Date and time handling in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of datetime timedelta addition

What is the output of this Python code snippet?

Python

from datetime import datetime, timedelta

start = datetime(2024, 6, 15, 12, 0, 0)
delta = timedelta(days=3, hours=4, minutes=30)
result = start + delta
print(result.strftime('%Y-%m-%d %H:%M:%S'))

A 2024-06-18 16:30:00

B 2024-06-18 04:30:00

C 2024-06-19 16:30:00

D 2024-06-15 16:30:00

💡 Hint

Add days, then hours and minutes to the start datetime.

Adding 3 days to June 15 gives June 18. Adding 4 hours and 30 minutes to 12:00 results in 16:30 on the same day.

❓ Predict Output

intermediate

Output of datetime string parsing and weekday

What is the output of this Python code?

Python

from datetime import datetime

date_str = '2024-06-20'
date_obj = datetime.strptime(date_str, '%Y-%m-%d')
print(date_obj.strftime('%A'))

A Wednesday

B Friday

C Thursday

D Saturday

💡 Hint

Check the weekday for June 20, 2024.

June 20, 2024 is a Thursday, but the code prints the full weekday name. The correct weekday is Thursday, so option C is correct.

❓ Predict Output

advanced

Output of timezone-aware datetime conversion

What is the output of this Python code snippet?

Python

from datetime import datetime, timezone, timedelta

utc_time = datetime(2024, 6, 15, 12, 0, 0, tzinfo=timezone.utc)
new_york_tz = timezone(timedelta(hours=-4))
ny_time = utc_time.astimezone(new_york_tz)
print(ny_time.strftime('%Y-%m-%d %H:%M:%S %Z%z'))

A 2024-06-15 08:00:00 UTC+0400

B 2024-06-15 12:00:00 UTC-0400

C 2024-06-15 16:00:00 UTC-0400

D 2024-06-15 08:00:00 UTC-0400

💡 Hint

New York is 4 hours behind UTC in this example.

The UTC time is 12:00. New York timezone is UTC-4, so the time is 8:00 with offset -0400.

❓ Predict Output

advanced

Output of datetime replace method

What is the output of this Python code?

Python

from datetime import datetime

now = datetime(2024, 6, 15, 14, 30, 45)
new_time = now.replace(hour=9, minute=0, second=0)
print(new_time.strftime('%Y-%m-%d %H:%M:%S'))

A 2024-06-15 09:30:45

B 2024-06-15 09:00:00

C 2024-06-15 14:30:45

D 2024-06-15 00:00:00

💡 Hint

The replace method changes only specified parts of the datetime.

The replace method sets hour, minute, and second to 9, 0, 0 respectively, keeping the date same.

🧠 Conceptual

expert

Error raised by invalid datetime string format

What error does this Python code raise?

Python

from datetime import datetime

invalid_date = '2024/06/15'
dt = datetime.strptime(invalid_date, '%Y-%m-%d')

A ValueError

B TypeError

C SyntaxError

D KeyError

💡 Hint

The string format does not match the expected format.

The string uses slashes but the format expects dashes, causing a ValueError during parsing.

Practice

1. Which Python module is commonly used to work with dates and times?

easy

A. os

B. math

C. random

D. datetime

Solution

1. Step 1: Recall Python modules for date/time

   The datetime module provides classes for manipulating dates and times.

2. Step 2: Identify unrelated modules

   math is for math functions, random for random numbers, os for operating system tasks.

3. Final Answer:

   datetime -> Option D

4. Quick Check:

   Module for date/time = datetime [OK]

Hint: Remember: datetime handles clocks and calendars [OK]

Common Mistakes:

• Confusing datetime with math or random modules
• Using os module for date/time
• Not importing datetime before use

2. Which of the following is the correct way to create a date object for January 1, 2024 using the datetime module?

easy

A. date = datetime(2024, 1, 1)

B. date = datetime.date('2024-01-01')

C. date = datetime.date(2024, 1, 1)

D. date = datetime.date(1, 1, 2024)

Solution

1. Step 1: Understand datetime.date constructor

   The date class constructor takes year, month, day as integers in that order.

2. Step 2: Check each option

   date = datetime.date(2024, 1, 1) uses correct syntax: datetime.date(2024, 1, 1). date = datetime(2024, 1, 1) misses .date. date = datetime.date('2024-01-01') passes a string, which is invalid. date = datetime.date(1, 1, 2024) has wrong argument order.

3. Final Answer:

   date = datetime.date(2024, 1, 1) -> Option C

4. Quick Check:

   date(year, month, day) = correct order [OK]

Hint: Use datetime.date(year, month, day) with integers [OK]

Common Mistakes:

• Passing date as string instead of integers
• Wrong argument order
• Missing .date after datetime

3. What will be the output of this code?

from datetime import date, timedelta
start = date(2024, 4, 25)
new_date = start + timedelta(days=10)
print(new_date)

medium

A. 2024-05-05

B. 2024-04-15

C. 2024-04-25

D. Error: unsupported operand

Solution

1. Step 1: Understand timedelta addition

   Adding timedelta(days=10) to April 25, 2024 adds 10 days.

2. Step 2: Calculate new date

   April 25 + 10 days = May 5, 2024.

3. Final Answer:

   2024-05-05 -> Option A

4. Quick Check:

   25 April + 10 days = 5 May [OK]

Hint: Add timedelta days to date to get new date [OK]

Common Mistakes:

• Subtracting days instead of adding
• Confusing timedelta with datetime
• Expecting string input for timedelta

4. Find the error in this code snippet:

from datetime import datetime
dt = datetime(2024, 2, 30)
print(dt)

medium

A. datetime() requires string arguments

B. February 30 is an invalid date

C. Missing import for timedelta

D. print() cannot display datetime objects

Solution

1. Step 1: Check date validity

   February 30 does not exist; February has max 29 days in leap years.

2. Step 2: Understand datetime constructor

   datetime() expects valid year, month, day integers; invalid dates cause ValueError.

3. Final Answer:

   February 30 is an invalid date -> Option B

4. Quick Check:

   Invalid date causes error [OK]

Hint: Check if date exists before creating datetime [OK]

Common Mistakes:

• Assuming all day numbers are valid
• Missing import errors
• Thinking print can't show datetime

5. You want to find how many days are between March 1, 2024 and April 15, 2024. Which code correctly calculates this?

hard

A. from datetime import date
start = date(2024, 3, 1)
end = date(2024, 4, 15)
days = (end - start).days
print(days)

B. from datetime import datetime
start = datetime(2024, 3, 1)
end = datetime(2024, 4, 15)
days = end + start
print(days)

C. from datetime import date
days = date(2024, 4, 15) - 45
print(days)

D. from datetime import timedelta
start = timedelta(days=2024)
end = timedelta(days=101)
days = end - start
print(days)

Solution

1. Step 1: Use date objects for subtraction

   Subtracting two date objects gives a timedelta representing the difference.

2. Step 2: Extract days from timedelta

   Access the .days attribute to get the number of days between dates.

3. Final Answer:

   45 -> Option A

4. Quick Check:

   April 15 - March 1 = 45 days [OK]

Hint: Subtract dates, then use .days to get difference [OK]

Common Mistakes:

• Adding dates instead of subtracting
• Subtracting integer from date
• Using timedelta incorrectly as date