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Commit 6fe46bc

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feat: add solutions to lc problem: No.0822 (doocs#4301)
No.0822.Card Flipping Game
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‎solution/0800-0899/0822.Card Flipping Game/README.md‎

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@@ -66,7 +66,7 @@ tags:
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### 方法一:哈希表
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我们注意到,对于位置 $i,ドル若 $fronts[i]$ 与 $backs[i]$ 元素相同,则一定不满足条件。
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我们注意到,对于位置 $i,ドル若 $\textit{fronts}[i]$ 与 $\textit{backs}[i]$ 元素相同,则一定不满足条件。
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因此,我们先找出正面与背面相同的元素,记录在哈希表 $s$ 中。
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}
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```
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#### Rust
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```rust
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use std::collections::HashSet;
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impl Solution {
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pub fn flipgame(fronts: Vec<i32>, backs: Vec<i32>) -> i32 {
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let n = fronts.len();
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let mut s: HashSet<i32> = HashSet::new();
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for i in 0..n {
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if fronts[i] == backs[i] {
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s.insert(fronts[i]);
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}
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}
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let mut ans = 9999;
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for &v in fronts.iter() {
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if !s.contains(&v) {
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ans = ans.min(v);
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}
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}
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for &v in backs.iter() {
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if !s.contains(&v) {
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ans = ans.min(v);
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}
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}
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if ans == 9999 {
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0
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} else {
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ans
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}
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}
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}
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```
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#### C#
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```cs

‎solution/0800-0899/0822.Card Flipping Game/README_EN.md‎

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@@ -59,7 +59,17 @@ There are no good integers no matter how we flip the cards, so we return 0.
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Hash Table
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We observe that for position $i,ドル if $\textit{fronts}[i]$ is equal to $\textit{backs}[i],ドル then it certainly does not satisfy the condition.
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Therefore, we first identify all elements that appear the same on both the front and back sides and record them in a hash set $s$.
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Next, we iterate through all elements in both the front and back arrays. For any element $x$ that is **not** in the hash set $s,ドル we update the minimum value of the answer.
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Finally, if we find any element that satisfies the condition, we return the minimum answer; otherwise, we return 0ドル$.
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The time complexity is $O(n)$ and the space complexity is $O(n),ドル where $n$ is the length of the arrays.
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<!-- tabs:start -->
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}
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```
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#### Rust
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```rust
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use std::collections::HashSet;
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impl Solution {
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pub fn flipgame(fronts: Vec<i32>, backs: Vec<i32>) -> i32 {
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let n = fronts.len();
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let mut s: HashSet<i32> = HashSet::new();
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for i in 0..n {
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if fronts[i] == backs[i] {
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s.insert(fronts[i]);
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}
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}
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let mut ans = 9999;
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for &v in fronts.iter() {
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if !s.contains(&v) {
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ans = ans.min(v);
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}
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}
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for &v in backs.iter() {
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if !s.contains(&v) {
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ans = ans.min(v);
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}
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}
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if ans == 9999 {
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0
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} else {
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ans
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}
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}
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}
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```
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#### C#
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```cs
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use std::collections::HashSet;
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impl Solution {
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pub fn flipgame(fronts: Vec<i32>, backs: Vec<i32>) -> i32 {
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let n = fronts.len();
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let mut s: HashSet<i32> = HashSet::new();
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for i in 0..n {
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if fronts[i] == backs[i] {
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s.insert(fronts[i]);
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}
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}
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let mut ans = 9999;
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for &v in fronts.iter() {
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if !s.contains(&v) {
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ans = ans.min(v);
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}
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}
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for &v in backs.iter() {
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if !s.contains(&v) {
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ans = ans.min(v);
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}
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}
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if ans == 9999 {
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0
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} else {
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ans
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}
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}
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}

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