300 Real Time SQL Interview practical Questions Asked at multiple companies
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Anyone who's preparing for an interview just reading theoretical concept will not help definitely you need to have practical hands on in #sql so create table with some data and try this queries running by your self so can help you to understand the logic of similar kind of queries

If you're preparing for an interview this doc will help a lot in the perpetration If you're experienced also freshers can also get hands on by practicing these queries and get confidence.

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⛓️ Tricky Python Interview Question

> What will this code output and why?

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

for f in funcs:
    f()

❓ Question: Why does this code print 2, 2, 2 instead of 0, 1, 2?

🔍 Explanation:

This is a classic example of late binding in Python closures. The lambda functions don't capture the value of i at each step of the loop. Instead, they all hold a reference to the same variable i.

🧠 What happens:

• The for loop completes. By the end of the loop, the variable i holds the value 2.
• The funcs list contains three functions, but each one is defined to print the value of whatever i is at the time of execution.
• When you call each function f(), it looks up the current value of i in its enclosing scope, which is 2.

👉 Result:

2
2
2

✅ How to fix:

You can force the lambda to capture the value of i at the time of definition by using a default argument.

funcs = []
for i in range(3):
    # The default argument captures i's value at definition time
    funcs.append(lambda num=i: print(num))

for f in funcs:
    f()

This trick works because default arguments are evaluated when the function is defined, not when it's called. This will correctly print:

0
1
2

This is a common interview question that tests your understanding of closures, scope, and late binding in Python.

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Interview question

What will this code output and why?

a = [1, 2, 3]
b = a

a += [4, 5]

print(a)
print(b)

a = (1, 2, 3)
b = a

a += (4, 5)

print(a)
print(b)

Try to answer yourself first, then check the explanation.

Part 1. Lists
a = [1, 2, 3]
b = a

a += [4, 5]

print(a)
print(b)

Step-by-step:

a = [1, 2, 3]

Variable a refers to the list [1, 2, 3].

b = a
Now b refers to the same list as a.
Not a copy, but the very same object in memory.

a += [4, 5]
Important: for lists, += works as an in-place modification:
this is roughly the same as:

a.extend([4, 5])


That is, elements 4 and 5 are added to the existing list that both a and b refer to.

Therefore:

print(a)  # [1, 2, 3, 4, 5]
print(b)  # [1, 2, 3, 4, 5]

Both point to the same modified list.

Part 2. Tuples
a = (1, 2, 3)
b = a

a += (4, 5)

print(a)
print(b)


Tuples are immutable, and here the magic begins.

a = (1, 2, 3)
a refers to the tuple (1, 2, 3).

b = a
b refers to the same tuple (1, 2, 3).

a += (4, 5)
For tuples, += cannot modify the object in place (they are immutable).
So Python does this:

a = a + (4, 5)


That is, a new tuple (1, 2, 3, 4, 5) is created and variable a is reassigned to it.

b still points to the old tuple (1, 2, 3).

Therefore:

print(a)  # (1, 2, 3, 4, 5)
print(b)  # (1, 2, 3)

The trick

For the list, a += [4, 5] mutates the object in place, visible through all variables referencing it.

For the tuple, a += (4, 5) creates a new object and reassigns only a.

In summary:

# List part:
[1, 2, 3, 4, 5]
[1, 2, 3, 4, 5]

# Tuple part:
(1, 2, 3, 4, 5)
(1, 2, 3)

If you want, I can explain another tricky question — about mutable default arguments in functions or about loops and closures.

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