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jan-2026-python-week-7-collections-1

683 words

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Collections 1

Course: Jan 2026 - Python

Collections 1

Submission deadline has passed for this assignment

Due Apr 01, 2026 at 11:59 PM IST

Instructions

Use "Test Run" to verify your code with public test cases.

Press "Submit" to have your assignment evaluated.

You can submit your assignment multiple times up until the deadline.

Make sure to submit your final code by the deadline to receive your score.

Summary

100 out of100

Score

Public Tests

3/3 Passed

Submitted on Apr 01, 2026 at 3:49 AM IST

Private Tests

3/3 Passed

Submitted on Apr 01, 2026 at 3:49 AM IST

**Change in eligibility criteria to write oppe2 exam: A5>=40/100 AND A6>=40/100 AND A7>=40/100 AND A8>=40/100. and becoming eligible to give the end term exam.

Reverse First half in an even length tuple**

Given an even-length tuple t, return a new tuple where the first half of the tuple is reversed, and the second half remains unchanged.

Example


t = (1, 2, 3, 4, 5, 6)

The first half (1, 2, 3) is reversed to (3, 2, 1), so the result is (3, 2, 1, 4, 5, 6).

Template Code(Click to Expand)



def reverse_first_half(t: tuple) -> tuple:
    '''
    Given an even-length tuple, return a new tuple where the first half
    is reversed, and the second half remains unchanged.

    Arguments:
    t: tuple - an even-length tuple.

    Return: tuple - a new tuple with the first half reversed.
    '''
    ...

NOTE: You can use the below tools for working out and debugging. Click to open them in new tab.

PythonTutor | Starboard Notebook | Pyodide Terminal

Public Tests ( 3/3 )

Case 1

Input:


text
t = (10, 20, 30, 40, 50, 60)
is_equal(
    reverse_first_half(t),
    (30, 20, 10, 40, 50, 60)
)

Case 1

Expected Output:


text
(30, 20, 10, 40, 50, 60)

Case 1

Actual Output:


text
(30, 20, 10, 40, 50, 60)

Case 2

Input:


text
t = ('a', 'b', 'c', 'd')
is_equal(
    reverse_first_half(t),
    ('b', 'a', 'c', 'd')
)

Case 2

Expected Output:


text
('b', 'a', 'c', 'd')

Case 2

Actual Output:


text
('b', 'a', 'c', 'd')

Case 3

Input:


text
t = (1, 2)
is_equal(
    reverse_first_half(t),
    (1, 2)
)

Case 3

Expected Output:


text
(1, 2)

Case 3

Actual Output:


text
(1, 2)

Private Tests ( 3/3 )

Case 1

Input:


text
t = (1, 2, 3, 4, 5, 6, 7, 8)
is_equal(
    reverse_first_half(t),
    (4, 3, 2, 1, 5, 6, 7, 8)
)

Case 1

Expected Output:


text
(4, 3, 2, 1, 5, 6, 7, 8)

Case 1

Actual Output:


text
(4, 3, 2, 1, 5, 6, 7, 8)

Case 2

Input:


text
t = (100, 200, 300, 400)
is_equal(
    reverse_first_half(t),
    (200, 100, 300, 400)
)

Case 2

Expected Output:


text
(200, 100, 300, 400)

Case 2

Actual Output:


text
(200, 100, 300, 400)

Case 3

Input:


text
t = (10, 20, 30, 40, 50, 60, 70, 80)
is_equal(
    reverse_first_half(t),
    (40, 30, 20, 10, 50, 60, 70, 80)
)

Case 3

Expected Output:


text
(40, 30, 20, 10, 50, 60, 70, 80)

Case 3

Actual Output:


text
(40, 30, 20, 10, 50, 60, 70, 80)

💻 IITM Official Solution


python

def reverse_first_half(t: tuple) -> tuple:
    '''
    Given an even-length tuple, return a new tuple where the first half
    is reversed, and the second half remains unchanged.

    Arguments:
    t: tuple - an even-length tuple.

    Return: tuple - a new tuple with the first half reversed.
    '''


    mid = len(t) // 2
    return t[:mid][::-1] + t[mid:]

💻 My Submitted Code


python
def reverse_first_half(t: tuple) -> tuple:
    # 1. Find the midpoint
    mid = len(t) // 2

    # 2. Slice the first half and reverse it using [::-1]
    first_half_reversed = t[:mid][::-1]

    # 3. Slice the second half (remains unchanged)
    second_half = t[mid:]

    # 4. Concatenate and return
    return first_half_reversed + second_half