String Traversal and Character Access in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to print each character in the string word using a for loop.

DSA Python

word = "hello"
for [1] in word:
    print([1])

Drag options to blanks, or click blank then click option'

Achar

Bi

Cword

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using the string variable name itself in the loop variable.

Using an index variable without accessing characters by index.

We use a variable name like char to represent each character in the string as we loop through it.

2fill in blank

medium

Complete the code to print the character at index 2 of the string word.

DSA Python

word = "world"
print(word[1])

Drag options to blanks, or click blank then click option'

A2

B[1]

C[2]

D[3]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets.

Using the wrong index number.

Indexing in strings starts at 0, so index 2 is the third character.

3fill in blank

hard

Fix the error in the code to print all characters of text using a while loop.

DSA Python

text = "data"
i = 0
while i < len(text):
    print(text[1])
    i += 1

Drag options to blanks, or click blank then click option'

A[i]

B(i)

C{i}

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses or braces instead of square brackets for indexing.

Not incrementing the index variable.

To access characters by index, use square brackets with the index variable.

4fill in blank

hard

Fill both blanks to create a dictionary that maps each index of word to its character.

DSA Python

word = "code"
index_map = { [1]: [2] for [1] in range(len(word)) }

Drag options to blanks, or click blank then click option'

Ai

Bword[i]

Cchar

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using the character variable as the loop variable instead of the index.

Swapping the key and value in the dictionary comprehension.

We use i as the index and word[i] as the character at that index.

5fill in blank

hard

Fill all three blanks to create a list of uppercase characters from word that come after the letter 'a'.

DSA Python

word = "alphabet"
result = [[1].upper() for [2] in word if [3] > 'a']

Drag options to blanks, or click blank then click option'

Achar

Dletter

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names in the list comprehension.

Not calling upper() on the character.

We use char as the variable for each character in the string and check if it is greater than 'a'.

Practice

1. What does the expression my_string[2] return when my_string = 'hello'?

easy

A. An error because index 2 is out of range

B. The first character 'h'

C. The third character 'l'

D. The last character 'o'

Solution

1. Step 1: Understand string indexing

   Strings in Python start indexing at 0, so index 2 means the third character.

2. Step 2: Identify the character at index 2

   In 'hello', index 0='h', index 1='e', index 2='l'.

3. Final Answer:

   The third character 'l' -> Option C

4. Quick Check:

   Index 2 of 'hello' is 'l' [OK]

Hint: Remember string indexing starts at zero [OK]

Common Mistakes:

• Thinking indexing starts at 1
• Confusing index 2 with last character
• Assuming index 2 is out of range

2. Which of the following is the correct way to loop through each character in the string text = 'abc'?

easy

A. for char in text: print(char)

B. for char in text[1]: print(char)

C. for i in range(text): print(i)

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

Solution

1. Step 1: Understand how to loop over a string

   Using for char in text iterates over each character directly.

2. Step 2: Check each option

   for i in range(text): print(i) tries to use range on a string (invalid). for char in text[1]: print(char) uses text[1] which is a single character, not iterable. for i in text.length: print(text[i]) uses text.length which is invalid in Python.

3. Final Answer:

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

4. Quick Check:

   Loop over string directly with for-in [OK]

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

Common Mistakes:

• Using range() directly on string
• Using incorrect string length property
• Trying to loop over a single character

3. What is the output of the following code?

word = 'python'
result = ''
for i in range(len(word)):
    if i % 2 == 0:
        result += word[i]
print(result)

medium

A. pyo

B. pto

C. pyhn

D. pthn

Solution

1. Step 1: Understand the loop and condition

   The loop runs from i=0 to i=5 (length 6). It adds characters at even indices (0,2,4).

2. Step 2: Identify characters at even indices

   word[0]='p', word[2]='t', word[4]='o'. Concatenate: 'pto'.

3. Final Answer:

   pto -> Option B

4. Quick Check:

   Even indices form 'pto' [OK]

Hint: Pick characters at even indexes using modulo [OK]

Common Mistakes:

• Including characters at odd indexes
• Confusing index counting starting at 1
• Adding characters incorrectly

4. Find the error in the following code that tries to print each character of text = 'data':

for i in range(len(text)):
    print(text(i))

medium

A. Using parentheses instead of square brackets for indexing

B. Missing colon after for loop

C. Using len() on string is invalid

D. Variable i is not defined

Solution

1. Step 1: Identify the indexing syntax error

   In Python, to access a character by index, square brackets [] are used, not parentheses ().

2. Step 2: Check the code line

   The code uses text(i) which is a function call, causing a TypeError.

3. Final Answer:

   Using parentheses instead of square brackets for indexing -> Option A

4. Quick Check:

   Use text[i] to access characters, not text(i) [OK]

Hint: Use square brackets [] for indexing strings [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Forgetting colon after for loop (not here)
• Misunderstanding len() usage

5. Given the string s = 'abracddabba', write code to create a new string containing only the characters at odd indices. What is the resulting string?

hard

A. bacdbr

B. brcdr

C. aracaa

D. badab

Solution

1. Step 1: Identify odd indices in the string

   Indices start at 0. Odd indices are 1,3,5,7,9.

2. Step 2: Extract characters at odd indices

   Characters: s[1]='b', s[3]='a', s[5]='d', s[7]='a', s[9]='b'. Concatenate: 'b'+'a'+'d'+'a'+'b' = 'badab'.

3. Final Answer:

   badab -> Option D

4. Quick Check:

   Characters at odd positions form 'badab' [OK]

Hint: Pick characters where index % 2 == 1 [OK]

Common Mistakes:

• Including even indices
• Starting index count at 1 instead of 0
• Mixing up characters order