String Traversal and Character Access in DSA Python

Mental Model

A string is a sequence of characters you can look at one by one from start to end.

Analogy: Like reading a sentence letter by letter to understand or find a specific word.

"H" -> "e" -> "l" -> "l" -> "o" -> null
 ↑

Dry Run Walkthrough

Input: string: "Hello", print each character one by one

Goal: To visit and print each character in the string in order

Step 1: Start at the first character 'H'

"H" -> "e" -> "l" -> "l" -> "o" -> null
 ↑

Why: We begin reading from the first character

Step 2: Print 'H' and move to next character 'e'

"H" -> "e" -> "l" -> "l" -> "o" -> null
    ↑

Why: We print current character and move forward to continue traversal

Step 3: Print 'e' and move to next character 'l'

"H" -> "e" -> "l" -> "l" -> "o" -> null
       ↑

Why: Continue printing each character in order

Step 4: Print 'l' and move to next character 'l'

"H" -> "e" -> "l" -> "l" -> "o" -> null
          ↑

Why: Keep moving forward to print all characters

Step 5: Print 'l' and move to next character 'o'

"H" -> "e" -> "l" -> "l" -> "o" -> null
             ↑

Why: Almost at the end, print each character

Step 6: Print 'o' and reach end of string

"H" -> "e" -> "l" -> "l" -> "o" -> null
                ↑

Why: Last character printed, traversal complete

Result:

"H" -> "e" -> "l" -> "l" -> "o" -> null
Printed characters: H e l l o

Annotated Code

DSA Python

class StringTraversal:
    def __init__(self, text: str):
        self.text = text

    def traverse_and_print(self):
        for i in range(len(self.text)):
            print(self.text[i], end=' ')

if __name__ == '__main__':
    s = StringTraversal("Hello")
    s.traverse_and_print()

for i in range(len(self.text)):

loop through each index from 0 to length-1 to access characters in order

print(self.text[i], end=' ')

print character at current index without newline to show sequence

Output

H e l l o 

Complexity Analysis

Time: O(n) because we visit each character once where n is string length

Space: O(1) because no extra space is used besides loop variables

vs Alternative: Direct indexing is simpler and faster than converting string to list first

Edge Cases

empty string ""

loop does not run, nothing is printed

DSA Python

for i in range(len(self.text)):

string with one character "A"

prints single character 'A' correctly

DSA Python

for i in range(len(self.text)):

When to Use This Pattern

When you need to process or examine each character in a string one by one, use string traversal to access characters by their position.

Common Mistakes

Mistake: Trying to modify characters directly in the string (strings are immutable)
Fix: Use a new string or list to store changes instead of changing original string

Mistake: Using incorrect index range causing index out of range error
Fix: Always loop from 0 to len(string)-1 using range(len(string))

Summary

It visits each character in a string one by one in order.

Use it when you want to read or process every character in a string.

Remember strings are like a chain of characters you can access by position starting at zero.

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