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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Python Coding Challenge - Question with Answer (01241125)

 Explanation:

1. Function Definition

def f(n):

You define a function named f.

It expects one argument n.

2. First Condition – Check if n is even

    if n % 2 == 0:

n % 2 gives the remainder when n is divided by 2.

If the remainder is 0, then n is even.

For n = 6:

6 % 2 = 0, so this condition is TRUE.

3. First Return

        return n // 2

Because the first condition is true, the function immediately returns integer division of n by 2.

6 // 2 = 3

4. Second Condition (Skipped)

    if n % 3 == 0:

        return n // 3

These lines are never reached for n = 6.

Even though 6 is divisible by 3, the function already returned at the previous line.

5. Final Return (Also Skipped)

    return n

This would run only if neither condition was true.

But in this case, the function has already returned earlier.

6. Function Call

print(f(6))

Calls f(6)

As shown above, f(6) returns 3

So the program prints:

3

Final Output

3

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Python Coding Challenge - Question with Answer (01231125)

 Explanation:

Initialization

i = 3

The variable i is set to 3.

This will be the starting value for the loop.

While Loop Condition

while i > 1:

The loop will keep running as long as i is greater than 1.

First iteration → i = 3 → condition TRUE

Second iteration → i = 2 → condition TRUE

Third time → i = 1 → condition FALSE → loop stops

Inner For Loop

for j in range(2):

range(2) gives values 0 and 1.

This means the inner loop runs 2 times for each while loop cycle.

Print Statement

print(i - j, end=" ")

This prints the result of i - j each time.

When i = 3:

j = 0 → prints 3

j = 1 → prints 2

When i = 2:

j = 0 → prints 2

j = 1 → prints 1

Decrease i

i -= 1

After each full inner loop, i is decreased by 1.

i = 3 → becomes 2

i = 2 → becomes 1

Now i = 1 → while loop stops.

Final Output

3 2 2 1

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Python Coding Challenge - Question with Answer (01221125)

 

Explanation:

1. Initialize the List

nums = [9, 8, 7, 6]

A list named nums is created.

Value: [9, 8, 7, 6].

2. Initialize the Sum Variable

s = 0

A variable s is created to store the running total.

Initial value: 0.

3. Start the Loop

for i in range(1, len(nums), 2):

range(1, len(nums), 2) generates numbers starting at 1, increasing by 2.

For nums of length 4, this produces:

i = 1, 3

So the loop will run two times.

4. Loop Iteration Details

Iteration 1 (i = 1)

s += nums[i-1] - nums[i]

nums[i-1] → nums[0] → 9

nums[i] → nums[1] → 8

Calculation: 9 - 8 = 1

Update s:

s = 0 + 1 = 1

Iteration 2 (i = 3)

s += nums[i-1] - nums[i]

nums[i-1] → nums[2] → 7

nums[i] → nums[3] → 6

Calculation: 7 - 6 = 1

Update s:

s = 1 + 1 = 2

5. Final Output

print(s)

The final value of s is 2.

Output: 2

Final Answer: 2

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Python Coding Challenge - Question with Answer (01211125)

 Explanation:

Initialize total

total = 0

A variable total is created.

It starts with the value 0.

This will accumulate the sum during the loops.

Start of the outer loop

for i in range(1, 4):

i takes values 1, 2, 3

(because range(1,4) includes 1,2,3)

We will analyze each iteration separately.

Outer Loop Iteration Details

When i = 1

j = i   →  j = 1

Enter the inner loop:

Condition: j > 0 → 1 > 0 → true

Step 1:

total += (i + j) = (1 + 1) = 2

total = 2

j -= 2 → j = -1

Next check:

j > 0 → -1 > 0 → false, inner loop stops.

When i = 2

j = i   →  j = 2

Step 1:

total += (2 + 2) = 4

total = 2 + 4 = 6

j -= 2 → j = 0

Next check:

j > 0 → 0 > 0 → false, inner loop ends.

When i = 3

j = i  → j = 3

Step 1:

total += (3 + 3) = 6

total = 6 + 6 = 12

j -= 2 → j = 1

Step 2:

Condition: j > 0 → 1 > 0 → true

total += (3 + 1) = 4

total = 12 + 4 = 16

j -= 2 → j = -1

Next check:

j > 0 → false → exit.

Final Output

print(total)

The final value collected from all iterations is:

Output: 16

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Python Coding Challenge - Question with Answer (01201125)

 

Explanation:

x = 0

Initializes the variable x with 0.

x will act as the counter for the loop.

while x < 5:

Starts a while loop that will run as long as x is less than 5.

Loop iterations will occur for x = 0, 1, 2, 3, 4.

if x % 2 == 0: print("E", end='')

Checks if x is even (x % 2 == 0).

If true, prints "E".

end='' ensures that the next output prints on the same line without a newline.

elif x % 2 != 0: print("O", end='')

If the if condition is false, checks if x is odd (x % 2 != 0).

If true, prints "O" on the same line.

x += 1

Increments x by 1 after each iteration.

This moves the loop forward and prevents an infinite loop.

Step-by-Step Execution Table

x Condition     True Output x after increment

0 x % 2 == 0 → True     E     1

1 x % 2 != 0 → Tru e     O     2

2 x % 2 == 0 → True         E     3

3 x % 2 != 0 → True         O     4

4 x % 2 == 0 → True         E     5

Final Output:

EOEOE

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Python Coding Challenge - Question with Answer (01191125)

 

Explanation:

lst = [1, 2, 3]

Creates a list lst containing three integers: 1, 2, 3.

This list will be used in the loops.

result = 0

Initializes a variable result to 0.

This variable will accumulate the count of times i == j is True.

for i in lst:

Starts the outer loop.

i will take each value in the list sequentially: 1, then 2, then 3.

for j in lst:

Starts the inner loop inside the outer loop.

