Skip to content

feat: update solutions to lc problem: No.1450 #3464

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 1 commit into from
Sep 1, 2024
Merged

feat: update solutions to lc problem: No.1450 #3464

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
143 changes: 20 additions & 123 deletions ...tion/1400-1499/1450.Number of Students Doing Homework at a Given Time/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -78,11 +78,13 @@ tags:

<!-- solution:start -->

### 方法一：遍历计数
### 方法一：直接遍历

同时遍历 $startTime$ 和 $endTime$，统计正在做作业的学生人数。
我们可以直接遍历两个数组，对于每个学生，判断 $\textit{queryTime}$ 是否在他们的作业时间区间内，若是，答案加一。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 是 $startTime$ 和 $endTime$ 的长度。
遍历结束后，返回答案即可。

时间复杂度 $O(n)$，其中 $n$ 为学生数量。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -93,7 +95,7 @@ class Solution:
def busyStudent(
self, startTime: List[int], endTime: List[int], queryTime: int
) -> int:
return sum(a <= queryTime <= b for a, b in zip(startTime, endTime))
return sum(x <= queryTime <= y for x, y in zip(startTime, endTime))
```

#### Java
Expand Down Expand Up @@ -130,15 +132,13 @@ public:
#### Go

```go
func busyStudent(startTime []int, endTime []int, queryTime int) int {
ans := 0
for i, a := range startTime {
b := endTime[i]
if a <= queryTime && queryTime <= b {
func busyStudent(startTime []int, endTime []int, queryTime int) (ans int) {
for i, x := range startTime {
if x <= queryTime && queryTime <= endTime[i] {
ans++
}
}
return ans
return
}
```

Expand All @@ -147,13 +147,13 @@ func busyStudent(startTime []int, endTime []int, queryTime int) int {
```ts
function busyStudent(startTime: number[], endTime: number[], queryTime: number): number {
const n = startTime.length;
let res = 0;
let ans = 0;
for (let i = 0; i < n; i++) {
if (startTime[i] <= queryTime && endTime[i] >= queryTime) {
res++;
if (startTime[i] <= queryTime && queryTime <= endTime[i]) {
ans++;
}
}
return res;
return ans;
}
```

Expand All @@ -162,13 +162,13 @@ function busyStudent(startTime: number[], endTime: number[], queryTime: number):
```rust
impl Solution {
pub fn busy_student(start_time: Vec<i32>, end_time: Vec<i32>, query_time: i32) -> i32 {
let mut res = 0;
let mut ans = 0;
for i in 0..start_time.len() {
if start_time[i] <= query_time && end_time[i] >= query_time {
res += 1;
ans += 1;
}
}
res
ans
}
}
```
Expand All @@ -177,116 +177,13 @@ impl Solution {

```c
int busyStudent(int* startTime, int startTimeSize, int* endTime, int endTimeSize, int queryTime) {
int res = 0;
int ans = 0;
for (int i = 0; i < startTimeSize; i++) {
if (startTime[i] <= queryTime && endTime[i] >= queryTime) {
res++;
}
}
return res;
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### 方法二：差分数组

差分数组可以 $O(1)$ 时间处理区间加减操作。例如，对区间 $[l, r]$ 中的每个数加上 $c$。

假设数组 $a$ 的所有元素分别为 $a[1], a[2], ... a[n]$，则差分数组 $b$ 的元素 $b[i]=a[i]-a[i-1]$。

$$
\begin{cases}
b[1]=a[1]\\
b[2]=a[2]-a[1]\\
b[3]=a[3]-a[2]\\
...\\
b[i]=a[i]-a[i-1]\\
\end{cases}
$$

那么 $a[i]=b[1]+b[2]+...+b[i]$，原数组 $a$ 是差分数组 $b$ 的前缀和。

在这道题中，我们定义差分数组 $c$，然后遍历两个数组中对应位置的两个数 $a$, $b$，则 $c[a]+=1$, $c[b+1]-=1$。

遍历结束后，对差分数组 $c$ 进行求前缀和操作，即可得到 $queryTime$ 时刻正在做作业的学生人数。

时间复杂度 $O(n+queryTime)$，空间复杂度 $O(1010)$。

<!-- tabs:start -->

#### Python3

```python
class Solution:
def busyStudent(
self, startTime: List[int], endTime: List[int], queryTime: int
) -> int:
c = [0] * 1010
for a, b in zip(startTime, endTime):
c[a] += 1
c[b + 1] -= 1
return sum(c[: queryTime + 1])
```

#### Java

```java
class Solution {
public int busyStudent(int[] startTime, int[] endTime, int queryTime) {
int[] c = new int[1010];
for (int i = 0; i < startTime.length; ++i) {
c[startTime[i]]++;
c[endTime[i] + 1]--;
}
int ans = 0;
for (int i = 0; i <= queryTime; ++i) {
ans += c[i];
}
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int busyStudent(vector<int>& startTime, vector<int>& endTime, int queryTime) {
vector<int> c(1010);
for (int i = 0; i < startTime.size(); ++i) {
c[startTime[i]]++;
c[endTime[i] + 1]--;
}
int ans = 0;
for (int i = 0; i <= queryTime; ++i) {
ans += c[i];
ans++;
}
return ans;
}
};
```

