String immutability in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of string modification attempt

What is the output of this Python code?

Python

s = "hello"
s[0] = 'H'
print(s)

A TypeError

B Hello

C hello

D SyntaxError

💡 Hint

Remember that strings cannot be changed by assigning to an index.

Strings in Python are immutable, so trying to assign a new character to an index causes a TypeError.

❓ Predict Output

intermediate

Result of string concatenation

What is the output of this code?

Python

s = "cat"
s = s + "s"
print(s)

A cat

B cats

C TypeError

D catss

💡 Hint

Concatenation creates a new string.

Concatenating strings creates a new string; the original string is not changed but reassigned.

❓ Predict Output

advanced

Output after string replace method

What is printed by this code?

Python

s = "banana"
s.replace('a', 'o')
print(s)

A banana

B bonono

C SyntaxError

D TypeError

💡 Hint

Does the replace method change the original string?

The replace method returns a new string but does not change the original string because strings are immutable.

❓ Predict Output

advanced

Value of string after slicing and concatenation

What is the output of this code?

Python

s = "python"
s = 'J' + s[1:]
print(s)

A nohtyJ

B python

C Jython

D TypeError

💡 Hint

Slicing creates a new string; concatenation creates a new string.

The code creates a new string by concatenating 'J' with the slice s[1:], resulting in 'Jython'.

❓ Predict Output

expert

Final string value after chained operations

What is the output of this code?

Python

s = "immutable"
s1 = s.upper()
s2 = s1.replace('A', '@')
s3 = s2.lower()
print(s3)

A elb@tummi

B immutable

C IMMUT@BLE

D immut@ble

💡 Hint

Check the order of operations and what each method returns.

The final string s3 is the lowercase version of s2, which replaced 'A' with '@' in uppercase s1. 'IMMUTABLE' contains 'A', so s2 is 'IMMUT@BLE', and s3 is 'immut@ble'.

Practice

1. What does it mean that strings in Python are immutable?

easy

A. Strings are stored as numbers internally.

B. You can change parts of a string directly.

C. Strings can be resized but not changed.

D. You cannot change a string after it is created.

Solution

1. Step 1: Understand the meaning of immutable

   Immutable means something cannot be changed after it is made.

2. Step 2: Apply this to strings in Python

   Strings cannot be changed once created; you must make a new string to change text.

3. Final Answer:

   You cannot change a string after it is created. -> Option D

4. Quick Check:

   Immutable = cannot change [OK]

Hint: Immutable means no changes allowed after creation [OK]

Common Mistakes:

• Thinking strings can be changed like lists
• Confusing immutability with fixed size
• Believing strings store numbers only

2. Which of the following is the correct way to create a new string by changing the first letter of word = 'hello' to uppercase?

easy

A. word[0] = 'H'

B. word = 'H' + word[1:]

C. word.upper()[0]

D. word.change(0, 'H')

Solution

1. Step 1: Recognize string immutability

   You cannot assign to a string index like word[0] = 'H'.

2. Step 2: Create a new string combining parts

   Use slicing and concatenation: 'H' + word[1:] makes a new string with first letter changed.

3. Final Answer:

   word = 'H' + word[1:] -> Option B

4. Quick Check:

   Use slicing + concat to change string [OK]

Hint: Use slicing and + to build new strings [OK]

Common Mistakes:

• Trying to assign directly to string index
• Using non-existent string methods like change()
• Misusing upper() which returns full uppercase

3. What will be the output of this code?

text = 'cat'
text[0] = 'b'
print(text)

medium

A. TypeError

B. cat

C. bat

D. SyntaxError

Solution

1. Step 1: Identify the operation on string index

   Trying to assign text[0] = 'b' attempts to change a character in the string.

2. Step 2: Recall string immutability rules

   Strings cannot be changed by index assignment; Python raises a TypeError.

3. Final Answer:

   TypeError -> Option A

4. Quick Check:

   Index assignment on string = TypeError [OK]

Hint: Assigning to string index causes TypeError [OK]

Common Mistakes:

• Expecting string to change without error
• Confusing TypeError with SyntaxError
• Thinking print shows changed string

4. The code below tries to replace the first character of word with 'J'. What is wrong and how to fix it?

word = 'python'
word[0] = 'J'
print(word)

medium

A. Strings are immutable; fix by creating a new string: word = 'J' + word[1:]

B. Syntax error; fix by adding parentheses around 'J'

C. Use word.replace(0, 'J') to fix

D. No error; code works fine

Solution

1. Step 1: Identify the error cause

   Assigning to word[0] is invalid because strings cannot be changed by index.

2. Step 2: Fix by creating a new string

   Build a new string with 'J' plus the rest: word = 'J' + word[1:].

3. Final Answer:

   Strings are immutable; fix by creating a new string: word = 'J' + word[1:] -> Option A

4. Quick Check:

   Fix immutability by new string creation [OK]

Hint: Fix by slicing and concatenating new string [OK]

Common Mistakes:

• Trying to assign directly to string index
• Using replace() incorrectly for index change
• Assuming no error occurs

5. Given a string s = 'banana', how can you create a new string where all 'a' characters are replaced by 'o' without modifying s directly?

hard

A. s = s[0] + 'o' + s[2:]

B. s[1] = 'o'

C. s.replace('a', 'o')

D. s.change('a', 'o')

Solution

1. Step 1: Understand string immutability

   You cannot change characters by index; must create a new string.

2. Step 2: Use string method to replace characters

   s.replace('a', 'o') returns a new string with all 'a' replaced by 'o'.

3. Final Answer:

   s.replace('a', 'o') -> Option C

4. Quick Check:

   Use replace() to get new string with changes [OK]

Hint: Use replace() to get new string with changes [OK]

Common Mistakes:

• Trying to assign to string index
• Using non-existent change() method
• Changing only one character manually