Python Coding challenge - Day 1162| What is the output of the following Python Code?

 

Code Explanation:

🔹 1. Importing asyncio

import asyncio

✅ Explanation:

Imports Python's asyncio module.

Used for asynchronous programming.

In this code, asyncio is imported but not actually used.

🔹 2. Defining an Async Function

async def func():

✅ Explanation:

async def creates an asynchronous function.

Also called a coroutine function.

⚠️ Important:

This is NOT a normal function.

Example:

def normal():

    return 10

returns value immediately.

But:

async def func():

    return 10

returns a coroutine when called.

🔹 3. Return Statement

return 10

✅ Explanation:

If the coroutine is executed,

it will eventually return:

10

But execution hasn't happened yet.

🔹 4. Calling the Async Function

x = func()

🔍 What most beginners think:

x = 10

❌ Wrong

✅ What actually happens:

Calling:

func()

creates a coroutine object.

So:

x

stores:

<coroutine object func at ...>

🔹 5. Why Function Doesn't Execute?

Because async functions must be:

await func()

or

asyncio.run(func())

to actually run.

Without that:

func()

only creates a coroutine object.

🔹 6. Checking Type

print(type(x))

✅ Explanation:

Python checks type of:

x

which is a coroutine object.

🔹 7. Result

Output becomes:

<class 'coroutine'>

🎯 Final Output

<class 'coroutine'>

Book:  700 Days Python Coding Challenges with Explanation

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🚀 Day 61/150 – Find String Length Without len() in Python

 

🚀 Day 61/150 – Find String Length Without len() in Python

Sometimes it’s useful to understand how Python counts characters internally.
Instead of using len(), we can count each character manually.

Example:
"python" → Length = 6

Let’s explore different ways 👇

🔹 Method 1 – Using for Loop

text = "python" count = 0 for ch in text: count += 1 print("Length:", count)

🔹 Method 2 – Using while Loop

text = "python" count = 0 while text[count:]: count += 1 print("Length:", count)

🔹 Method 3 – Taking User Input

text = input("Enter a string: ") count = 0 for ch in text: count += 1 print("Length:", count)

🔹 Method 4 – Using Recursion

def string_length(s): if s == "": return 0 return 1 + string_length(s[1:]) print(string_length("python"))

💡 Key Takeaways

• Strings are iterable, so you can count characters one by one
• for loop is the easiest manual way
• while and recursion help understand string behavior
• Great exercise for learning loops and indexing

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Python Coding Challenge - Question with Answer (ID -070626)

 

 Code Explanation:

🔹 Step 1: Create Variable

x = 0

Variable x is assigned:

0

Current memory:

x → 0

🔹 Step 2: Evaluate First Print Statement

print(x or (x := 5))

Python first evaluates:

x

Current value:

0

🔹 Step 3: Check or Operator

Expression:

0 or (x := 5)

Remember:

0

is a falsy value.

For or:

If left side is falsy,

evaluate the right side.

So Python moves to:

(x := 5)

🔹 Step 4: Execute Walrus Operator

x := 5

Walrus operator does two things:

1️⃣ Assigns value

x = 5

2️⃣ Returns value

5

Now memory becomes:

x → 5

and the expression returns:

5

🔹 Step 5: Complete First Print

Expression becomes:

print(5)

Output:

5

🔹 Step 6: Execute Second Print

print(x)

Current value of x:

5

So Python executes:

print(5)

Output:

5

Final Output:

5

5

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Python Coding Challenge - Question with Answer (ID -060626)

 

Code Expkanation:

🔹 Step 1: Create a List

x = [1,2,3]

A list is created:

[1, 2, 3]

🔹 Step 2: Start Pattern Matching

match x:

Python checks the value of:

x

which is:

[1,2,3]

Now Python tries to match it against the available case patterns.

🔹 Step 3: Check the Pattern

case [1, *a]:

This pattern means:

First element must be 1

and

Store all remaining elements in a

🔹 Step 4: Match First Element

List:

[1,2,3]

Pattern:

[1, *a]

Comparison:

1 == 1

✅ Match successful

🔹 Step 5: Capture Remaining Elements

After matching the first element:

1

remaining elements are:

[2,3]

These are assigned to:

a

So:

a = [2,3]

🔹 Step 6: Execute Print Statement

print(a)

becomes:

print([2,3])

Output:

[2, 3]

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🚀 Day 60/150 – Find Second Largest Element in Python

 

🚀 Day 60/150 – Find Second Largest Element in Python

The second largest element is the number that is just smaller than the largest number in the list.

Example:
[10, 20, 4, 45, 99] → Largest = 99, Second Largest = 45

Let’s explore different ways to find it 👇

🔹 Method 1 – Using Sorting

numbers = [10, 20, 4, 45, 99]

numbers.sort()

print("Second Largest:", numbers[-2])

🔹 Method 2 – Using set() + max()

numbers = [10, 20, 4, 45, 99] numbers = list(set(numbers)) numbers.remove(max(numbers)) print("Second Largest:", max(numbers))

🔹 Method 3 – Using Loop

numbers = [10, 20, 4, 45, 99] largest = second = float('-inf') for num in numbers: if num > largest: second = largest largest = num elif num > second and num != largest: second = num print("Second Largest:", second)

🔹 Method 4 – Taking User Input

numbers = list(map(int, input("Enter numbers: ").split())) numbers = sorted(set(numbers)) print("Second Largest:", numbers[-2])

💡 Key Takeaways

• Sorting is the easiest way
• set() helps remove duplicates
• Loop method is efficient because it scans only once
• Always consider duplicate values when finding the second largest

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🚀 Day 59/150 – Rotate a List in Python

Rotating a list means shifting its elements either to the left or to the right.

Example:
[1, 2, 3, 4, 5]

Rotate right by 2 → [4, 5, 1, 2, 3]
Rotate left by 2 → [3, 4, 5, 1, 2]

Let’s explore different ways to rotate a list 👇

🔹 Method 1 – Right Rotation Using Slicing

numbers = [1, 2, 3, 4, 5] k = 2 rotated = numbers[-k:] + numbers[:-k] print("Right Rotated:", rotated)

🔹 Method 2 – Left Rotation Using Slicing

numbers = [1, 2, 3, 4, 5] k = 2 rotated = numbers[k:] + numbers[:k] print("Left Rotated:", rotated)

🔹 Method 3 – Using Loop (Right Rotation by One)

numbers = [1, 2, 3, 4, 5] last = numbers[-1] for i in range(len(numbers) - 1, 0, -1): numbers[i] = numbers[i - 1] numbers[0] = last print("Rotated List:", numbers)

🔹 Method 4 – Taking User Input

numbers = list(map(int, input("Enter numbers: ").split())) k = int(input("Enter rotation count: ")) k = k % len(numbers) rotated = numbers[-k:] + numbers[:-k] print("Rotated List:", rotated)

💡 Key Takeaways

• Slicing is the easiest way to rotate a list
• Use k % len(list) to handle large rotation values
• Right rotation uses [-k:] +[:-k]
• Left rotation uses [k:] +[:k]

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