String Traversal and Character Access in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does string traversal mean?

String traversal means going through each character in a string one by one, usually from the first character to the last.

beginner

How do you access the third character in a string named text in Python?

You use text[2] because counting starts at zero. So the first character is text[0], second is text[1], and third is text[2].

beginner

What happens if you try to access a character at an index that is outside the string length?

You get an error called IndexError because the position does not exist in the string.

intermediate

Why is zero-based indexing used in string traversal?

Zero-based indexing means counting starts at 0. It helps computers calculate positions easily and is a common standard in programming.

beginner

How can you loop through each character in a string s in Python?

You can use a for loop like this:
for char in s:
print(char)
This prints each character one by one.

What is the index of the first character in a string?

A Depends on the string length

B 1

C -1

D 0

In most programming languages including Python, string indexing starts at 0.

What will print('hello'[1]) output?

A l

B h

C e

D o

Index 1 is the second character, which is 'e' in 'hello'.

Which of these will cause an error?

A Accessing <code>string[0]</code> in a non-empty string

B Accessing <code>string[len(string)]</code>

C Looping through string with <code>for char in string</code>

D Using <code>string[-1]</code> to get last character

Index equal to string length is out of range and causes IndexError.

How do you get the last character of a string s in Python?

A <code>s[-1]</code>

B <code>s[len(s)]</code>

C <code>s[0]</code>

D <code>s[len(s)-2]</code>

Negative index -1 accesses the last character in Python strings.

What does this code do?
for i in range(len(s)):
print(s[i])

A Prints each character in string s one by one

B Prints the length of s

C Prints the string s reversed

D Causes an error

It loops over each index and prints the character at that index.

Explain how to access characters in a string and what happens if you use an invalid index.

Think about how counting starts and what happens if you go beyond the string length.

Describe two ways to traverse a string in Python and print each character.

One way uses characters directly, the other uses indexes.

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