String indexing (positive and negative) in Python - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of positive and negative indexing

What is the output of this Python code?

Python

s = "hello"
print(s[1], s[-1])

A h o

B e o

C e h

D o e

💡 Hint

Remember, positive index starts at 0 from left, negative index starts at -1 from right.

s[1] is the second character 'e', s[-1] is the last character 'o'. So output is 'e o'.

❓ Predict Output

intermediate

Indexing with negative values

What will this code print?

Python

word = "python"
print(word[-3])

A h

B t

C o

D n

💡 Hint

Count backwards from the end starting at -1.

word[-1] is 'n', word[-2] is 'o', word[-3] is 'h'.

❓ Predict Output

advanced

Mixing positive and negative indexes

What is the output of this code snippet?

Python

text = "abcdef"
print(text[2], text[-4])

A c d

B d d

C d c

D c c

💡 Hint

Remember positive index 2 is third character, negative index -4 counts from right.

text[2] is 'c', text[-4] is 'c'. So output is 'c c'.

❓ Predict Output

advanced

Index error with out-of-range negative index

What error does this code raise?

Python

s = "abc"
print(s[-4])

A IndexError

B No error, prints 'a'

C TypeError

D KeyError

💡 Hint

Negative index must be within the string length.

Index -4 is out of range for string of length 3, so IndexError is raised.

🧠 Conceptual

expert

Length and indexing relationship

Given a string s of length 7, which index accesses the last character?

A s[7]

B s[-6]

C s[6]

D s[-7]

💡 Hint

Positive indexes start at 0, so last index is length minus one.

For length 7, last character is at index 6 (0-based). s[7] is out of range. s[-7] is first character.

Practice

1.

What does the index 0 represent in the string "hello"?

easy

A. The last letter 'o'

B. The middle letter 'l'

C. The first letter 'h'

D. An error because indexing starts at 1

Solution

1. Step 1: Understand positive indexing

   In Python, string indexes start at 0 for the first character.

2. Step 2: Apply to the string "hello"

   The character at index 0 is the first letter 'h'.

3. Final Answer:

   The first letter 'h' -> Option C

4. Quick Check:

   Index 0 = first letter 'h' [OK]

Hint: Index 0 always means the first letter [OK]

Common Mistakes:

• Thinking indexing starts at 1
• Confusing index 0 with last letter
• Assuming negative indexes start at 0

2.

Which of the following is the correct way to get the last character of a string s in Python?

easy

A. s[1]

B. s[-1]

C. s[0]

D. s[len(s)]

Solution

1. Step 1: Recall negative indexing

   Negative index -1 refers to the last character in a string.

2. Step 2: Check each option

   s[1] is second character, s[0] is first, s[len(s)] causes error (index out of range).

3. Final Answer:

   s[-1] -> Option B

4. Quick Check:

   Last char = s[-1] [OK]

Hint: Use -1 to get last character quickly [OK]

Common Mistakes:

• Using s[len(s)] causes IndexError
• Confusing s[1] as last character
• Forgetting negative indexes start at -1

3.

What is the output of this code?

word = "python"
print(word[2], word[-3])

medium

A. t h

B. t o

C. t t

D. h n

Solution

1. Step 1: Find character at index 2

   Index 0='p', 1='y', 2='t'. So word[2] = 't'.

2. Step 2: Find character at index -3

   Negative indexes count from end: -1='n', -2='o', -3='h'. So word[-3] = 'h'.

3. Final Answer:

   t h -> Option A

4. Quick Check:

   word[2] = 't', word[-3] = 'h' [OK]

Hint: Count from start with positive, from end with negative [OK]

Common Mistakes:

• Mixing positive and negative indexes
• Counting negative indexes starting at 0
• Confusing letters at indexes 2 and -3

4.

Find the error in this code snippet:

text = "example"
print(text[-0])

medium

A. No error, prints 'e' (same as text[0])

B. SyntaxError due to invalid index

C. IndexError because -0 is out of range

D. IndexError: -0 is invalid

Solution

1. Step 1: Understand -0 as index

   In Python, -0 is the same as 0, so text[-0] equals text[0].

2. Step 2: Check output

   text[0] is 'e', the first letter, so it prints 'e' without error.

3. Final Answer:

   No error, prints 'e' (same as text[0]) -> Option A

4. Quick Check:

   -0 = 0 index, prints first letter 'e' [OK]

Hint: -0 equals 0, so no error and prints first letter [OK]

Common Mistakes:

• Thinking -0 is invalid index
• Expecting IndexError for -0
• Confusing negative zero with positive zero

5.

Given the string s = "programming", which expression returns the substring "ing" using only negative indexes?

hard

A. s[-3:-0]

B. s[-4:-1]

C. s[-3:-1]

D. s[-3:]

Solution

1. Step 1: Understand slicing with negative indexes

   s[-3:] means start at third last character to end.

2. Step 2: Apply to "programming"

   Third last character is 'i', so s[-3:] = 'ing'. Other options exclude last character or cause empty slice.

3. Final Answer:

   s[-3:] -> Option D

4. Quick Check:

   Negative slice from -3 to end = 'ing' [OK]

Hint: Use s[-3:] to get last 3 letters easily [OK]

Common Mistakes:

• Using s[-3:-1] misses last letter
• Using s[-4:-1] gives 'min' not 'ing'
• Using s[-3:-0] is invalid slice