Solution Explanation for Card Flipping Game

The problem asks to find the minimum "good" integer after flipping some cards. A good integer is defined as a number that appears on the back of at least one card but not on the front of any card.

The solution uses a set s to store numbers that appear on both the front and back of at least one card. These numbers cannot be good integers because they will always be visible on at least one side, regardless of flipping.

The algorithm iterates through both fronts and backs arrays. For each number, it checks if it's present in the s set. If a number is not in s, it means it's a potential candidate for a good integer. The algorithm maintains a minimum ans variable to track the smallest good integer found so far.

After checking all numbers, the minimum good integer (ans) is returned. If no good integer exists (ans remains the initial large value), 0 is returned.

Time Complexity Analysis

The algorithm involves:

1. Creating the set s: This takes O(n) time, where n is the length of the input arrays.
2. Iterating through fronts and backs: This takes O(n) time for each array, totaling O(n) time.
3. Checking for membership in s: Checking if an element is in a set takes O(1) on average (using a hash set).

Therefore, the overall time complexity of the algorithm is O(n).

Space Complexity Analysis

The space complexity is dominated by the set s. In the worst case, s could store up to n elements (if all cards have the same number on both sides). Therefore, the space complexity is O(n).

Code Examples and Explanations (in multiple languages)

The provided code solutions efficiently implement this approach:

Python:

class Solution:
    def flipgame(self, fronts: List[int], backs: List[int]) -> int:
        s = {a for a, b in zip(fronts, backs) if a == b} #Efficiently creates the set using set comprehension
        return min((x for x in chain(fronts, backs) if x not in s), default=0) #Uses generator expression for efficient iteration and min function to find the minimum

Java:

class Solution {
    public int flipgame(int[] fronts, int[] backs) {
        Set<Integer> s = new HashSet<>();
        int n = fronts.length;
        for (int i = 0; i < n; ++i) {
            if (fronts[i] == backs[i]) {
                s.add(fronts[i]);
            }
        }
        int ans = 9999; //Initialize with a large value
        for (int v : fronts) {
            if (!s.contains(v)) {
                ans = Math.min(ans, v);
            }
        }
        for (int v : backs) {
            if (!s.contains(v)) {
                ans = Math.min(ans, v);
            }
        }
        return ans % 9999; //Handles potential integer overflow
    }
}

C++:

class Solution {
public:
    int flipgame(vector<int>& fronts, vector<int>& backs) {
        unordered_set<int> s;
        int n = fronts.size();
        for (int i = 0; i < n; ++i) {
            if (fronts[i] == backs[i]) {
                s.insert(fronts[i]);
            }
        }
        int ans = 9999;
        for (int& v : fronts) {
            if (!s.count(v)) {
                ans = min(ans, v);
            }
        }
        for (int& v : backs) {
            if (!s.count(v)) {
                ans = min(ans, v);
            }
        }
        return ans % 9999;
    }
};

The other languages (Go, TypeScript, C#) follow similar structures, utilizing their respective sets and min/minimum functions for optimal performance. The % 9999 in some solutions is a safeguard against potential integer overflow, although unlikely given the problem constraints.