String Traversal and Character Access in DSA Python - Time & Space Complexity

Time Complexity: String Traversal and Character Access

O(n)

Understanding Time Complexity

When we look at how fast a program runs that goes through a string, we want to know how the time changes as the string gets longer.

We ask: How does the time to check each letter grow when the string grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

def print_chars(s):
    for i in range(len(s)):
        print(s[i])

This code goes through each character in the string and prints it one by one.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing each character in the string one by one.
• How many times: Exactly once for each character in the string.

How Execution Grows With Input

As the string gets longer, the number of times we print characters grows the same way.

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

Pattern observation: The operations grow directly with the string length; double the string, double the work.

Final Time Complexity

Time Complexity: O(n)

This means the time to run the code grows in a straight line with the size of the string.

Common Mistake

[X] Wrong: "Accessing each character is instant and does not add to time as the string grows."

[OK] Correct: Each character must be checked one by one, so more characters mean more time.

Interview Connect

Understanding how going through a string works helps you explain how your code handles data step by step, a skill useful in many coding questions.

Self-Check

"What if we used a while loop with a pointer instead of a for loop? How would the time complexity change?"

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