# Solving Leetcode Interviews in Seconds with AI: Number of Students Doing Homework at a Given Time


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1450" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> 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.   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.    Constraints:  startTime.length == endTime.length 1 <= startTime.length <= 100 1 <= startTime[i] <= endTime[i] <= 1000 1 <= queryTime <= 1000  

	# Explanation
	Here's the solution to the problem:

*   **Approach:** Iterate through the `startTime` and `endTime` arrays. For each student `i`, check if `queryTime` falls within the inclusive range `[startTime[i], endTime[i]]`. If it does, increment a counter.
*   **Complexity:**
    *   Runtime Complexity: O(n), where n is the number of students (length of `startTime` or `endTime`).
    *   Storage Complexity: O(1) (constant extra space)
*   **Code:**

	
	# Code
	```python
	def busyStudent(startTime, endTime, queryTime):
    """
    Counts the number of students doing homework at queryTime.

    Args:
        startTime (list[int]): Start times of homework.
        endTime (list[int]): End times of homework.
        queryTime (int): The time to check.

    Returns:
        int: The number of students doing homework at queryTime.
    """
    count = 0
    for i in range(len(startTime)):
        if startTime[i] <= queryTime <= endTime[i]:
            count += 1
    return count
	```