# 822. Card Flipping Game

On a table are `N` cards, with a positive integer printed on the front and back of each card (possibly different).

We flip any number of cards, and after we choose one card.

If the number `X` on the back of the chosen card is not on the front of any card, then this number X is good.

What is the smallest number that is good?  If no number is good, output `0`.

Here, `fronts[i]` and `backs[i]` represent the number on the front and back of card `i`

A flip swaps the front and back numbers, so the value on the front is now on the back and vice versa.

Example:

``````Input: fronts = [1,2,4,4,7], backs = [1,3,4,1,3]
Output: 2
Explanation: If we flip the second card, the fronts are [1,3,4,4,7] and the backs are [1,2,4,1,3].
We choose the second card, which has number 2 on the back, and it isn't on the front of any card, so 2 is good.
``````

Note:

1. `1 <= fronts.length == backs.length <= 1000`.
2. `1 <= fronts[i] <= 2000`.
3. `1 <= backs[i] <= 2000`.

``````class Solution {
public:
int flipgame(vector<int>& fronts, vector<int>& backs) {
int res = INT_MAX, n = fronts.size();
unordered_set<int> same;
for (int i = 0; i < n; ++i) {
if (fronts[i] == backs[i]) same.insert(fronts[i]);
}
for (int front : fronts) {
if (!same.count(front)) res = min(res, front);
}
for (int back : backs) {
if (!same.count(back)) res = min(res, back);
}
return (res == INT_MAX) ? 0 : res;
}
};
``````

https://leetcode.com/problems/card-flipping-game/

https://leetcode.com/problems/card-flipping-game/discuss/125791/C%2B%2BJavaPython-Easy-and-Concise-with-Explanation

LeetCode All in One 题目讲解汇总(持续更新中…)

 微信打赏 Venmo 打赏
（欢迎加入博主的知识星球，博主将及时答疑解惑，并分享刷题经验与总结，试运营期间前五十位可享受半价优惠～）

×

Help us with donation