String as character array in Data Structures Theory - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to access the first character of the string.

Data Structures Theory

char firstChar = str[1];

Drag options to blanks, or click blank then click option'

A{0}

B(0)

C[0]

D<0>

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets.

Starting index at 1 instead of 0.

In many programming languages, strings can be treated like arrays of characters. To access the first character, you use square brackets with index 0.

2fill in blank

medium

Complete the code to get the length of the string.

Data Structures Theory

int length = str[1];

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A.index()

B.size()

C.count()

D.length()

Attempts:

3 left

💡 Hint

Common Mistakes

Using .size() which may work in some languages but not all.

Using .count() which counts occurrences of a character.

The .length() method returns the number of characters in a string.

3fill in blank

hard

Fix the error in accessing the last character of the string.

Data Structures Theory

char lastChar = str[1];

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A[str.length() + 1]

B[str.length() - 1]

C[str.length()]

D[str.length() * 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using str.length() as index which is out of bounds.

Adding 1 to length causing an error.

Since string indices start at 0, the last character is at index length - 1.

4fill in blank

hard

Fill both blanks to create a substring from index 2 to 5.

Data Structures Theory

String sub = str[1]([2], 4);

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A.substring

B2

C3

D.substr

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong method name.

Starting at the wrong index.

The .substr(start, length) method extracts a part of the string starting at index start with the given length.

5fill in blank

hard

Fill all three blanks to create a dictionary of characters and their counts from the string.

Data Structures Theory

charCount = [1](c: str[[2]] for [3] in range(len(str)))

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Adict

Bi

Dlist

Attempts:

3 left

💡 Hint

Common Mistakes

Using a list instead of a dictionary.

Using a variable name other than the loop index.

This code creates a dictionary where each character from the string is a key. It uses a loop variable i to access each character by index.

Practice

1. What does it mean to treat a string as a character array?

easy

A. Changing the string into a list of words

B. Storing numbers instead of letters

C. Combining multiple strings into one

D. Accessing each letter individually by its position

Solution

1. Step 1: Understand string as sequence of characters

   A string is made up of letters arranged in order, like beads on a string.

2. Step 2: Accessing letters by position

   Treating a string as a character array means you can get each letter by its place number (index).

3. Final Answer:

   Accessing each letter individually by its position -> Option D

4. Quick Check:

   String as character array = access letters by position [OK]

Hint: Think of string as a list of letters you can pick one by one [OK]

Common Mistakes:

• Confusing string with list of words
• Thinking string stores numbers only
• Believing strings cannot be accessed by position

2. Which of the following is the correct way to access the third character of a string s in most programming languages?

easy

A. s[2]

B. s(3)

C. s[3]

D. s{2}

Solution

1. Step 1: Recall zero-based indexing

   Most languages start counting positions from 0, so first character is s[0], second s[1], third s[2].

2. Step 2: Identify correct syntax for indexing

   Square brackets [] are used to access characters by index, so s[2] is correct.

3. Final Answer:

   s[2] -> Option A

4. Quick Check:

   Third character index = 2, use s[2] [OK]

Hint: Remember counting starts at zero, so third char is index 2 [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Using 1-based indexing (s[3])
• Using curly braces for indexing

3. Given the string word = "hello", what is the output of word[1]?

medium

A. "h"

B. "e"

C. "l"

D. "o"

Solution

1. Step 1: Identify the string characters and their indexes

   The string "hello" has characters: h(0), e(1), l(2), l(3), o(4).

2. Step 2: Access character at index 1

   Index 1 corresponds to the second character, which is "e".

3. Final Answer:

   "e" -> Option B

4. Quick Check:

   word[1] = "e" [OK]

Hint: Index 1 means second letter in the string [OK]

Common Mistakes:

• Choosing first character instead of second
• Confusing index 1 with index 2
• Thinking indexing starts at 1

4. What is wrong with this code snippet if text = "world" and we try to print text[5]?

medium

A. It will cause an error because index 5 is out of range

B. It will print a space character

C. It will print the last character

D. It will print the first character

Solution

1. Step 1: Check string length and valid indexes

   The string "world" has length 5, with valid indexes 0 to 4.

2. Step 2: Understand index out of range error

   Accessing index 5 is beyond the last character, causing an error in most languages.

3. Final Answer:

   It will cause an error because index 5 is out of range -> Option A

4. Quick Check:

   Index 5 invalid for length 5 string [OK]

Hint: Indexes go from 0 to length-1; 5 is too big here [OK]

Common Mistakes:

• Assuming index 5 returns last character
• Expecting a space or empty character
• Thinking indexing starts at 1

5. You want to change the first letter of the string name = "alice" to uppercase by treating it as a character array. Which approach is correct?

hard

A. Directly assign name[0] = 'A' (works in all languages)

B. Use name.upper()[0] to change first letter only

C. Convert string to list, change first element, then join back

D. Strings are immutable; you cannot change any character

Solution

1. Step 1: Understand string immutability

   In many languages, strings cannot be changed directly because they are immutable.

2. Step 2: Use list conversion to modify characters

   Convert the string to a list (array) of characters, change the first element, then join back to a string.

3. Final Answer:

   Convert string to list, change first element, then join back -> Option C

4. Quick Check:

   Modify string by list conversion and join [OK]

Hint: Strings can't change directly; convert to list first [OK]

Common Mistakes:

• Trying to assign directly to string index
• Using upper() on whole string but expecting partial change
• Believing strings are mutable