# Leetcode 1450. Number of Students Doing Homework at a Given Time 練習
> Difficulty : `Medium`
> `Java` `演算法Algorithm`

> 撰寫人[name=KVJK_2125] [time=Fri, Mar 29, 2024 22:30]

## 題目
### 說明 Description
Given two integer arrays `startTime` and `endTime` and given an integer `queryTime`.

The i^th^ student started doing their homework at the time `startTime[i]` and finished it at time `endTime[i]`.

Return the number of students doing their homework at time queryTime. More formally, return the number of students where `queryTime` lays in the interval `[startTime[i], endTime[i]]` inclusive.

翻譯：
給兩個整數陣列 startTime （開始時間）和 endTime （結束時間），並給一個整數 queryTime。

第 i 位學生在 startTime[i] 時開始寫作業，在 endTime[i] 時完成作業。

回傳在 queryTime 正在做作業的學生人數。更正式的說，回傳能夠使 queryTime 位於區間 [startTime[i], endTime[i]] 內的學生人數。

### 範例 Example
Example 1:
>Input: startTime = [1,2,3], endTime = [3,2,7], queryTime = 4
Output: 1

>Explanation: We have 3 students where:
The first student started doing homework at time 1 and finished at time 3 and wasn't doing anything at time 4.
The second student started doing homework at time 2 and finished at time 2 and also wasn't doing anything at time 4.
The third student started doing homework at time 3 and finished at time 7 and was the only student doing homework at time 4.

Example 2:
>Input: startTime = [4], endTime = [4], queryTime = 4
Output: 1

>Explanation: The only student was doing their homework at the queryTime.

### 限制條件 Constraints
- `startTime.length == endTime.length`
- `1 <= startTime.length <= 100`
- `1 <= startTime[i] <= endTime[i] <= 1000`
- `1 <= queryTime <= 1000`

## 解題
### 思路 Intuition/Approach
- Step1.<br> 建立一個整數變數`int ans = startTime.length`，設定成和開始時間一樣多。<br>

- Step2.<br> 判斷從`startTime`到`endTime`中，必須經過`queryTime`。<br>因此，在`startTime > queryTime`或`endTime < queryTime`時，均不經過`queryTime`，故`ans--`。維持直到計算結束，回傳`ans`。

### 程式碼 Code（加註解）

class Solution {
    public int busyStudent(int[] startTime, int[] endTime, int queryTime) {
        //建立一個整數變數，設定為多少個startTime就多少
        int ans = startTime.length;
        //以startTime.length作為迴圈條件
        for(int i = 0; i < startTime.length; i ++){
            //確定是不是沒有經過queryTime
            if(startTime[i] > queryTime || endTime[i] < queryTime)
                //沒有經過的話，將ans--
                ans--;
        }
        //回傳ans
        return ans;
    }
}