String indexing (positive and negative) in Python - Time & Space Complexity

Time Complexity: String indexing (positive and negative)

O(1)

Understanding Time Complexity

When we access characters in a string by their position, it is important to know how fast this operation is.

We want to understand how the time to get a character changes as the string gets longer.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

text = "hello world"
char1 = text[2]      # positive index
char2 = text[-3]     # negative index
print(char1, char2)
    

This code gets characters from a string using positive and negative positions.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing a single character by index in the string.
• How many times: Twice, but each access is independent and direct.

How Execution Grows With Input

Getting a character by index does not depend on the string length; it is a direct jump.

Input Size (n)	Approx. Operations
10	2
100	2
1000	2

Pattern observation: The number of operations stays the same no matter how long the string is.

Final Time Complexity

Time Complexity: O(1)

This means accessing a character by index takes the same short time regardless of string length.

Common Mistake

[X] Wrong: "Accessing a character by index takes longer if the string is longer."

[OK] Correct: Strings are stored so that each position can be reached directly, so length does not slow down access.

Interview Connect

Knowing that string indexing is fast helps you understand how to write efficient code when working with text.

Self-Check

"What if we tried to find a character by searching through the string instead of indexing? How would the time complexity change?"

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