feat: add solutions to lc problem: No.2534

No.2534.Time Taken to Cross the Door

Author: yanglbme

solution/2500-2599/2534.Time Taken to Cross the Door/README.md

@@ -83,7 +83,19 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：队列 + 模拟
+
+我们定义两个队列，其中 $q[0]$ 存放想要进入的人的编号，而 $q[1]$ 存放想要离开的人的编号。
+
+我们维护一个时间 $t$，表示当前时间，一个状态 $st$，表示当前门的状态，当 $st = 1$ 表示门没使用或者上一秒有人离开，当 $st = 0$ 表示上一秒有人进入。初始时 $t = 0$，而 $st = 1$。
+
+我们遍历数组 $\textit{arrival}$，对于每个人，如果当前时间 $t$ 小于等于该人到达门前的时间 $arrival[i]$，我们将该人的编号加入对应的队列 $q[\text{state}[i]]$ 中。
+
+然后我们判断当前队列 $q[0]$ 和 $q[1]$ 是否都不为空，如果都不为空，我们将 $q[st]$ 队列的队首元素出队，并将当前时间 $t$ 赋值给该人的通过时间；如果只有一个队列不为空，我们根据哪个队列不为空，更新 $st$ 的值，然后将该队列的队首元素出队，并将当前时间 $t$ 赋值给该人的通过时间；如果两个队列都为空，我们将 $st$ 的值更新为 $1$，表示门没使用。
+
+接下来，我们将时间 $t$ 自增 $1$，继续遍历数组 $\textit{arrival}$，直到所有人都通过门。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 表示数组 $\textit{arrival}$ 的长度。
 
 <!-- tabs:start -->
 
@@ -245,6 +257,45 @@ function timeTaken(arrival: number[], state: number[]): number[] {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::VecDeque;
+
+impl Solution {
+    pub fn time_taken(arrival: Vec<i32>, state: Vec<i32>) -> Vec<i32> {
+        let n = arrival.len();
+        let mut q = vec![VecDeque::new(), VecDeque::new()];
+        let mut t = 0;
+        let mut i = 0;
+        let mut st = 1;
+        let mut ans = vec![-1; n];
+
+        while i < n || !q[0].is_empty() || !q[1].is_empty() {
+            while i < n && arrival[i] <= t {
+                q[state[i] as usize].push_back(i);
+                i += 1;
+            }
+
+            if !q[0].is_empty() && !q[1].is_empty() {
+                ans[*q[st].front().unwrap()] = t;
+                q[st].pop_front();
+            } else if !q[0].is_empty() || !q[1].is_empty() {
+                st = if q[0].is_empty() { 1 } else { 0 };
+                ans[*q[st].front().unwrap()] = t;
+                q[st].pop_front();
+            } else {
+                st = 1;
+            }
+
+            t += 1;
+        }
+
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2500-2599/2534.Time Taken to Cross the Door/README_EN.md

@@ -82,7 +82,19 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Queue + Simulation
+
+We define two queues, where $q[0]$ stores the indices of people who want to enter, and $q[1]$ stores the indices of people who want to exit.
+
+We maintain a variable $t$ to represent the current time, and a variable $st$ to represent the current state of the door. When $st = 1$, it means the door is not in use or someone exited in the previous second. When $st = 0$, it means someone entered in the previous second. Initially, $t = 0$ and $st = 1$.
+
+We traverse the array $\textit{arrival}$. For each person, if the current time $t$ is less than or equal to the time the person arrives at the door $\textit{arrival}[i]$, we add the person's index to the corresponding queue $q[\text{state}[i]]$.
+
+Then we check if both queues $q[0]$ and $q[1]$ are not empty. If both are not empty, we dequeue the front element from the queue $q[st]$ and assign the current time $t$ to the person's passing time. If only one queue is not empty, we update the value of $st$ based on which queue is not empty, then dequeue the front element from that queue and assign the current time $t$ to the person's passing time. If both queues are empty, we update the value of $st$ to $1$, indicating the door is not in use.
+
+Next, we increment the time $t$ by $1$ and continue traversing the array $\textit{arrival}$ until all people have passed through the door.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ represents the length of the array $\textit{arrival}$.
 
 <!-- tabs:start -->
 
@@ -244,6 +256,45 @@ function timeTaken(arrival: number[], state: number[]): number[] {
 }
 ```
 
+#### Rust
+
+```rust
+use std::collections::VecDeque;
+
+impl Solution {
+    pub fn time_taken(arrival: Vec<i32>, state: Vec<i32>) -> Vec<i32> {
+        let n = arrival.len();
+        let mut q = vec![VecDeque::new(), VecDeque::new()];
+        let mut t = 0;
+        let mut i = 0;
+        let mut st = 1;
+        let mut ans = vec![-1; n];
+
+        while i < n || !q[0].is_empty() || !q[1].is_empty() {
+            while i < n && arrival[i] <= t {
+                q[state[i] as usize].push_back(i);
+                i += 1;
+            }
+
+            if !q[0].is_empty() && !q[1].is_empty() {
+                ans[*q[st].front().unwrap()] = t;
+                q[st].pop_front();
+            } else if !q[0].is_empty() || !q[1].is_empty() {
+                st = if q[0].is_empty() { 1 } else { 0 };
+                ans[*q[st].front().unwrap()] = t;
+                q[st].pop_front();
+            } else {
+                st = 1;
+            }
+
+            t += 1;
+        }
+
+        ans
+    }
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2500-2599/2534.Time Taken to Cross the Door/Solution.rs

@@ -0,0 +1,34 @@
+use std::collections::VecDeque;
+
+impl Solution {
+    pub fn time_taken(arrival: Vec<i32>, state: Vec<i32>) -> Vec<i32> {
+        let n = arrival.len();
+        let mut q = vec![VecDeque::new(), VecDeque::new()];
+        let mut t = 0;
+        let mut i = 0;
+        let mut st = 1;
+        let mut ans = vec![-1; n];
+
+        while i < n || !q[0].is_empty() || !q[1].is_empty() {
+            while i < n && arrival[i] <= t {
+                q[state[i] as usize].push_back(i);
+                i += 1;
+            }
+
+            if !q[0].is_empty() && !q[1].is_empty() {
+                ans[*q[st].front().unwrap()] = t;
+                q[st].pop_front();
+            } else if !q[0].is_empty() || !q[1].is_empty() {
+                st = if q[0].is_empty() { 1 } else { 0 };
+                ans[*q[st].front().unwrap()] = t;
+                q[st].pop_front();
+            } else {
+                st = 1;
+            }
+
+            t += 1;
+        }
+
+        ans
+    }
+}