#### Go

```go
func busyStudent(startTime []int, endTime []int, queryTime int) int {
c := make([]int, 1010)
for i, a := range startTime {
b := endTime[i]
c[a]++
c[b+1]--
}
ans := 0
for i := 0; i <= queryTime; i++ {
ans += c[i]
}
return ans
return ans;
}
```

Expand Down
121 changes: 22 additions & 99 deletions ...n/1400-1499/1450.Number of Students Doing Homework at a Given Time/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -60,7 +60,13 @@ The third student started doing homework at time 3 and finished at time 7 and wa

<!-- solution:start -->

### Solution 1
### Solution 1: Direct Traversal

We can directly traverse the two arrays. For each student, we check if $\textit{queryTime}$ is within their homework time interval. If it is, we increment the answer by one.

After the traversal, we return the answer.

The time complexity is $O(n)$, where $n$ is the number of students. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand All @@ -71,7 +77,7 @@ class Solution:
def busyStudent(
self, startTime: List[int], endTime: List[int], queryTime: int
) -> int:
return sum(a <= queryTime <= b for a, b in zip(startTime, endTime))
return sum(x <= queryTime <= y for x, y in zip(startTime, endTime))
```

#### Java
Expand Down Expand Up @@ -108,15 +114,13 @@ public:
#### Go

```go
func busyStudent(startTime []int, endTime []int, queryTime int) int {
ans := 0
for i, a := range startTime {
b := endTime[i]
if a <= queryTime && queryTime <= b {
func busyStudent(startTime []int, endTime []int, queryTime int) (ans int) {
for i, x := range startTime {
if x <= queryTime && queryTime <= endTime[i] {
ans++
}
}
return ans
return
}
```

Expand All @@ -125,13 +129,13 @@ func busyStudent(startTime []int, endTime []int, queryTime int) int {
```ts
function busyStudent(startTime: number[], endTime: number[], queryTime: number): number {
const n = startTime.length;
let res = 0;
let ans = 0;
for (let i = 0; i < n; i++) {
if (startTime[i] <= queryTime && endTime[i] >= queryTime) {
res++;
if (startTime[i] <= queryTime && queryTime <= endTime[i]) {
ans++;
}
}
return res;
return ans;
}
```

Expand All @@ -140,13 +144,13 @@ function busyStudent(startTime: number[], endTime: number[], queryTime: number):
```rust
impl Solution {
pub fn busy_student(start_time: Vec<i32>, end_time: Vec<i32>, query_time: i32) -> i32 {
let mut res = 0;
let mut ans = 0;
for i in 0..start_time.len() {
if start_time[i] <= query_time && end_time[i] >= query_time {
res += 1;
ans += 1;
}
}
res
ans
}
}
```
Expand All @@ -155,94 +159,13 @@ impl Solution {

```c
int busyStudent(int* startTime, int startTimeSize, int* endTime, int endTimeSize, int queryTime) {
int res = 0;
int ans = 0;
for (int i = 0; i < startTimeSize; i++) {
if (startTime[i] <= queryTime && endTime[i] >= queryTime) {
res++;
ans++;
}
}
return res;
}
```

<!-- tabs:end -->

<!-- solution:end -->

<!-- solution:start -->

### Solution 2

<!-- tabs:start -->

#### Python3

```python
class Solution:
def busyStudent(
self, startTime: List[int], endTime: List[int], queryTime: int
) -> int:
c = [0] * 1010
for a, b in zip(startTime, endTime):
c[a] += 1
c[b + 1] -= 1
return sum(c[: queryTime + 1])
```

#### Java

```java
class Solution {
public int busyStudent(int[] startTime, int[] endTime, int queryTime) {
int[] c = new int[1010];
for (int i = 0; i < startTime.length; ++i) {
c[startTime[i]]++;
c[endTime[i] + 1]--;
}
int ans = 0;
for (int i = 0; i <= queryTime; ++i) {
ans += c[i];
}
return ans;
}
}
```

#### C++

```cpp
class Solution {
public:
int busyStudent(vector<int>& startTime, vector<int>& endTime, int queryTime) {
vector<int> c(1010);
for (int i = 0; i < startTime.size(); ++i) {
c[startTime[i]]++;
c[endTime[i] + 1]--;
}
int ans = 0;
for (int i = 0; i <= queryTime; ++i) {
ans += c[i];
}
return ans;
}
};
```

#### Go

```go
func busyStudent(startTime []int, endTime []int, queryTime int) int {
c := make([]int, 1010)
for i, a := range startTime {
b := endTime[i]
c[a]++
c[b+1]--
}
ans := 0
for i := 0; i <= queryTime; i++ {
ans += c[i]
}
return ans
return ans;
}
```

Expand Down
8 changes: 4 additions & 4 deletions solution/1400-1499/1450.Number of Students Doing Homework at a Given Time/Solution.c
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,9 +1,9 @@
int busyStudent(int* startTime, int startTimeSize, int* endTime, int endTimeSize, int queryTime) {
int res = 0;
int ans = 0;
for (int i = 0; i < startTimeSize; i++) {
if (startTime[i] <= queryTime && endTime[i] >= queryTime) {
res++;
ans++;
}
}
return res;
}
return ans;
}
Loading
Loading