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jan-2026-python-week-7-collections-3

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Collections 3

Course: Jan 2026 - Python

Collections 3

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 3:50 AM IST

Private Tests

2/2 Passed

Submitted on Apr 01, 2026 at 3:50 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.

Number of unique common digits between two integers**

Given two integers, return the number of unique digits that are common in both numbers.

Eg, 287498,295424 - 2, 4 and 9 are common to both nums so answer is 3

Template Code(Click to Expand)



def number_of_unique_common_digits(n1: int, n2: int) -> int:
    '''
    Given two integers, return the number of unique digits that are common in both numbers.
    Eg, 287498,295424 - 2, 4 and 9 are common to both nums so answer is 3
    Arguments:
    n1: int - the first number
    n2: int - the second number

    Return:
    int - the number of unique common digits.
    '''
    ...

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
is_equal(number_of_unique_common_digits(12345, 54321), 5)

Case 1

Expected Output:


text
5

Case 1

Actual Output:


text
5

Case 2

Input:


text
is_equal(number_of_unique_common_digits(287498, 295424), 3)

Case 2

Expected Output:


text
3

Case 2

Actual Output:


text
3

Case 3

Input:


text
is_equal(number_of_unique_common_digits(67890, 9876), 4)

Case 3

Expected Output:


text
4

Case 3

Actual Output:


text
4

Private Tests ( 2/2 )

Case 1

Input:


text
is_equal(
    number_of_unique_common_digits(98765, 56789),
    5
)
is_equal(
    number_of_unique_common_digits(42245, 49860),
    1
)
is_equal(
    number_of_unique_common_digits(1234567890, 9876543210),
    10
)

Case 1

Expected Output:


text
5

1

10

Case 1

Actual Output:


text
5

1

10

Case 2

Input:


text
is_equal(
    number_of_unique_common_digits(1234567890, 9876543210),
    10
)
is_equal(
    number_of_unique_common_digits(1234567890, 1111155555),
    2
)

Case 2

Expected Output:


text
10

2

Case 2

Actual Output:


text
10

2

💻 IITM Official Solution


python

def number_of_unique_common_digits(n1: int, n2: int) -> int:
    '''
    Given two integers, return the number of unique digits that are common in both numbers.
    Eg, 287498,295424 - 2, 4 and 9 are common to both nums so answer is 3
    Arguments:
    n1: int - the first number
    n2: int - the second number

    Return:
    int - the number of unique common digits.
    '''

    return len(set(str(n1)) & set(str(n2)))

💻 My Submitted Code


python
def number_of_unique_common_digits(n1: int, n2: int) -> int:
    # 1. Convert integers to strings, then to sets to get unique digits
    set1 = set(str(n1))
    set2 = set(str(n2))

    # 2. Use the set intersection operator (&) to find common digits
    common_digits = set1 & set2

    # 3. Return the count
    return len(common_digits)