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feat: update solutions to lc problem: No.1450 (#3464)
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No.1450.Number of Students Doing Homework at a Given Time
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@yanglbme
yanglbme authored Sep 1, 2024
1 parent d207f72 commit 30ca343
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143 changes: 20 additions & 123 deletions ...tion/1400-1499/1450.Number of Students Doing Homework at a Given Time/README.md
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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
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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
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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;
}
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