Python Coding Challenge - Question with Answer (ID -151225)

 

Explanation:

1. List Initialization

a = [1, 2]

b = [3, 4]

Two lists are created:

a contains values 1 and 2

b contains values 3 and 4

2. Outer Loop

for i in a:

Iterates over each element in list a

First iteration: i = 1

Second iteration: i = 2

3. Inner Loop with zip()

for x, y in zip(a, b):

zip(a, b) pairs elements from both lists:

First pair: (1, 3)

Second pair: (2, 4)

The loop runs twice for each value of i

4. Calculation and Printing

print(i + x + y)

Adds:

i from the outer loop

x from list a

y from list b

Prints the result in each iteration

5. Execution Flow

When i = 1:

1 + 1 + 3 = 5

1 + 2 + 4 = 7

When i = 2:

2 + 1 + 3 = 6

2 + 2 + 4 = 8

6. Final Output

5

7

6

8

Total print statements = 4

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Python Coding Challenge - Question with Answer (ID -141225)

 

Key Idea (Very Important )

• The for x in a loop iterates over the original elements of the list, not the updated ones.

• Changing a[0] does not affect the sequence of values that x will take.

---

Step-by-step Execution

Initial values:

a = [1, 2, 3, 4]
total = 0

Iteration 1

x = 1 

total = 0 + 1 = 1 

a[0] = 1

a = [1, 2, 3, 4]

Iteration 2

x = 2 

total = 1 + 2 = 3 

a[0] = 3

a = [3, 2, 3, 4]

Iteration 3

x = 3 

total = 3 + 3 = 6 

a[0] = 6

a = [6, 2, 3, 4]

Iteration 4

x = 4 

total = 6 + 4 = 10 

a[0] = 10

a = [10, 2, 3, 4]

---

Final Output

[10, 2, 3, 4]

---

 Takeaway

• Modifying a list inside a for loop does not change the iteration values.

• The loop uses an internal index to fetch the next element.

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Python Coding Challenge - Question with Answer (ID -131225)

 

Explanation:

Initial List Creation

a = [0, 1, 2]

A list a is created with three elements.

Length of a = 3

Loop Setup

for i in range(len(a)):

len(a) is evaluated once at the start → 3

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

Loop will run 3 times, even though a changes later

First Iteration (i = 0)

a = list(map(lambda x: x + a[i], a))

a[i] → a[0] → 0

map() adds 0 to every element of a

Calculation:

[0+0, 1+0, 2+0] → [0, 1, 2]

Updated list:

a = [0, 1, 2]

Second Iteration (i = 1)

a[i] → a[1] → 1 (from updated list)

Calculation:

[0+1, 1+1, 2+1] → [1, 2, 3]

Updated list:

a = [1, 2, 3]

Third Iteration (i = 2)

a[i] → a[2] → 3 (value has changed)

Calculation:

[1+3, 2+3, 3+3] → [4, 5, 6]

Updated list:

a = [4, 5, 6]

Final Output

print(a)

Output

[4, 5, 6]

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Python Coding Challenge - Question with Answer (ID -121225)

 

Step-by-step Explanation

1️⃣ lst = [10, 20, 30]

You start with a list of three numbers.

2️⃣ for i in range(len(lst)):

len(lst) = 3, so range(3) generates:

i = 0
i = 1
i = 2

You will loop three times, using each index of the list.

---

3️⃣ Inside the loop: lst[i] += i

This means:

lst[i] = lst[i] + i

Now update values step-by-step:

---

▶ When i = 0

lst[0] = lst[0] + 0
lst[0] = 10 + 0 = 10

List becomes:

[10, 20, 30]

---

▶ When i = 1

lst[1] = lst[1] + 1
lst[1] = 20 + 1 = 21

List becomes:

[10, 21, 30]
▶ When i = 2

lst[2] = lst[2] + 2
lst[2] = 30 + 2 = 32

List becomes:

[10, 21, 32]

---

Final Output

[10, 21, 32]

---

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✅ Summary

This program adds each element’s index to its value.

Index (i)	Original Value	Added	New Value
0	10	+0	10
1	20	+1	21
2	30	+2	32

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Python Coding Challenge - Question with Answer (ID -111225)

 

Final Output

[1 2 3]

👉 The array does NOT change.

