Python Coding Challenge - Question with Answer (ID -300126)

 

Explanation:

1️⃣ Variable Initialization

x = 1

A variable x is created.

Its initial value is 1.

This value will be updated repeatedly inside the loop.

2️⃣ Loop Declaration

for i in range(1, 6):

This loop runs with i taking values:

1, 2, 3, 4, 5

The loop will execute 5 times.

3️⃣ Loop Body (Core Logic)

x *= i - x

This line is equivalent to:

x = x * (i - x)

Each iteration:

First computes (i - x)

Then multiplies the current value of x with that result

Stores the new value back into x

4️⃣ Iteration-by-Iteration Execution

🔹 Iteration 1 (i = 1)

Current x = 1

Calculation:

x = 1 × (1 − 1) = 0

Updated x = 0

🔹 Iteration 2 (i = 2)

Current x = 0

Calculation:

x = 0 × (2 − 0) = 0

Updated x = 0

🔹 Iteration 3 (i = 3)

Current x = 0

Calculation:

x = 0 × (3 − 0) = 0

Updated x = 0

🔹 Iteration 4 (i = 4)

Current x = 0

Calculation:

x = 0 × (4 − 0) = 0

Updated x = 0

🔹 Iteration 5 (i = 5)

Current x = 0

Calculation:

x = 0 × (5 − 0) = 0

Updated x = 0

5️⃣ Final Output

print(x)

After the loop ends, x is still 0.

So the output printed is:

0

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Python Coding Challenge - Question with Answer (ID -290126)

 

🔍 What does index() do?

list.index(value, start)
It returns the index of the first occurrence of value,
starting the search from position start.

---

Step-by-step execution:

1. arr = [10, 20, 30]
   Index positions:

   0 → 10
   1 → 20
   2 → 30

2. arr.index(20, 1) means:
   👉 Find the value 20, but start searching from index 1.

3. At index 1, the value is 20 → match found.

---

✅ Output:

1

---

⚠️ Tricky Case (important for quizzes)

arr.index(10, 1)

This will raise:

ValueError: 10 is not in list

Because it starts searching from index 1, and 10 is only at index 0.

---

Mental Rule (easy to remember):

index(x, i)

 = "Find x after index i-1"

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Python Coding Challenge - Question with Answer (ID -280126)

 

What’s happening?

1️⃣ t is a tuple

t = (1, 2, 3)

Tuples are immutable sequences.

---

2️⃣ __mul__() is the magic method for *

• t * n internally calls:

t.__mul__(n)

So this:

t * 0

and this:

t.__mul__(0)

are exactly the same.

---

3️⃣ Multiplying a tuple by 0

Rule:

Any sequence × 0 → empty sequence

So:

(1, 2, 3) * 0

becomes:

()

---

✅ Final Output

()

---

⚠️ Important Concept (Interview Favorite)

• The tuple itself is not modified

• A new empty tuple is created

• This works for:

  • lists

  • tuples

  • strings

Example:

print("abc" * 0) # ''
print([1,2] * 0) # []

---

 One-line takeaway

__mul__(0)

 returns an empty sequence of the same type

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Python Coding Challenge - Question with Answer (ID -270126)

 

1️⃣ async def foo()

async def foo(): 
     return 5

• This defines an asynchronous function.

• Calling foo() does NOT execute it immediately.

• It returns a coroutine object.

---

2️⃣ asyncio.run(foo())

asyncio.run(foo())

• asyncio.run():

  • Creates an event loop

  • Executes the coroutine foo()

  • Waits until it finishes

  • Returns the final result

✅ Since foo() returns 5, this line evaluates to:

5

---

3️⃣ + 1

asyncio.run(foo()) + 1

• Now it becomes:

5 + 1

• Result:

6

---

4️⃣ print(...)

print(6)

---

✅ Final Output

6

---

⚠️ Important Trick

If you wrote this instead:

print(foo() + 1)

❌ It would raise an error because:

• foo() returns a coroutine, not an integer

---

💡 Key Takeaway

• async def → returns a coroutine

• asyncio.run() → executes it and returns the result

• You can only use the returned value after awaiting/running

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Python Coding Challenge - Question with Answer (ID -230126)

 

Explanation:

Line 1: Global Variable Initialization

x = 1

A global variable x is created.

