Structure arrays in MATLAB - Time & Space Complexity

Time Complexity: Structure arrays

O(n)

Understanding Time Complexity

When working with structure arrays in MATLAB, it is important to understand how the time to access or modify data grows as the array gets bigger.

We want to know how the number of operations changes when we work with more elements in the structure array.

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

% Create a structure array with n elements
for i = 1:n
    students(i).name = sprintf('Student%d', i);
    students(i).score = randi(100);
end

% Access all scores and sum them
totalScore = 0;
for i = 1:n
    totalScore = totalScore + students(i).score;
end
    

This code creates a structure array of students with names and scores, then sums all the scores.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping through the structure array to assign fields and then to access the scores.
• How many times: Each loop runs exactly n times, where n is the number of elements in the structure array.

How Execution Grows With Input

As the number of students (n) increases, the total number of operations grows proportionally.

Input Size (n)	Approx. Operations
10	About 20 (10 for creation + 10 for summing)
100	About 200 (100 + 100)
1000	About 2000 (1000 + 1000)

Pattern observation: The operations double when the input size doubles, showing a steady, linear growth.

Final Time Complexity

Time Complexity: O(n)

This means the time it takes grows directly in proportion to the number of elements in the structure array.

Common Mistake

[X] Wrong: "Accessing fields in a structure array is instant and does not depend on the number of elements."

[OK] Correct: Accessing each element's field requires going through each element, so the time grows with the number of elements.

Interview Connect

Understanding how structure arrays scale helps you write efficient MATLAB code and answer questions about data handling in interviews confidently.

Self-Check

"What if we used vectorized operations or built-in functions instead of loops? How would the time complexity change?"

Practice

1. What is a structure array in MATLAB used for?

easy

A. To perform matrix multiplication

B. To store only numeric arrays

C. To group related data with different types under named fields

D. To create loops for iteration

Solution

1. Step 1: Understand the purpose of structure arrays

   Structure arrays allow grouping of different types of data under named fields.

2. Step 2: Compare options with the definition

   Only To group related data with different types under named fields correctly describes this use; others describe unrelated concepts.

3. Final Answer:

   To group related data with different types under named fields -> Option C

4. Quick Check:

   Structure arrays = group data by fields [OK]

Hint: Remember: structures group different data types by names [OK]

Common Mistakes:

• Thinking structures only hold numeric data
• Confusing structures with loops or matrices

2. Which of the following is the correct way to create a structure array with fields 'Name' and 'Age' in MATLAB?

easy

A. person = struct('Name', 'Alice', 'Age', 30);

B. person = struct(Name = 'Alice', Age = 30);

C. person = struct['Name', 'Alice', 'Age', 30];

D. person = struct{'Name', 'Alice', 'Age', 30};

Solution

1. Step 1: Recall MATLAB struct syntax

   MATLAB uses struct('field1', value1, 'field2', value2) to create structures.

2. Step 2: Check each option's syntax

   person = struct('Name', 'Alice', 'Age', 30); uses correct parentheses and quotes. Options B, C, D use invalid syntax.

3. Final Answer:

   person = struct('Name', 'Alice', 'Age', 30); -> Option A

4. Quick Check:

   struct uses parentheses and quotes [OK]

Hint: Use parentheses and quotes with struct() [OK]

Common Mistakes:

• Using square or curly brackets instead of parentheses
• Omitting quotes around field names

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

student(1).Name = 'John';
student(1).Score = 85;
student(2).Name = 'Jane';
student(2).Score = 92;
disp(student(2).Name);

medium

A. Jane

B. John

C. 85

D. 92

Solution

1. Step 1: Understand structure array indexing

   student(2) accesses the second element; .Name accesses the 'Name' field.

2. Step 2: Identify the value stored in student(2).Name

   It is 'Jane' as assigned in the code.

3. Final Answer:

   Jane -> Option A

4. Quick Check:

   student(2).Name = 'Jane' [OK]

Hint: Use index and field name to get value [OK]

Common Mistakes:

• Confusing index 1 and 2
• Accessing Score instead of Name

4. Identify the error in this MATLAB code snippet:

car.Make = 'Toyota';
car(1).Model = 'Corolla';
car(1).Year = 2020;

medium

A. Field names must be numeric

B. Missing semicolon after first line

C. Mixing scalar and array structure assignment causes error

D. No error, code runs fine

Solution

1. Step 1: Analyze structure assignment styles

   First line creates scalar structure car with field Make. Next lines assign to car(1), valid for scalar structures.

2. Step 2: Understand MATLAB structure array rules

   MATLAB allows indexing scalar structures and dynamically adding fields; no error occurs.

3. Final Answer:

   No error, code runs fine -> Option D

4. Quick Check:

   Scalar struct indexing + field addition works [OK]

Hint: Scalar structures can be indexed like arrays [OK]

Common Mistakes:

• Thinking scalar and array assignments cannot mix
• Believing field names must be numeric
• Assuming missing semicolons cause errors

5. Given a structure array books with fields Title and Pages, how do you create a new structure array shortBooks containing only books with less than 100 pages?

hard

A. shortBooks = books(books.Pages < 100);

B. shortBooks = books([books.Pages] < 100);

C. shortBooks = books{books.Pages < 100};

D. shortBooks = books(books.Pages > 100);

Solution

1. Step 1: Extract numeric array of Pages

   Use [books.Pages] to get all Pages values as a numeric array.

2. Step 2: Use logical indexing to filter

   Compare with < 100 to get logical indices, then index books with these.

3. Step 3: Check syntax correctness

   shortBooks = books([books.Pages] < 100); correctly uses parentheses and logical indexing; others are incorrect.

4. Final Answer:

   shortBooks = books([books.Pages] < 100); -> Option B

5. Quick Check:

   Logical indexing with [books.Pages] works [OK]

Hint: Use [struct.field] for numeric indexing [OK]

Common Mistakes:

• Using curly braces {} instead of parentheses ()
• Forgetting to use brackets [] around field extraction
• Using wrong comparison operator