Iterating over strings in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of iterating and modifying a string

What is the output of this Python code?

text = "hello"
result = ""
for ch in text:
    if ch == 'l':
        result += 'L'
    else:
        result += ch
print(result)

Python

text = "hello"
result = ""
for ch in text:
    if ch == 'l':
        result += 'L'
    else:
        result += ch
print(result)

A heLLo

B hello

C heLlo

D heLLoL

💡 Hint

Remember that the loop checks each character and replaces 'l' with 'L'.

The loop goes through each character in 'hello'. When it finds 'l', it adds 'L' to result. Otherwise, it adds the original character. So 'l' becomes 'L', resulting in 'heLLo'.

❓ Predict Output

intermediate

Counting vowels in a string

What is the output of this code?

text = "Programming"
vowels = 'aeiouAEIOU'
count = 0
for char in text:
    if char in vowels:
        count += 1
print(count)

Python

text = "Programming"
vowels = 'aeiouAEIOU'
count = 0
for char in text:
    if char in vowels:
        count += 1
print(count)

A 5

B 4

C 3

D 2

💡 Hint

Check each letter if it is a vowel.

The vowels in 'Programming' are 'o', 'a', and 'i', so the count is 3.

❓ Predict Output

advanced

Output of nested iteration over string

What does this code print?

word = "abc"
result = ""
for i in word:
    for j in word:
        if i == j:
            result += i.upper()
        else:
            result += j
print(result)

Python

word = "abc"
result = ""
for i in word:
    for j in word:
        if i == j:
            result += i.upper()
        else:
            result += j
print(result)

A AbcBaCcaB

B AbcaBcabC

C AbcBacCab

D ABCabcabc

💡 Hint

Look at how the inner loop builds the string for each outer loop character.

For each character i in 'abc', the inner loop goes through 'abc'. When i equals j, it adds uppercase i, else adds j. So for i='a', it adds 'A', 'b', 'c' → 'Abc'. For i='b', adds 'a', 'B', 'c' → 'aBc'. For i='c', adds 'a', 'b', 'C' → 'abC'. Combined: 'AbcaBcabC'. Option B matches this.

❓ Predict Output

advanced

Result of modifying string characters conditionally

What is the output of this code?

text = "Data123"
result = ""
for ch in text:
    if ch.isalpha():
        result += ch.lower()
    else:
        result += '*'
print(result)

Python

text = "Data123"
result = ""
for ch in text:
    if ch.isalpha():
        result += ch.lower()
    else:
        result += '*'
print(result)

A data***

B DATA123

C data123

D Data***

💡 Hint

Check how letters and digits are handled differently.

Letters are converted to lowercase, digits replaced by '*'. So 'Data123' becomes 'data***'.

🧠 Conceptual

expert

Why strings are immutable in Python?

Which of the following is the main reason strings are immutable in Python?

A To prevent strings from being garbage collected

B To make string concatenation faster

C Because strings are stored in a special memory area called stack

D To allow strings to be used as dictionary keys and set elements safely

💡 Hint

Think about what properties dictionary keys need.

Strings are immutable so they can be hashed and used safely as dictionary keys and set elements without risk of changing their value.

Practice

1. What does the following code do?
for letter in "hello":
print(letter)

easy

A. Prints each letter of the word 'hello' on a new line

B. Prints the whole word 'hello' once

C. Prints the length of the word 'hello'

D. Causes an error because strings can't be looped

Solution

1. Step 1: Understand the for loop over a string

   The loop goes through each character in the string "hello" one by one.

2. Step 2: Print each character

   Inside the loop, each letter is printed on its own line.