Current value of x → 1

Line 2–5: Function Definition

def f():

    global x

    x += 2

def f():

Defines a function named f.

global x

Tells Python that x inside this function refers to the global variable, not a local one.

x += 2

Updates the global x.

When f() runs:

x becomes 1 + 2 = 3

Important:

The function does NOT return anything, so by default it returns:

None

Line 6: Function Call Inside Expression (TRICKY)

x = x + f()

Step-by-step execution order:

f() is called first

Inside f():

x becomes 3

returns None

Expression becomes:

x = 1 + None

Line 6 Result: ERROR

TypeError: unsupported operand type(s) for +: 'int' and 'NoneType'

Why?

You cannot add:

an int (1)

with None

f() returned None, not a number

Line 7: Print Statement

print(x)

This line never runs, because the program crashes on the previous line.

Final Outcome

Output:

TypeError

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Python Coding Challenge - Question with Answer (ID -220126)

 

🔹 Step 1: Create Dictionary

d = {'a': 1}

Dictionary d has one key:

{'a': 1}

---

🔹 Step 2: Get Keys View

k = d.keys()

• d.keys() does NOT return a list

• It returns a dictionary view object: dict_keys

• This view is dynamic (live)

So k is linked to d, not a snapshot.

---

🔹 Step 3: Modify Dictionary

d['b'] = 2

Now dictionary becomes:

{'a': 1, 'b': 2}

Because k is a live view, it automatically reflects this change.

---

🔹 Step 4: Print Keys

print(list(k))

k now sees both keys:

['a', 'b']

---

✅ Final Output

['a', 'b']

---

🔥 Key Takeaways (Important for Interviews)

• dict.keys(), dict.values(), dict.items() are dynamic views

• They update automatically when the dictionary changes

• To freeze keys, use:

list(d.keys())

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Python Coding Challenge - Question with Answer (ID -210126)

 

Explanation:

Create an empty list

x = []

x is an empty list

Memory-wise, x is a mutable object

Create a single-element tuple

t = (x,)

t is a tuple containing one element

That element is the same list x

Important: the comma , makes it a tuple

Structure so far:

t → ( x )

x → []

 Append the tuple to the list

x.append(t)

You append t inside the list x

Now x contains t

Since t already contains x, this creates a circular reference

Circular structure formed:

t → ( x )

x → [ t ]

Visually:

t

└── x

    └── t

        └── x

            └── ...

Access deeply nested elements

print(len(t[0][0][0]))

Let’s break it down step by step:

🔹 t[0]

First element of tuple t

This is x

🔹 t[0][0]

First element of list x

This is t

🔹 t[0][0][0]

First element of tuple t

This is again x

So:

t[0][0][0] == x

Final Step: len()

len(x)

x contains exactly one element

That element is t

Therefore:

print(len(t[0][0][0]))  # → 1

Final Output

1

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Python Coding Challenge - Question with Answer (ID -200126)

 

🔹Step 1 — Understand range(len(arr))

len(arr) = 4, so:

range(4) → 0, 1, 2, 3

So the loop runs 4 times with i = 0, 1, 2, 3.

---

🔹 Step 2 — Understand negative indexing

In Python:

Index	0	1	2	3
Value	1	2	3	4
Negative index	-4	-3	-2	-1

---

🔹 Step 3 — Iterate and modify the list

Iteration 1: i = 0

arr[0] = arr[-0]

But -0 == 0, so:

arr[0] = arr[0] = 1

Array remains:
👉 [1, 2, 3, 4]

---

Iteration 2: i = 1

arr[1] = arr[-1] 

arr[-1] = 4

So:

arr[1] = 4

Array becomes:
👉 [1, 4, 3, 4]

---

Iteration 3: i = 2

arr[2] = arr[-2] 

arr[-2] = 3

So:

arr[2] = 3

Array becomes:
👉 [1, 4, 3, 4] (no change here)

---

Iteration 4: i = 3

arr[3] = arr[-3]

arr[-3] = 4 (because arr is now [1,4,3,4])

So:

arr[3] = 4

Final array:
👉 [1, 4, 3, 4]

---

✅ Final Output

[1, 4, 3, 4]

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Python Coding Challenge - Question with Answer (ID -190126)

 

Let’s break this down line by line 👇

lst = [1, 2, 3]

Here, you create a list named lst with three elements: 1, 2, and 3.