---

Why the Array Doesn't Change

🔹 1. for i in a: gives you a copy of each element

• i is just a temporary variable

• It does NOT modify the original array a

So this line:

i = i * 2

✔️ Only changes i
❌ Does NOT change a

---

🔹 2. What Actually Happens Internally

Iteration steps:

Loop Step	i Value	i * 2	a (unchanged)
1	1	2	[1 2 3]
2	2	4	[1 2 3]
3	3	6	[1 2 3]

You never assign the new values back into a.

---

✅ Correct Way to Modify the NumPy Array

✅ Method 1: Using Index

for i in range(len(a)):
    a[i] = a[i] * 2
print(a)

✅ Output:

[2 4 6]

---

✅ Method 2: Vectorized NumPy Way (Best & Fastest)

a = a * 2
print(a)

✅ Output:

[2 4 6]

---

Key Concept (Exam + Interview Favorite)

❗ Looping directly over a NumPy array does NOT change the original array unless you assign back using the index.

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Python Coding Challenge - Question with Answer (ID -101225)

 

Step-by-Step Explanation

✅ Step 1: Function Definition

def f(x): 
    return x + 1

This function:

• Takes one number x

• Returns x + 1

✅ Example:
f(1) → 2
f(5) → 6

---

✅ Step 2: First map()

m = map(f, [1, 2, 3])

This applies f() to each element:

Original	After f(x)
1	2
2	3
3	4

 Important:

• map() does NOT execute immediately

• It creates a lazy iterator

So at this point:

m → (2, 3, 4) # not yet computed

---

✅ Step 3: Second map()

m = map(f, m)

Now we apply f() again on the result of the first map:

First map	Second map
2	3
3	4
4	5

So the final values become:

(3, 4, 5)

---

✅ Step 4: Convert to List

print(list(m))

This executes the lazy iterator and prints:

[3, 4, 5]

---

✅ Final Output

[3, 4, 5]

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Python Coding Challenge - Question with Answer (ID -091225)

 

Step-by-Step Execution

✅ Initial Values:

clcoding = [1, 2, 3, 4]
total = 0

---

1st Iteration

x = 1 

total = 0 + 1 = 1

          clcoding[0] = 1

 ✅ (no visible change)

clcoding = [1, 2, 3, 4]

---

2nd Iteration

x = 2 

total = 1 + 2 = 3
clcoding[0] = 3 ✅

clcoding = [3, 2, 3, 4]

---

3rd Iteration

x = 3

total = 3 + 3 = 6

clcoding[0] = 6 ✅

clcoding = [6, 2, 3, 4]

---

4th Iteration

x = 4

total = 6 + 4 = 10

clcoding[0] = 10 ✅

clcoding = [10, 2, 3, 4]

---

✅ Final Output

[10, 2, 3, 4]

---

Why This Is Tricky

• ✅ x comes from the original iteration sequence

• ✅ But you are modifying the same list during iteration

• ✅ Only index 0 keeps changing

• ✅ The loop still reads values 1, 2, 3, 4 safely

---

Key Concept

 Changing list values during iteration is allowed
 But changing list size can cause unexpected behavior

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Python Coding Challenge - Question with Answer (ID -081225)

 

Step 1: Initial List

clcoding = [1, 2, 3]

List length = 3

---

 Step 2: Understanding the Lambda

f = lambda x: (clcoding.append(0), len(clcoding))[1]

This line does two things at once using a tuple:

Part	What it Does
clcoding.append(0)	Adds 0 to the list
len(clcoding)	Gets updated length
[1]	Returns second value only

✅ So each time f(x) runs → list grows by 1 → new length is returned

---

 Step 3: map() is Lazy

m = map(f, clcoding)

 map() does NOT run immediately.
It runs only when next(m) is called.

---

 Step 4: Execution Loop (3 Times)

▶ First next(m)

• List before: [1, 2, 3]

• append(0) → [1, 2, 3, 0]

• len() → 4

• ✅ Prints: 4

---

▶ Second next(m)

• List before: [1, 2, 3, 0]

• append(0) → [1, 2, 3, 0, 0]

• len() → 5

• ✅ Prints: 5

---

▶ Third next(m)

• List before: [1, 2, 3, 0, 0]

• append(0) → [1, 2, 3, 0, 0, 0]

• len() → 6

• ✅ Prints: 6

---

 Final Output

4
5
6

---

Key Concepts Used (Interview Important)

• ✅ map() is lazy

• ✅ Mutable list modified during iteration

• ✅ Tuple execution trick inside lambda

• ✅ Side-effects inside functional calls

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Python Coding Challenge - Question with Answer (ID -061225)

 

Step-by-Step Explanation

1️⃣ Create the list

clcoding = [1, 2, 3, 4]

---

2️⃣ Apply filter

f = filter(lambda x: x % 2 == 0, clcoding)

✅ This creates a filter object (iterator) that will return only even numbers
✅ Important: filter() does NOT run immediately — it waits until you convert it to a list or loop over it.