3. Final Answer:

   Prints each letter of the word 'hello' on a new line -> Option A

4. Quick Check:

   Loop over string prints letters individually [OK]

Hint: For loops over strings print letters one by one [OK]

Common Mistakes:

• Thinking the whole string prints at once
• Confusing string length with letters
• Believing strings can't be looped

2. Which of the following is the correct syntax to loop over each character in the string text?

easy

A. for char to text: print(char)

B. for char in range(text): print(char)

C. for char in text: print(char)

D. for i in text.length: print(i)

Solution

1. Step 1: Identify correct for loop syntax for strings

   In Python, to loop over characters in a string, use for variable in string:.

2. Step 2: Check each option

   for char in text:
print(char) uses correct syntax. The other options use invalid syntax like 'to' instead of 'in', range() on a string, or non-existent .length attribute.

3. Final Answer:

   for char in text:
   print(char) -> Option C

4. Quick Check:

   Correct for loop syntax over string is for char in text:
   print(char) [OK]

Hint: Use 'for char in string:' to loop letters [OK]

Common Mistakes:

• Using range() on a string directly
• Trying to use .length instead of len()
• Using incorrect loop keywords

3. What is the output of this code?

word = "code"
result = ""
for ch in word:
    result += ch.upper()
print(result)

medium

A. code

B. CODE

C. Code

D. cODe

Solution

1. Step 1: Loop through each letter in 'code'

   The loop takes each character: 'c', 'o', 'd', 'e'.

2. Step 2: Convert each letter to uppercase and add to result

   Each letter is changed to uppercase ('C', 'O', 'D', 'E') and added to the empty string result.

3. Final Answer:

   CODE -> Option B

4. Quick Check:

   Uppercase each letter and join = CODE [OK]

Hint: Uppercase letters inside loop to build new string [OK]

Common Mistakes:

• Not converting letters to uppercase
• Printing original string instead of result
• Using += without initializing result

4. Find the error in this code:

text = "hello"
for i in text:
    print(text[i])

medium

A. Variable 'text' is not defined

B. Missing colon after for loop

C. Loop should be 'for i in range(text):'

D. Using 'i' as index causes TypeError

Solution

1. Step 1: Understand the loop variable 'i'

   Here, 'i' takes each character from 'text', so 'i' is a letter, not an index.

2. Step 2: Using 'text[i]' causes error

   Since 'i' is a letter, using it as an index causes a TypeError because string indices must be integers.

3. Final Answer:

   Using 'i' as index causes TypeError -> Option D

4. Quick Check:

   Loop variable is letter, not index, so indexing fails [OK]

Hint: Loop variable is letter, not index; use range(len(text)) for indices [OK]

Common Mistakes:

• Assuming loop variable is index
• Trying to index string with a character
• Confusing loop variable with range

5. Write code to count how many vowels are in the string sentence. Which code correctly does this?

hard

A. count = 0
for ch in sentence:
if ch in 'aeiouAEIOU':
count += 1
print(count)

B. count = 0
for i in range(len(sentence)):
if sentence[i] == 'aeiou':
count += 1
print(count)

C. count = 0
for ch in sentence:
if ch == 'a' or 'e' or 'i' or 'o' or 'u':
count += 1
print(count)

D. count = 0
for ch in sentence:
if ch in ['a','e','i','o','u']:
count = count + 1
print(count)

Solution

1. Step 1: Check vowel membership correctly

   count = 0 for ch in sentence: if ch in 'aeiouAEIOU': count += 1 print(count) checks if each character is in the string of vowels (both lowercase and uppercase), which is correct.

2. Step 2: Verify counting logic

   count = 0 for ch in sentence: if ch in 'aeiouAEIOU': count += 1 print(count) increments count correctly when a vowel is found and prints the total count.

3. Final Answer:

   count = 0 for ch in sentence: if ch in 'aeiouAEIOU': count += 1 print(count) -> Option A

4. Quick Check:

   Use 'in' with vowel string and increment count [OK]

Hint: Use 'if ch in "aeiouAEIOU"' to check vowels [OK]

Common Mistakes:

• Using '==' to compare with multiple vowels
• Checking membership incorrectly with list without uppercase
• Not incrementing count properly