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jan-2026-python-week-7-strings-2

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Strings 2

Course: Jan 2026 - Python

Strings 2

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 3:48 AM IST

Private Tests

2/2 Passed

Submitted on Apr 01, 2026 at 3:48 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.

Check if an even-length string has A or a in the second half**

Given an even-length string s, check if the second half contains the character "a" or "A". Return True if it does, otherwise return False.

Example


s = "abcDef"

The second half is "Def", which does not contain "a" or "A", so the result is False.

Template Code(Click to Expand)




def has_a_in_second_half(s: str) -> bool:
    '''
    Given an even-length string, check if the second half contains
    the character "a" or "A".

    Arguments:
    s: str - an even-length string.

    Return: bool - True if "a" or "A" is found in the second half, else False.
    '''
    ...

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
s = "HelloWorld"
is_equal(
    has_a_in_second_half(s),
    False
)

Case 1

Expected Output:


text
False

Case 1

Actual Output:


text
False

Case 2

Input:


text
s = "whenwhat"
is_equal(
    has_a_in_second_half(s),
    True
)

Case 2

Expected Output:


text
True

Case 2

Actual Output:


text
True

Case 3

Input:


text
s = "whenWHAT"
is_equal(
    has_a_in_second_half(s),
    True
)

Private Tests ( 2/2 )

Case 1

Input:


text
s = "first-half"
is_equal(
    has_a_in_second_half(s),
    True
)
s = "half-first"
is_equal(
    has_a_in_second_half(s),
    False
)
s = "thisthAt"
is_equal(
    has_a_in_second_half(s),
    True
)

Case 1

Expected Output:


text
True

False

True

Case 1

Actual Output:


text
True

False

True

Case 2

Input:


text
s = "concatenation"
is_equal(
    has_a_in_second_half(s),
    True
)
s = "BANANA"
is_equal(
    has_a_in_second_half(s),
    True
)
s = "APPLETON"
is_equal(
    has_a_in_second_half(s),
    False
)

Case 2

Expected Output:


text
True

True

False

Case 2

Actual Output:


text
True

True

False

💻 IITM Official Solution


python


def has_a_in_second_half(s: str) -> bool:
    '''
    Given an even-length string, check if the second half contains
    the character "a" or "A".

    Arguments:
    s: str - an even-length string.

    Return: bool - True if "a" or "A" is found in the second half, else False.
    '''


    mid = len(s) // 2
    return 'a' in s[mid:] or 'A' in s[mid:]
    # alternate way
    # return 'a' in s[mid:].lower()d

💻 My Submitted Code


python
def has_a_in_second_half(s: str) -> bool:
    # Find the midpoint
    mid = len(s) // 2
    # Slice the second half
    second_half = s[mid:]

    # Check for 'a' or 'A' in the sliced part
    # A more concise way: 'a' in second_half.lower()
    return 'a' in second_half or 'A' in second_half