For each value of i, j will iterate over the list [1, 2, 3].

result += (i == j)

(i == j) is a comparison: returns True if i equals j, else False.

In Python, True is treated as 1 and False as 0 when used in arithmetic.

Adds 1 to result if i == j, otherwise adds 0.

Step-by-step calculation:

i j i == j Added to result Cumulative result

1 1 True                 1                         1

1 2 False         0                         1

1 3 False         0                         1

2 1 False         0                         1

2 2 True                 1                         2

2 3 False         0                         2

3 1 False 0 2

3 2 False 0 2

3 3 True 1 3

Correction: Actually, let's carefully check:

Wait, outer loop iterates i=1,2,3 and inner loop j=1,2,3.

Count True values: (1==1),(2==2),(3==3) once for each combination?

Actually, let's compute carefully:

i=1: j=1 → True → +1 → result=1

i=1: j=2 → False → +0 → result=1

i=1: j=3 → False → +0 → result=1

i=2: j=1 → False → +0 → result=1

i=2: j=2 → True → +1 → result=2

i=2: j=3 → False → +0 → result=2

i=3: j=1 → False → +0 → result=2

i=3: j=2 → False → +0 → result=2

i=3: j=3 → True → +1 → result=3

Total result = 3

Ah! So in my previous message I mistakenly said 9. The correct answer is 3. Thanks for double-checking!

print(result)

Prints the final value of result.

Output:

3

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Python Coding Challenge - Question with Answer (01181125)

 

Explanation:

1. Dictionary Creation

d = {'apple': 2, 'banana': 3, 'cherry': 4}

Creates a dictionary d with keys as fruit names and values as numbers.

Example content:

'apple': 2

'banana': 3

'cherry': 4

2. Initialize Counter

count = 0

Initializes a variable count to store the running total.

Starts at 0.

3. Loop Over Dictionary Items

for k, v in d.items():

Loops over each key-value pair in the dictionary.

k = key (fruit name), v = value (number).

.items() gives pairs: ('apple', 2), ('banana', 3), ('cherry', 4).

4. Check for Letter 'a' in Key

if 'a' in k:

Checks if the key contains the letter 'a'.

Only keys with 'a' are processed.

'apple' → True, 'banana' → True, 'cherry' → False.

5. Add Value to Counter

count += v

Adds the value v to count if the key has 'a'.

Step by step:

'apple': count = 0 + 2 → 2

'banana': count = 2 + 3 → 5

'cherry': skipped

6. Print Final Count

print(count)

Prints the final total of values where keys contain 'a'.

Output:

5

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Python Coding Challenge - Question with Answer (01171125)

 

Explanation:

Assign a value to num

num = 5

We set num = 5.

We will use this value for checking divisibility.

Create the list a

a = [5, 10, 11, 13]

This list contains four numbers:

5, 10, 11, 13

Initialize sum variable

s = 0

We create s to store the total of selected numbers.

Starting value = 0

Start loop — go through each value in list

for v in a:

The loop takes each value v from the list:

First: v = 5

Then: v = 10

Then: v = 11

Then: v = 13

Check divisibility using IF + continue

    if v % num == 0:

        continue

We check:

If v is divisible by num (5), skip it.

Check each:

v v % 5 Divisible? Action

5 0             Yes         Skip

10 0             Yes         Skip

11 1              No         Add

13 3              No         Add

continue means: skip the rest of the loop and go to the next number.

Add the non-skipped values

    s += v

Only numbers NOT divisible by 5 get added:

s = 11 + 13 = 24

Print the final answer

print(s)

So the output is:

24

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Python Coding Challenge - Question with Answer (01161125)

 Explanation:

List Initialization

a = [1, 4, 7, 10]

You create a list a that contains four numbers:

1, 4, 7, 10

Variable Initialization

s = 0

You create a variable s to store the sum.

Initially, sum = 0.

Loop Through Each Value

for v in a:

This loop will pick each value from list a, one by one.

So v becomes:

1 → 4 → 7 → 10

Check Condition

if v % 3 == 1:

You check if the number leaves remainder 1 when divided by 3.

Let’s check each:

v v % 3 Remainder Condition v%3==1?

1 1             Yes             ✔ Add

4 1             Yes             ✔ Add

7 1             Yes             ✔ Add

10 1             Yes             ✔ Add

All numbers satisfy the condition.

Add to Sum

s += v

Add all numbers that passed the condition.

s = 1 + 4 + 7 + 10 = 22

Print the Final Sum

print(s)

This prints:

22

Final Output: 22

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Python Coding Challenge - Question with Answer (01151125)

Explanation:

Line 1: Initialization

count = 0

A variable count is created and initialized to 0.

This variable will act as the loop counter.

Line 2: Start of While Loop

while count < 5:

The loop will continue as long as count is less than 5.

The condition is checked before each iteration.

Line 3: Increment Counter

count += 1

In each iteration, count is increased by 1.

Equivalent to count = count + 1.

So count will take values: 1, 2, 3, 4, 5 during the iterations.

Line 4: Continue Condition

if count == 4:

    continue

If count equals 4, the continue statement executes.

continue means: skip the rest of this iteration and move to the next iteration.

Here, print(count) will be skipped when count == 4.

Line 5: Print Statement

print(count, end=" ")

If count is not 4, it will be printed.

end=" " ensures that the output is printed on the same line separated by a space.

Execution Flow

Iteration count value if count==4? Action Output so far

1             1                     No         print 1         1

2             2                     No         print 2         1 2

3             3                     No         print 3 1 2 3

4             4                     Yes continue (skip) 1 2 3

5             5                     No print 5 1 2 3 5

Final Output

1 2 3 5

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