---

3️⃣ Modify the original list

clcoding.append(6)

Now the list becomes:

[1, 2, 3, 4, 6]

---

4️⃣ Convert filter to list

print(list(f))

Now the filter actually runs, using the updated list.

✅ Even numbers from [1, 2, 3, 4, 6] are:

[2, 4, 6]

---

✅ Final Output

[2, 4, 6]

---

⭐ Key Concept (Trick)

✅ filter() is lazy — it evaluates only when needed,
✅ So it always uses the latest version of the list.

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Python Coding Challenge - Question with Answer (ID -051225)

 

1. What reduce() Does

reduce() combines all elements of a list into a single value using a function.

Example logic:

reduce(lambda x, y: x + y, [1, 2, 3])
# works like this:
# 1 + 2 = 3
# 3 + 3 = 6

---

🔹 2. Your List Has Only ONE Element

a = [5]

Since the list has only one element, there is:

• ❌ No second value to add

• ✅ So reduce() simply returns the same single value

So:

reduce(lambda x, y: x + y, [5]) → 5

---

🔹 3. The Loop Runs 3 Times

for i in range(3):

This runs for:

• i = 0

• i = 1

• i = 2
  👉 Total = 3 times

Each time, this runs:

print(reduce(lambda x, y: x + y, a))

And each time it prints:

5

---

✅ Final Output:

5
5
5

---

 Key Trick in This Code

• reduce() with a single-element list always returns that element

• The loop just repeats the same result

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Python Coding Challenge - Question with Answer (ID -041225)

 

Line-by-Line Explanation

✅ 1. Dictionary Created

d = {"x": 5, "y": 15}

• A dictionary with:

  • Key "x" → Value 5

  • Key "y" → Value 15

---

✅ 2. Initialize Sum Variable

s = 0

• s will store the final total.

---

✅ 3. Loop Through Values

for v in d.values():

• .values() returns only the values:

  5, 15

---

✅ 4. Conditional Addition (Ternary If-Else)

s += v if v > 10 else 2

This means:

• If v > 10 → add v

• Else → add 2

---

Loop Execution

Iteration	v	Condition v > 10	Added to s	New s
1st	5	❌ False	+2	2
2nd	15	✅ True	+15	17

---

✅ 5. Final Output

print(s)

✅ Output:

17

---

Key Concepts Used

✅ Dictionary
✅ Loop
✅ .values()
✅ Ternary if-else
✅ Accumulator variable

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Python Coding Challenge - Question with Answer (ID - 031225)

 

✅ Actual Output

[10 20 30]

---

Why didn’t the array change?

Even though we write:

i = i + 5

👉 This DOES NOT modify the NumPy array.

 What really happens:

Step	Explanation
for i in x:	i receives a copy of each value, not the original element
i = i + 5	Only changes the local variable i
x	The original NumPy array stays unchanged

So this loop works like:

i = 10 → i = 15 (x unchanged)
i = 20 → i = 25 (x unchanged)
i = 30 → i = 35 (x unchanged)

But x never updates, so output is still:

[10 20 30]

---

✅ Correct Way to Modify the NumPy Array

✔ Method 1: Using Index

for idx in range(len(x)):
    x[idx] = x[idx] + 5
print(x)

✅ Output:

[15 25 35]

---

✔ Method 2: Best & Fastest Way (Vectorized)

x = x + 5
print(x)

✅ Output:

[15 25 35]

---

Key Concept (IMPORTANT for Interviews)

Loop variable in NumPy gives a copy, not a reference.

 To change NumPy arrays, use indexing or vectorized operations.

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Python Coding Challenge - Question with Answer (ID -021225)

 

Step-by-Step Execution

Initial list

arr = [1, 2, 3]

---

✅ 1st Iteration

i = 1
arr.remove(1)

List becomes:

[2, 3]

Now Python moves to the next index (index 1).

