README.md

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.

Constraints:

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

Solutions (Rust)

1. Solution

use std::collections::HashSet;

impl Solution {
    pub fn flipgame(fronts: Vec<i32>, backs: Vec<i32>) -> i32 {
        let mut good = HashSet::new();
        let mut ban = HashSet::new();

        for i in 0..backs.len() {
            if fronts[i] == backs[i] {
                good.remove(&backs[i]);
                ban.insert(backs[i]);
            } else {
                if !ban.contains(&fronts[i]) {
                    good.insert(fronts[i]);
                }
                if !ban.contains(&backs[i]) {
                    good.insert(backs[i]);
                }
            }
        }

        good.into_iter().min().unwrap_or(0)
    }
}