String Basics and Memory Representation in DSA Python - Time & Space Complexity

Time Complexity: String Basics and Memory Representation

O(n²)

Understanding Time Complexity

We want to understand how the time to work with strings changes as the string gets longer.

Specifically, how operations like reading or copying a string take more time with bigger strings.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

text = "hello world"
new_text = ""
for char in text:
    new_text += char
print(new_text)
    

This code copies each character from one string to another by adding characters one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through each character in the string.
• How many times: Once for every character in the original string (n times).

How Execution Grows With Input

As the string gets longer, the time to copy it grows faster than just the length.

Input Size (n)	Approx. Operations
10	About 55 operations
100	About 5050 operations
1000	About 500500 operations

Pattern observation: The operations grow roughly with the square of the string length.

Final Time Complexity

Time Complexity: O(n²)

This means copying a string by adding characters one by one takes much more time as the string grows, roughly the square of its length.

Common Mistake

[X] Wrong: "Adding characters one by one to a string is always fast and takes time proportional to the string length (O(n))."

[OK] Correct: Each time you add a character, a new string is created copying all previous characters, so the total work adds up to much more than just n.

Interview Connect

Understanding how string operations cost time helps you write efficient code and explain your choices clearly in interviews.

Self-Check

"What if we used a list to collect characters and joined them at the end? How would the time complexity change?"

Practice

1. Which of the following statements about Python strings is true?

easy

A. Strings are stored as linked lists in memory.

B. Strings are immutable, meaning they cannot be changed after creation.

C. Strings are mutable and can be changed character by character.

D. Strings always use 4 bytes per character regardless of content.

Solution

1. Step 1: Understand string mutability in Python

   Python strings cannot be changed after they are created. Any modification creates a new string.

2. Step 2: Analyze other options

   Strings are not mutable, not stored as linked lists, and use variable bytes depending on encoding.

3. Final Answer:

   Strings are immutable, meaning they cannot be changed after creation. -> Option B

4. Quick Check:

   Strings are immutable [OK]

Hint: Remember: Python strings cannot be changed once made [OK]

Common Mistakes:

• Thinking strings can be changed in place
• Assuming fixed byte size per character
• Confusing string storage with lists

2. Which of the following is the correct way to create a string in Python?

easy

A. str = 'Hello World'

B. str = Hello World

C. str = "Hello World

D. str = 'Hello World"

Solution

1. Step 1: Check string syntax rules

   Strings must be enclosed in matching single or double quotes.

2. Step 2: Validate each option

   str = 'Hello World' uses matching single quotes correctly. Options B, C, and D have missing or mismatched quotes causing syntax errors.

3. Final Answer:

   str = 'Hello World' -> Option A

4. Quick Check:

   Matching quotes create strings correctly [OK]

Hint: Strings need matching quotes around text [OK]

Common Mistakes:

• Forgetting quotes around string text
• Using mismatched quotes
• Confusing string with variable names

3. What will be the output of the following code?

text = 'abc'
print(id(text))
text += 'd'
print(id(text))

medium

A. Two different numbers printed

B. Same number printed twice

C. SyntaxError

D. RuntimeError

Solution

1. Step 1: Understand string immutability and id()

   Strings are immutable, so adding 'd' creates a new string object with a new id.

2. Step 2: Trace the code output

   First print shows id of 'abc', second print shows id of new string 'abcd', which is different.

3. Final Answer:

   Two different numbers printed -> Option A

4. Quick Check:

   String immutability changes id after modification [OK]

Hint: String id changes after modification due to immutability [OK]

Common Mistakes:

• Assuming string id stays same after += operation
• Confusing id() with value
• Expecting syntax or runtime errors here

4. Find the error in the following code snippet:

my_str = 'hello'
my_str[0] = 'H'
print(my_str)

medium

A. Output: hello

B. SyntaxError: invalid syntax

C. TypeError: 'str' object does not support item assignment

D. Output: Hello

Solution

1. Step 1: Identify string mutability rules

   Strings cannot be changed by assigning to an index because they are immutable.

2. Step 2: Analyze the error raised

   Assigning to my_str[0] causes a TypeError about item assignment not supported.

3. Final Answer:

   TypeError: 'str' object does not support item assignment -> Option C

4. Quick Check:

   Immutable strings cause TypeError on index assignment [OK]

Hint: Strings cannot be changed by index assignment [OK]

Common Mistakes:

• Expecting string to change after index assignment
• Thinking code runs without error
• Confusing TypeError with SyntaxError

5. Given the code below, what will be the output?

str1 = 'abc'
str2 = 'abc'
print(str1 is str2)
str3 = ''.join(['a', 'b', 'c'])
print(str1 is str3)

hard

A. False\nTrue

B. False\nFalse

C. True\nTrue

D. True\nFalse

Solution

1. Step 1: Understand string interning for literals

   Python reuses the same object for identical string literals, so str1 and str2 point to the same object.

2. Step 2: Understand string creation with join()

   str3 is created at runtime and is a new object, so str1 is not str3.

3. Final Answer:

   True\nFalse -> Option D

4. Quick Check:

   Literal strings share object, runtime strings do not [OK]

Hint: Literal strings share memory; constructed strings do not [OK]

Common Mistakes:

• Assuming all identical strings share same object
• Confusing 'is' with '=='
• Ignoring how strings are created