---

✅ 2nd Iteration

Now the list is:

[2, 3]

Index 1 has:

i = 3

So:

arr.remove(3)

List becomes:

[2]

---

✅ Loop Ends

There is no next element to continue.

---

✅ Final Output

[2]

---

Why This Happens (Important Concept)

• You are modifying the list while looping over it

• When an element is removed:

  • The list shifts to the left

  • The loop skips elements

• This causes unexpected results

---

✅ Correct Way to Remove All Elements

✔️ Option 1: Loop on a copy

for i in arr[:]: 
    arr.remove(i)

✔️ Option 2: Use clear()

arr.clear()

---

Key Interview Point

❌ Never modify a list while iterating over it directly.

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Python Coding Challenge - Question with Answer (ID -011225)

 

Step-by-Step Execution

✅ range(3) generates:

0, 1, 2

✅ Initial value:

x = 0
🔁 Loop Iterations:

➤ 1st Iteration:

i = 0 
x = x + i = 0 + 0 = 0

➤ 2nd Iteration:

i = 1
x = x + i = 0 + 1 = 1

➤ 3rd Iteration:

i = 2
x = x + i = 1 + 2 = 3

---

✅ After the Loop Ends

• i remains stored as the last value → 2

x = 3

✅ Final Output:

2 3

---

Important Concept (Interview Tricky Point)

✅ In Python, loop variables (i) stay alive after the loop ends, unlike some other languages.

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Python Coding Challenge - Question with Answer (ID -301125)

 

What is happening?

🔹 1. map() creates an iterator

res = map(lambda x: x + 1, nums)

This does NOT create a list immediately. It creates a lazy iterator:

res → (2, 3, 4)

…but nothing is calculated yet.

---

🔹 2. The for loop consumes the iterator

for i in res: 
    pass

• This loop runs through all values: 2, 3, 4

• But since we used pass, nothing is printed

• Important: After this loop, the iterator is now EXHAUSTED

---

🔹 3. Printing after exhaustion

print(list(res))

• res is already empty

• So converting it to a list gives:

[]

---

✅ Final Output

[]

---

Key Tricky Rule

 A map() object can be used ONLY ONCE.
Once it is looped through, it becomes empty forever.

---

✅ Correct Way (If you want reuse)

res = list(map(lambda x: x + 1, nums))

for i in res:
    pass
print(res) # [2, 3, 4]

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Python Coding Challenge - Question with Answer (ID -291125)

 

Step-by-Step Explanation

1️⃣ Lists Creation

a = [1, 2, 3]
b = [10, 20, 30]

• a contains: 1, 2, 3

• b contains: 10, 20, 30

---

2️⃣ zip(a, b)

zip(a, b)

This pairs elements from both lists:

(1, 10)
(2, 20)
(3, 30)

---

3️⃣ Loop Execution

for i, j in zip(a, b):

• i gets values from list a

• j gets values from list b

Iteration	i	j
1st	1	10
2nd	2	20
3rd	3	30

---

4️⃣ Inside Loop

i = i + 100

• This changes only the loop variable i,

• ✅ It does NOT change the original list a

Example:

• 1 → becomes 101

• 2 → becomes 102

• 3 → becomes 103

But this updated i is never printed or used further.

---

5️⃣ Print Statement

print(j)

• Only j is printed

• j comes from list b

So it prints:

10
20
30

---

✅ Final Output

10
20
30

---

🔥 Key Learning Points

✅ Changing i does not modify list a
✅ zip() pairs values but does not link memory
✅ Only j is printed, so i + 100 has no visible effect

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Python Coding Challenge - Question with Answer (ID -281125)

 

Step-by-Step Explanation

✅ 1. Create an Empty List

funcs = []

This list will store functions (lambdas).

---

✅ 2. Loop Runs 3 Times

for i in range(3):

The values of i will be:

0 → 1 → 2

---

✅ 3. Lambda Is Appended Each Time

funcs.append(lambda: i)

⚠️ Important:
You are NOT storing the VALUE of i,
You are storing a function that will return i later.

So after the loop:

• funcs[0] → returns i

• funcs[1] → returns i

• funcs[2] → returns i

All three lambdas refer to the SAME variable i.