So right now:

lst = [1, 2, 3]

Length = 3

---

lst.append([4, 5])

Now you use append(). Important point:

👉 append() adds the ENTIRE item as a single element

You are NOT adding 4 and 5 separately.
You are adding a list [4, 5] as one single element.

So after append:

lst = [1, 2, 3, [4, 5]]

Now the list contains 4 elements:

1 

2

 3

[4, 5] → (this is one nested list, counted as ONE element)

---

print(len(lst))

len(lst) counts total elements in the outer list.

So the output is:

4

---

 Common Confusion (Important!)

If you had used extend() instead of append(), the result would be different:

lst = [1, 2, 3]
lst.extend([4, 5])
print(len(lst))

Now:

lst = [1, 2, 3, 4, 5]

Output would be 5

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Python Coding Challenge - Question with Answer (ID -180126)

 

Step 1: Creating the tuple

t = (1, 2, [3, 4])

Here, t is a tuple containing:

• 1 → integer (immutable)

• 2 → integer (immutable)

• [3, 4] → a list (mutable)

So the structure is:

t = (immutable, immutable, mutable)

Step 2: Understanding t[2] += [5]

t[2] += [5]

This line is equivalent to:

t[2] = t[2] + [5]

Now two things happen internally:

👉 First: The list inside the tuple is modified

Because lists are mutable, t[2] (which is [3, 4]) gets updated in place to:

[3, 4, 5]

👉 Second: Python tries to assign back to the tuple

After modifying the list, Python tries to do:

t[2] = [3, 4, 5]

But this fails because tuples are immutable — you cannot assign to an index in a tuple.

❌ Result: Error occurs

You will get this error:

TypeError: 'tuple' object does not support item assignment

❗ Important Observation (Tricky Part)

Even though an error occurs, the internal list actually gets modified before the error.

So if you check the tuple in memory after the error, it would conceptually be:

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

But print(t) never runs because the program crashes before that line.

---

✅ If you want this to work without error:

Use this instead:

t[2].append(5)
print(t)

Output:

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

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Python Coding Challenge - Question with Answer (ID -170126)

 

Code Explanation:

1. Global Variable Definition

x = 10

A global variable x is created.

Value of x is 10.

This variable is accessible outside functions.

2. Function Definition (outer)

def outer():

A function named outer is defined.

No code runs at this point.

Function execution starts only when called.

3. Local Variable Inside Function

    x = 5

A local variable x is created inside outer.

This shadows the global x.

This x exists only within outer().

4. Lambda Function and Closure

    return map(lambda y: y + x, range(3))

range(3) produces values: 0, 1, 2

The lambda function:

Uses variable x

Captures x from outer’s local scope

This behavior is called a closure

map() is lazy, so no calculation happens yet.

A map object is returned.

5. Global Variable Reassignment

x = 20

The global x is updated from 10 to 20.

This does not affect the lambda inside outer.

Closures remember their own scope, not global changes.

6. Function Call and Map Evaluation

result = list(outer())

outer() is called.

Inside outer():

x = 5 is used

list() forces map() execution:

y Calculation Result

0 0 + 5 5

1 1 + 5 6

2 2 + 5 7

Final list becomes:

[5, 6, 7]

7. Output

print(result)

Output:

[5, 6, 7]

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Python Coding Challenge - Question with Answer (ID -160126)

Code Explanation:

Tuple Initialization

t = (5, -5, 10)

A tuple named t is created.

It contains both positive and negative integers.

Tuples are immutable, meaning their values cannot be changed.

Creating a Tuple of Absolute Values

u = tuple(abs(i) for i in t)

Each element of tuple t is passed to the abs() function.

Negative numbers are converted to positive.

A new tuple u is formed:

u = (5, 5, 10)

Reversing the Tuple

v = u[::-1]

Tuple slicing with step -1 is used.

