String Traversal and Character Access in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of character access in a loop

What is the output of the following code snippet?

DSA Python

s = "hello"
result = []
for i in range(len(s)):
    if i % 2 == 0:
        result.append(s[i])
print(''.join(result))

A hlo

B el

C hello

D heo

💡 Hint

Look at characters at even indexes only.

The loop picks characters at indexes 0, 2, 4 which are 'h', 'l', 'o'. Joining them gives 'hlo'.

❓ Predict Output

intermediate

Output of negative indexing in string

What will be printed by this code?

DSA Python

s = "world"
print(s[-1] + s[-2] + s[-5])

A dlo

B dlw

C dwl

D wrd

💡 Hint

Negative indexes count from the end starting at -1.

s[-1] is 'd', s[-2] is 'l', s[-5] is 'w'. Concatenated: 'dlw'.

🔧 Debug

advanced

Identify the error in string character access

What error does this code raise when executed?

DSA Python

s = "data"
for i in range(len(s)+1):
    print(s[i])

A IndexError

B TypeError

C SyntaxError

D No error, prints all characters

💡 Hint

Check the range and valid indexes for string s.

The loop tries to access s[4] which is out of range for string of length 4, causing IndexError.

❓ Predict Output

advanced

Output of slicing with step in string

What is the output of this code?

DSA Python

s = "algorithm"
print(s[1:8:2])

A lorm

B lghm

C lgih

D loih

💡 Hint

Slice from index 1 to 7 with step 2.

Indexes 1,3,5,7 correspond to 'l', 'r', 'i', 'h'. Joining gives 'lrih'.

🧠 Conceptual

expert

Count of unique characters accessed in loop

Consider the code below. How many unique characters are printed?

DSA Python

s = "abracadabra"
seen = set()
for i in range(len(s)):
    if s[i] not in seen:
        print(s[i], end='')
        seen.add(s[i])

A 6

B 7

C 5

D 8

💡 Hint

Count distinct characters in the string.

The unique characters are 'a', 'b', 'r', 'c', 'd'. Counting carefully, there are 5 unique letters but the code prints them as they appear first time. The string has 'a','b','r','c','d' only 5 unique letters, but the code prints them once each, so 5 characters printed.

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