String searching (contains, strfind) in MATLAB - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to check if the string 'hello' contains the substring 'll'.

MATLAB

result = [1]('hello', 'll');

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

Aindex

Bstrfind

Cfindstr

Dcontains

Attempts:

3 left

💡 Hint

Common Mistakes

Using strfind returns indices, not a logical value.

The contains function returns true if the substring is found.

2fill in blank

medium

Complete the code to find the starting index of substring 'cat' in the string 'concatenate'.

MATLAB

index = [1]('concatenate', 'cat');

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

Acontains

Bstrfind

Cfindstr

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using contains returns logical, not index.

The strfind function returns the starting index of the substring.

3fill in blank

hard

Fix the error in the code to correctly find if 'dog' is in 'hotdog'.

MATLAB

found = [1]('hotdog', 'dog');
if found
    disp('Found!')
end

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

Afindstr

Bstrfind

Ccontains

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Using strfind returns indices, not a logical value.

contains returns a logical value suitable for if conditions.

4fill in blank

hard

Fill both blanks to create a dictionary of words and their starting indices if the word length is greater than 3.

MATLAB

words = {'apple', 'bat', 'cat', 'doge'};
indices = containers.Map();
for i = 1:length(words)
    word = words{i};
    if length(word) [1] 3
        indices(word) = [2]('concatenate', word);
    end
end

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

A>

Bstrfind

C<

Dcontains

Attempts:

3 left

💡 Hint

Common Mistakes

Using contains instead of strfind for index.

Use > to check length and strfind to find the index.

5fill in blank

hard

Fill all three blanks to create a map of uppercase words and their indices if the word contains 'at'.

MATLAB

words = {'cat', 'dog', 'bat', 'rat'};
indices = containers.Map();
for i = 1:length(words)
    word = words{i};
    if [1](word, 'at')
        indices([2]) = [3]('concatenate', word);
    end
end

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

Acontains

Bword

Cstrfind

Dupper(word)

Attempts:

3 left

💡 Hint

Common Mistakes

Using word instead of upper(word) as key.

Use contains to check substring, upper(word) as key, and strfind for index.

Practice

1. Which MATLAB function returns a logical true or false indicating if a substring exists within a string?

easy

A. contains

B. strfind

C. findstr

D. substring

Solution

1. Step 1: Understand the purpose of contains

   contains returns true or false if a substring is found inside a string.

2. Step 2: Compare with strfind

   strfind returns positions, not logical true/false.

3. Final Answer:

   contains -> Option A

4. Quick Check:

   Logical check for substring = contains [OK]

Hint: Use contains for true/false substring check [OK]

Common Mistakes:

• Confusing strfind with contains
• Thinking strfind returns true/false
• Using findstr which is outdated

2. Which of the following is the correct syntax to find the starting indices of substring 'cat' in string str = 'concatenate' using strfind?

easy

A. contains(str, 'cat')

B. strfind('cat', str)

C. strfind(str, 'cat')

D. strfind(str, cat)

Solution

1. Step 1: Recall strfind syntax

   The first argument is the main string, second is the substring to find.

2. Step 2: Check each option

   strfind(str, 'cat') correctly uses strfind(str, 'cat'). strfind('cat', str) reverses arguments. contains(str, 'cat') uses contains, not strfind. strfind(str, cat) misses quotes around 'cat'.

3. Final Answer:

   strfind(str, 'cat') -> Option C

4. Quick Check:

   strfind(mainString, substring) = strfind(str, 'cat') [OK]

Hint: First argument main string, second substring in strfind [OK]

Common Mistakes:

• Swapping arguments in strfind
• Forgetting quotes around substring
• Using contains instead of strfind for indices

3. What is the output of the following MATLAB code?

str = 'abracadabra';
pos = strfind(str, 'ra');
disp(pos);

medium

A. [3 10]

B. [2 9]

C. [4 11]

D. []

Solution

1. Step 1: Identify substring positions in 'abracadabra'

   Look for 'ra' in the string: 'abracadabra'. The substring 'ra' starts at index 3 and 10.

2. Step 2: Confirm output of strfind

   strfind returns starting indices of matches, so output is [3 10].

3. Final Answer:

   [3 10] -> Option A

4. Quick Check:

   Positions of 'ra' = [3 10] [OK]

Hint: Count string indices carefully for strfind output [OK]

Common Mistakes:

• Counting indices starting at 0 instead of 1
• Confusing substring length with position
• Expecting logical output instead of indices

4. Identify the error in this MATLAB code snippet:

str = 'hello world';
pos = strfind('world', str);
disp(pos);

medium

A. disp cannot display arrays

B. Arguments of strfind are reversed

C. strfind cannot search for 'world'

D. Missing semicolon after strfind

Solution

1. Step 1: Check strfind argument order

   strfind expects the main string first, then the substring. Here, arguments are reversed.

2. Step 2: Understand impact of reversed arguments

   Searching 'world' inside 'hello world' is correct, but reversed arguments cause wrong or empty output.

3. Final Answer:

   Arguments of strfind are reversed -> Option B

4. Quick Check:

   strfind(mainString, substring) correct order [OK]

Hint: Main string first, substring second in strfind [OK]

Common Mistakes:

• Swapping arguments in strfind
• Assuming disp can't show arrays
• Thinking semicolon is mandatory for correctness

5. You want to find all positions of the substring 'ana' in str = 'banana bandana' and check if it exists. Which code correctly does this?

hard

A. positions = strfind('ana', str); exists = contains('ana', str);

B. positions = contains(str, 'ana'); exists = strfind(str, 'ana');

C. positions = find(str, 'ana'); exists = contains(str, 'ana');

D. positions = strfind(str, 'ana'); exists = contains(str, 'ana');

Solution

1. Step 1: Use strfind to get positions

   strfind(str, 'ana') returns all starting indices of 'ana' in the string.

2. Step 2: Use contains to check existence

   contains(str, 'ana') returns true if 'ana' is found anywhere in str.

3. Step 3: Verify options

   positions = strfind(str, 'ana'); exists = contains(str, 'ana'); correctly uses both functions. positions = contains(str, 'ana'); exists = strfind(str, 'ana'); swaps roles. positions = strfind('ana', str); exists = contains('ana', str); reverses arguments. positions = find(str, 'ana'); exists = contains(str, 'ana'); uses invalid function find for strings.

4. Final Answer:

   positions = strfind(str, 'ana'); exists = contains(str, 'ana'); -> Option D

5. Quick Check:

   Positions with strfind, existence with contains [OK]

Hint: Use strfind for positions, contains for true/false [OK]

Common Mistakes:

• Swapping arguments in strfind or contains
• Using find instead of strfind
• Confusing output types of contains and strfind