This reverses the order of elements in u.

Resulting tuple:

v = (10, 5, 5)

Summing Selected Elements Using Index

x = sum(v[i] for i in range(1, len(v)))

range(1, len(v)) generates indices 1 and 2.

Elements selected:

v[1] = 5

v[2] = 5

These values are added together.

Result:

x = 10

Displaying the Output

print(x)

Prints the final calculated value.

Final Output

10

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Python Coding Challenge - Question with Answer (ID -150126)

 

Explanation:

1. Create a dictionary of scores

scores = {"A": 85, "B": 72, "C": 90}

This stores names as keys and their scores as values.

2. Set the minimum score option

opt = {"min": 80}

This option decides the minimum score required to pass the filter.

3. Filter dictionary keys

res = list(filter(lambda k: scores[k] >= opt["min"], scores))

filter loops over the dictionary keys ("A", "B", "C").

scores[k] gets each key’s value.

Keeps only keys whose score is ≥ 80.

list() converts the result into a list.

4. Print the filtered result

print(res)

Displays the keys that passed the condition.

Output

['A', 'C']

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Python Coding Challenge - Question with Answer (ID -140126)

 

Step-by-step execution

1. lst = [10, 20, 30]
   A list with three elements is created.

2. for i in lst:
   The loop iterates over each element in the list one by one.

---

▶️ Iteration 1

i = 10

• Check: if i == 20 → False

• So print(i) runs → prints 10

---

▶️ Iteration 2

i = 20

• Check: if i == 20 → True

• break executes → the loop stops immediately

• print(i) is not executed for 20

---

▶️ Iteration 3

• Never runs, because the loop already stopped at break.

---

✅ Final Output

10

---

Key Concepts

Keyword	Meaning
for	Loops through each element in a list
if	Checks a condition
break	Stops the loop immediately
print(i)	Prints the current value

---

Summary

• The loop starts printing values from the list.

• When it encounters 20, break stops the loop.

• So only 10 is printed.

👉 Output: 10

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Python Coding Challenge - Question with Answer (ID -130126)

 

Let’s explain it step-by-step:

s = {1, 2, 3}
s.add([4, 5])
print(s)

---

 What happens here?

Line 1

s = {1, 2, 3}

A set is created with three integers.

---

Line 2

s.add([4, 5])

Here you are trying to add a list [4, 5] into a set.

👉 But sets can only store hashable (immutable) objects.

• int, float, str, tuple → hashable ✅

• list, dict, set → not hashable ❌

A list is mutable, so Python does not allow it inside a set.

So this line raises an error:

TypeError: unhashable type: 'list'

---

Line 3

print(s)

This line is never executed, because the program stops at the error in line 2.

---

❌ Final Output

There is no output — instead you get:

TypeError: unhashable type: 'list'

---

✅ How to fix it?

If you want to store 4 and 5 as a group inside the set, use a tuple:

s = {1, 2, 3}
s.add((4, 5))
print(s)

Output (order may vary):

{1, 2, 3, (4, 5)}

Or if you want to add them as individual elements:

s = {1, 2, 3}
s.update([4, 5])
print(s)

Output:

{1, 2, 3, 4, 5}

---

🔑 Key Concept

Sets only allow immutable (hashable) elements.
That’s why lists can’t go inside sets — but tuples can 

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Python Coding Challenge - Question with Answer (ID -120126)

 

 Step 1: Create the tuple

t = (1, 2, 3)

tuple is immutable → its values cannot be changed.

---

🔹 Step 2: Loop through the tuple

for i in t:

Each element of t is assigned one by one to the variable i:

Loop iteration	i value
1st	1
2nd	2
3rd	3

---

🔹 Step 3: Modify i

i = i * 2

This only changes the local variable i, not the tuple.

