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jan-2026-python-week-7-numbers-2

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Numbers 2

Course: Jan 2026 - Python

Numbers 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

2/2 Passed

Submitted on Apr 01, 2026 at 3:46 AM IST

Private Tests

2/2 Passed

Submitted on Apr 01, 2026 at 3:46 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.

Create a dictionary with elements in the list as values and the indices as keys.**

Given a list of items, create a dictionary with the indices as keys and the items as items.

Template Code(Click to Expand)



def create_indexed_dict(items: list) -> dict:
    '''
    Given a list of items, create a dictionary with the indices as keys and the items as items.

    Args:
        items (list): A list of items.

    Returns:
        dict: A dictionary with indices as keys and items as items.
    '''
    ...

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Public Tests ( 2/2 )

Case 1

Input:


text
names1 = ["Alice", "Bob", "Charlie", "David"]
expected1 = {0: "Alice", 1: "Bob", 2: "Charlie", 3: "David"}
is_equal(create_indexed_dict(names1), expected1)

Case 1

Expected Output:


text
{0: 'Alice', 1: 'Bob', 2: 'Charlie', 3: 'David'}

Case 1

Actual Output:


text
{0: 'Alice', 1: 'Bob', 2: 'Charlie', 3: 'David'}

Case 2

Input:


text
names1 = ["Apple", "Banana", "Cherry", "Date"]
expected1 = {0: "Apple", 1: "Banana", 2: "Cherry", 3: "Date"}
is_equal(create_indexed_dict(names1), expected1)

Case 2

Expected Output:


text
{0: 'Apple', 1: 'Banana', 2: 'Cherry', 3: 'Date'}

Case 2

Actual Output:


text
{0: 'Apple', 1: 'Banana', 2: 'Cherry', 3: 'Date'}

Private Tests ( 2/2 )

Case 1

Input:


text
names = ["Eve", "Frank", "Grace", "Hank", "Ivy"]
expected = {0: "Eve", 1: "Frank", 2: "Grace", 3: "Hank", 4: "Ivy"}
is_equal(create_indexed_dict(names), expected)

Case 1

Expected Output:


text
{0: 'Eve', 1: 'Frank', 2: 'Grace', 3: 'Hank', 4: 'Ivy'}

Case 1

Actual Output:


text
{0: 'Eve', 1: 'Frank', 2: 'Grace', 3: 'Hank', 4: 'Ivy'}

Case 2

Input:


text
names = ["Jack", "Kathy", "Leo"]
expected = {0: "Jack", 1: "Kathy", 2: "Leo"}
is_equal(create_indexed_dict(names), expected)

Case 2

Expected Output:


text
{0: 'Jack', 1: 'Kathy', 2: 'Leo'}

Case 2

Actual Output:


text
{0: 'Jack', 1: 'Kathy', 2: 'Leo'}

💻 IITM Official Solution


python

def create_indexed_dict(items: list) -> dict:
    '''
    Given a list of items, create a dictionary with the indices as keys and the items as items.

    Args:
        items (list): A list of items.

    Returns:
        dict: A dictionary with indices as keys and items as items.
    '''


    # basic approach
    d = {}
    for i in range(len(items)):
        d[i] = items[i]
    return d

    # other approaches

    # using comprehensions
    # return {i: item for i, item in enumerate(items)}

    # functional
    # return dict(enumerate(items))

💻 My Submitted Code


python
def create_indexed_dict(items: list) -> dict:
    # Option 1: Dictionary Comprehension
    # return {i: items[i] for i in range(len(items))}

    # Option 2: Using enumerate (Cleaner/Pythonic)
    return dict(enumerate(items))