n) space
• The generator expression and chain() use O(1) additional space as they create iterators
• The min() function processes elements one at a time without storing them all

Overall space complexity: O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Misunderstanding What Makes a Number "Good"

The Mistake: Many people initially think that a number is good if it appears on ANY back side and doesn't appear on ANY front side. This leads to incorrect solutions like:

# WRONG approach
def flipgame(self, fronts, backs):
    front_set = set(fronts)
    back_set = set(backs)
    # Find numbers only in backs but not in fronts
    good_numbers = back_set - front_set
    return min(good_numbers) if good_numbers else 0

Why It's Wrong: The problem allows you to flip cards! After flipping, what was on the back becomes visible (face-up), and what was on the front becomes hidden (face-down). A number can be good even if it initially appears on the front, as long as you can flip cards to hide all occurrences of it.

The Fix: Understand that you have control over which side is up for each card, except when both sides have the same number.

Pitfall 2: Forgetting About Cards with Identical Sides

The Mistake: Trying to track which specific cards to flip without recognizing the fundamental constraint:

# WRONG approach - overly complex and misses the key insight
def flipgame(self, fronts, backs):
    min_good = float('inf')
    for target in set(fronts + backs):
        can_hide_all = True
        for i in range(len(fronts)):
            if fronts[i] == target and backs[i] == target:
                can_hide_all = False
                break
            # Trying to figure out complex flipping logic...
        if can_hide_all:
            min_good = min(min_good, target)
    return min_good if min_good != float('inf') else 0

Why It's Wrong: This approach overcomplicates the problem. The key insight is that if a number appears on both sides of the same card, it's impossible to make it good. For all other numbers, you can always arrange the cards to make them good.

The Fix: Simply identify numbers that appear on both sides of the same card - these are the only numbers that cannot be good. All other numbers that appear on any card are valid candidates.

Pitfall 3: Not Considering All Numbers on Cards

The Mistake: Only checking numbers from the fronts array or only from the backs array:

# WRONG approach
def flipgame(self, fronts, backs):
    same = {a for a, b in zip(fronts, backs) if a == b}
    # Only checking fronts array - missing potential answers from backs
    return min((x for x in fronts if x not in same), default=0)

Why It's Wrong: A good number could be any number that appears on any card (either front or back), as long as it doesn't appear on both sides of the same card.

The Fix: Check all numbers from both fronts and backs arrays when finding the minimum valid number, which is why the correct solution uses chain(fronts, backs).