So:

• When i = 1 → i becomes 2

• When i = 2 → i becomes 4

• When i = 3 → i becomes 6

But t never changes because:

• You are not assigning back to t

• And tuples cannot be modified in place

---

🔹 Step 4: Print the tuple

print(t)

Since t was never changed, output is:

(1, 2, 3)

---

✅ Final Answer

Output:

(1, 2, 3)

---

Key Concept

Point	Explanation
Tuples are immutable	Their values cannot be changed
i is just a copy	Changing i does not affect t
No assignment to t	So t stays the same

---

🔹 If you actually want to double the values:

t = tuple(i * 2 for i in t)
print(t)

Output:

(2, 4, 6)

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Python Coding Challenge - Question with Answer (ID -090126)

 

Step 1: Understand range(3)

range(3) → 0, 1, 2

So the loop runs with i = 0, then 1, then 2.

---

🔹 Step 2: Loop execution

i	Condition i > 1	Action
0	False	Go to else → print(0)
1	False	Go to else → print(1)
2	True	continue → skip print

---

🔹 What continue does

continue skips the rest of the loop body and jumps to the next iteration.

So when i == 2, continue is executed → print(i) is skipped.

---

✅ Final Output

0
1

---

🔹 Key Points

• else runs only when if condition is False.

• continue skips the remaining code in the loop for that iteration.

• So 2 is never printed.

---

In one line:

This code prints all values of i less than or equal to 1 and skips values greater than 1.

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Python Coding Challenge - Question with Answer (ID -080126)

 

Step-by-Step Explanation

1️⃣ Create the original list

arr = [1, 2, 3]

arr points to a list in memory:

arr → [1, 2, 3]

---

2️⃣ Make a copy of the list

arr3 = arr.copy()

• .copy() creates a new list object with the same values.

• arr3 is a separate list, not a reference to arr.

arr → [1, 2, 3]
arr3 → [1, 2, 3] (different memory location)

---

3️⃣ Modify the copied list

arr3.append(4)

This only changes arr3:

arr → [1, 2, 3]
arr3 → [1, 2, 3, 4]

---

4️⃣ Print original list

print(arr)

✅ Output

[1, 2, 3]

---

Key Concept

Operation	Effect
.copy()	Creates a new list
=	Only creates a reference
append()	Modifies list in place

---

 Compare with reference assignment:

arr = [1, 2, 3]
arr3 = arr # reference, not copy
arr3.append(4)
print(arr)

Output:

[1, 2, 3, 4]

Because both variables point to the same list.

---

Final Answer

👉 arr remains unchanged because arr3 is a separate copy, not a reference.

Printed Output:

[1, 2, 3]

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Python Coding Challenge - Question with Answer (ID -070126)

 

Step 1

t = (1, 2, 3)

A tuple t is created with values:
(1, 2, 3)

---

🔹 Step 2

t[0] = 10

Here you are trying to change the first element of the tuple.

❗ Problem:
Tuples are immutable, meaning once created, their elements cannot be changed.

So Python raises an error:

TypeError: 'tuple' object does not support item assignment

---

🔹 Step 3

print(t)

This line is never executed because the program stops when the error occurs in the previous line.

---

✅ Final Result

The code produces an error, not output:

TypeError: 'tuple' object does not support item assignment

---

Key Concept

Data Type	Mutable?
List ([])	Yes
Tuple (())	❌ No

---

✔️ Correct Way (if you want to modify it)

Convert tuple to list, modify, then convert back:

t = (1, 2, 3)
lst = list(t)
lst[0] = 10
t = tuple(lst)
print(t)

Output:

(10, 2, 3)

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Python Coding Challenge - Question with Answer (ID -060126)

 

Step-by-step Explanation

Line 1

lst = [1, 2, 3]

You create a list:

lst → [1, 2, 3]

---

Line 2

lst = lst.append(4)

This is the key tricky part.

• lst.append(4) modifies the list in-place

• But append() returns None

So this line becomes:

lst = None

Because:

lst.append(4) → None

The list is updated internally, but you overwrite lst with None.

---

Line 3

print(lst)

Since lst is now None, it prints:

None

---

✅ Final Output

None

---

 Why this happens

Function	Modifies list	Returns value
append()	Yes	None

So you should not assign it back.

---

✅ Correct way to do it

lst = [1, 2, 3]
lst.append(4)
print(lst)

Output:

[1, 2, 3, 4]

---

 Summary

• append() changes the list but returns None

• Writing lst = lst.append(4) replaces your list with None

• Always use lst.append(...) without assignment

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