822. Card Flipping Game

Description

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

Solutions

Solution 1: Hash Table

We observe that for position \(i\), if \(\textit{fronts}[i]\) is equal to \(\textit{backs}[i]\), then it certainly does not satisfy the condition.

Therefore, we first identify all elements that appear the same on both the front and back sides and record them in a hash set \(s\).

Next, we iterate through all elements in both the front and back arrays. For any element \(x\) that is not in the hash set \(s\), we update the minimum value of the answer.

Finally, if we find any element that satisfies the condition, we return the minimum answer; otherwise, we return \(0\).

The time complexity is \(O(n)\) and the space complexity is \(O(n)\), where \(n\) is the length of the arrays.

Python3JavaC++GoTypeScriptRustC#

class Solution:
    def flipgame(self, fronts: List[int], backs: List[int]) -> int:
        s = {a for a, b in zip(fronts, backs) if a == b}
        return min((x for x in chain(fronts, backs) if x not in s), default=0)

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;
        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;
    }
}

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;
    }
};

func flipgame(fronts []int, backs []int) int {
    s := map[int]struct{}{}
    for i, a := range fronts {
        if a == backs[i] {
            s[a] = struct{}{}
        }
    }
    ans := 9999
    for _, v := range fronts {
        if _, ok := s[v]; !ok {
            ans = min(ans, v)
        }
    }
    for _, v := range backs {
        if _, ok := s[v]; !ok {
            ans = min(ans, v)
        }
    }
    return ans % 9999
}

function flipgame(fronts: number[], backs: number[]): number {
    const s: Set<number> = new Set();
    const n = fronts.length;
    for (let i = 0; i < n; ++i) {
        if (fronts[i] === backs[i]) {
            s.add(fronts[i]);
        }
    }
    let ans = 9999;
    for (const v of fronts) {
        if (!s.has(v)) {
            ans = Math.min(ans, v);
        }
    }
    for (const v of backs) {
        if (!s.has(v)) {
            ans = Math.min(ans, v);
        }
    }
    return ans % 9999;
}

use std::collections::HashSet;

impl Solution {
    pub fn flipgame(fronts: Vec<i32>, backs: Vec<i32>) -> i32 {
        let n = fronts.len();
        let mut s: HashSet<i32> = HashSet::new();

        for i in 0..n {
            if fronts[i] == backs[i] {
                s.insert(fronts[i]);
            }
        }

        let mut ans = 9999;
        for &v in fronts.iter() {
            if !s.contains(&v) {
                ans = ans.min(v);
            }
        }
        for &v in backs.iter() {
            if !s.contains(&v) {
                ans = ans.min(v);
            }
        }

        if ans == 9999 {
            0
        } else {
            ans
        }
    }
}

public class Solution {
    public int Flipgame(int[] fronts, int[] backs) {
        var s = new HashSet<int>();
        int n = fronts.Length;
        for (int i = 0; i < n; ++i) {
            if (fronts[i] == backs[i]) {
                s.Add(fronts[i]);
            }
        }
        int ans = 9999;
        for (int i = 0; i < n; ++i) {
            if (!s.Contains(fronts[i])) {
                ans = Math.Min(ans, fronts[i]);
            }
            if (!s.Contains(backs[i])) {
                ans = Math.Min(ans, backs[i]);
            }
        }
        return ans % 9999;
    }
}

822. Card Flipping Game

Description

Solutions

Solution 1: Hash Table

Comments