This post is part of the Algorithms Problem Solving series.

## Problem description

This is the number of students problem. The description looks like this:

Given two integer arrays `startTime`

and `endTime`

and given an integer `queryTime`

.

The `ith`

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.

## Examples

```
Input: startTime = [1,2,3], endTime = [3,2,7], queryTime = 4
Output: 1
Input: startTime = [4], endTime = [4], queryTime = 4
Output: 1
Input: startTime = [4], endTime = [4], queryTime = 5
Output: 0
Input: startTime = [1,1,1,1], endTime = [1,3,2,4], queryTime = 7
Output: 0
Input: startTime = [9,8,7,6,5,4,3,2,1], endTime = [10,10,10,10,10,10,10,10,10], queryTime = 5
Output: 5
```

## Solution

The idea is just to iterate through the lists and compare them with the `query_time`

to see if it is in the interval between start and end time. If it is, just increment the `number_of_students`

counter. After the for loop finishes, return this value.

```
def busy_student(start_time, end_time, query_time):
number_of_students = 0
for index in range(len(start_time)):
start, end = start_time[index], end_time[index]
if query_time >= start and query_time <= end:
number_of_students += 1
return number_of_students
```

But we could also use the `zip`

function to iterate through the list simultaneously:

```
def busy_student(start_time, end_time, query_time):
number_of_students = 0
for start, end in zip(start_time, end_time):
if query_time >= start and query_time <= end:
number_of_students += 1
return number_of_students
```

The runtime complexity is `O(N)`

where `N`

is the number of integers in the `start_time`

and `end_time`

.

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