Python Coding Challenge - Question with Answer (ID -100126)

 

Explanation:

1. Function Definition

def make_filter(limit, seen=[]):

What happens:

Defines a function make_filter.

Takes:

limit → a threshold value.

seen → a mutable default list shared across calls.

Important: seen is created once when the function is defined, not each time it is called.

2. Returning a Lambda Function

return lambda x: x > limit and not seen.append(x)

This returns a function that:

Checks if x > limit

Appends x to seen

Uses not to invert the return value of seen.append(x)

Since:

seen.append(x) → returns None

not None → True

So the lambda returns:

True if x > limit

False otherwise

But it also mutates the seen list.

3. Creating the Filter Function

f = make_filter(2)

This means:

limit = 2

seen = [] (shared list)

f is now:

lambda x: x > 2 and not seen.append(x)

4. Applying filter

filter(f, [1,2,3,4,3,5])

5. Why duplicates are not removed

Because:

seen.append(x) is always executed for x > limit.

No check is done to prevent duplicates.

The lambda only tests x > limit.

So every value > 2 passes, including repeated 3.

6. Final Output

print(list(filter(f, [1,2,3,4,3,5])))

Output:

[3, 4, 3, 5]

Final Answer

[3, 4, 3, 5]

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