Iterator pattern in LLD - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare the Iterator interface method.

LLD

interface Iterator {
    boolean [1]();
}

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

AhasNext

Bnext

Cremove

Dreset

Attempts:

3 left

💡 Hint

Common Mistakes

Choosing next instead of hasNext

Using remove which is optional

The hasNext method checks if there are more elements to iterate over.

2fill in blank

medium

Complete the code to return the next element in the iterator.

LLD

public Object [1]() {
    if (!hasNext()) {
        throw new NoSuchElementException();
    }
    return collection.get(currentIndex++);
}

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

Aremove

BhasNext

Cnext

Dreset

Attempts:

3 left

💡 Hint

Common Mistakes

Using hasNext instead of next

Trying to remove element here

The next method returns the next element and advances the iterator.

3fill in blank

hard

Fix the error in the iterator implementation to correctly check for more elements.

LLD

public boolean hasNext() {
    return currentIndex [1] collection.size();
}

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

A<

B<=

C>

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= or > which causes errors

Using <= which is off by one

The hasNext method should return true if currentIndex is less than collection size.

4fill in blank

hard

Fill both blanks to implement a simple iterator for an array.

LLD

class ArrayIterator implements Iterator {
    private Object[] items;
    private int [1] = 0;

    public ArrayIterator(Object[] items) {
        this.items = items;
    }

    public boolean hasNext() {
        return [2] < items.length;
    }
}

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

AcurrentIndex

Bindex

Cpointer

Dposition

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variable names inconsistently

Not initializing the index variable

The variable currentIndex tracks the current position in the array for iteration.

5fill in blank

hard

Fill all three blanks to complete the iterator's next method.

LLD

public Object [1]() {
    if (!hasNext()) {
        throw new NoSuchElementException();
    }
    return items[[2]++];
}

public void [3]() {
    [2] = 0;
}

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

Anext

BcurrentIndex

Creset

DhasNext

Attempts:

3 left

💡 Hint

Common Mistakes

Confusing method names

Not incrementing index properly

Missing reset method

The next method returns the current item and increments the index. The reset method sets the index back to zero.

Practice

1.

What is the main purpose of the Iterator pattern in system design?

easy

A. To manage user authentication and authorization

B. To store data in a database efficiently

C. To create multiple copies of an object

D. To provide a way to access elements of a collection sequentially without exposing its underlying structure

Solution

1. Step 1: Understand the role of Iterator pattern

   The Iterator pattern is designed to provide a way to access elements of a collection one by one without revealing the internal structure of the collection.

2. Step 2: Compare with other options

   Options B, C, and D describe unrelated design patterns or system functions such as data storage, object cloning, and security management.

3. Final Answer:

   To provide a way to access elements of a collection sequentially without exposing its underlying structure -> Option D

4. