String Traversal and Character Access in DSA Python - Build from Scratch

String Traversal and Character Access

📖 Scenario: Imagine you are creating a simple program that reads a word and checks each letter one by one. This is like reading a word slowly to understand each letter.

🎯 Goal: You will build a program that stores a word, sets up a counter, goes through each letter in the word, and prints each letter on its own line.

📋 What You'll Learn

Create a string variable with the exact value 'hello'

Create an integer variable called index and set it to 0

Use a while loop with the condition index < len(word)

Inside the loop, print the character at position index in word

Increase index by 1 inside the loop

Print each character on its own line

💡 Why This Matters

🌍 Real World

Reading and processing text one character at a time is useful in text editors, spell checkers, and simple parsers.

💼 Career

Understanding string traversal is a basic skill needed for programming tasks like data cleaning, input validation, and building text-based applications.

1

Create the word variable

Create a string variable called word and set it to the exact value 'hello'.

DSA Python

# Create the string variable called word
# Your code here

Hint

Use single quotes around the word hello.

2

Set up the index counter

Create an integer variable called index and set it to 0.

DSA Python

word = 'hello'
# Create the index variable and set it to 0
# Your code here

Hint

This variable will help us move through each letter in the word.

3

Traverse the word using a while loop

Use a while loop with the condition index < len(word). Inside the loop, print the character at position index in word. Then increase index by 1.

DSA Python

word = 'hello'
index = 0
# Use a while loop to go through each character in word
# Your code here

Hint

Remember to increase the index inside the loop to avoid an infinite loop.

4

Print each character on its own line

Run the program to print each character of word on its own line using the print statement inside the loop.

DSA Python

word = 'hello'
index = 0
while index < len(word):
    print(word[index])
    index += 1
# Run the program to see the output

Hint

The output should show each letter of 'hello' on a new line.

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