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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