---

✅ 4. Final Value of i

After the loop finishes:

i = 2

---

✅ 5. Calling All Functions

print([f() for f in funcs])

This executes:

f() → returns i → which is 2

So all functions return:

[2, 2, 2]

---

✅ ✅ Final Output

[2, 2, 2]

---

Why This Happens? (Late Binding)

Python lambdas:

• Do NOT remember the value

• They remember the variable reference

• Value is looked up only when the function is called

This behavior is called Late Binding.

---

✅ ✅ ✅ Correct Way (Fix This Problem)

✅ Solution 1: Use Default Argument

funcs = []

for i in range(3):
    funcs.append(lambda i=i: i)
print([f() for f in funcs])

✅ Output:

[0, 1, 2]

Why this works:

• i=i captures the value immediately

• Each lambda stores its own copy

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Python Coding Challenge - Question with Answer (ID -271125)

 

✅ Step-by-Step Explanation

🔹 1. This is your list:

nums = [0, 1, 2, 3, 4]

It contains both falsy and truthy values.

---

🔹 2. This is the filter with lambda:

result = filter(lambda x: x, nums)

• lambda x: x means:
  👉 Return the value itself.

• filter() keeps only values that are truthy.

• In Python, these are falsy values:

  • 0, None, False, ""

So 0 is removed, and all non-zero numbers remain.

---

🔹 3. Convert result to list:

print(list(result))

Since filter() returns an iterator, we convert it to a list to display it.

---

✅ FINAL OUTPUT

[1, 2, 3, 4]

---

 Key Concept

👉 This line:

filter(lambda x: x, nums)

means:

“Keep only the values that are True in a boolean sense.”

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Python Coding Challenge - Question with Answer (01261125)

 

Explanation:

🔹 Line 1: Creating a Nested List

data = [[1,2], [3,4], [5,6]]

Explanation

data is a list of lists (nested list).

It contains three separate sublists.

Each sublist holds two numbers.

We will later flatten these sublists into one single list.

🔹 Line 2: Creating an Empty Output List

flat = []

Explanation

flat is an empty list.

This list will store all numbers from the nested structure after flattening.

It starts empty and will be filled step by step.

🔹 Line 3: Flattening Using map() With Lambda

list(map(lambda x: flat.extend(x), data))

Explanation

This is the most important line. Here’s how it works:

A. map()

map() loops over each sublist inside data.

B. lambda x

x represents each sublist from data, one at a time:

First: [1, 2]

Second: [3, 4]

Third: [5, 6]

C. flat.extend(x)

.extend() adds each element of x into flat.

It does not add the sublist — it adds the individual numbers.

D. How flat changes

Step Value of x flat after extend

1 [1, 2] [1, 2]

2 [3, 4] [1, 2, 3, 4]

3 [5, 6] [1, 2, 3, 4, 5, 6]

E. Why list(...) ?

map() doesn’t run immediately.

Wrapping it with list() forces all iterations to execute.

🔹 Line 4: Printing the Final Output

print(flat)

Explanation

Prints the fully flattened list.

All nested values are now in one single list.

🎉 Final Output

[1, 2, 3, 4, 5, 6]

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Python Coding Challenge - Question with Answer (01251125)

 

Explanation:

Initialize i

i = 0

The variable i is set to 0.

It will be used as the counter for the while loop.

Create an empty list

funcs = []

funcs is an empty list.

We will store lambda functions inside this list.

Start the while loop

while i < 5:

The loop runs as long as i is less than 5.

So the loop will execute for: i = 0, 1, 2, 3, 4.

Append a lambda that captures the current value of i

funcs.append(lambda i=i: i)

Why is i=i important?

i=i is a default argument.

Default arguments in Python are evaluated at the moment the function is created.

So each lambda stores the current value of i during that specific loop iteration.

What values get stored?

When i = 0 → lambda stores 0

When i = 1 → lambda stores 1

When i = 2 → lambda stores 2

When i = 3 → lambda stores 3

When i = 4 → lambda stores 4

So five different lambdas are created, each holding a different number.

Increment i

i += 1

After each iteration, i increases by 1.

This moves the loop to the next number.

Call all lambda functions and print their outputs

print([f() for f in funcs])

What happens here?

A list comprehension calls each stored lambda f() in funcs.

Each lambda returns the value it captured earlier.

Final Output:

[0, 1, 2, 3, 4]

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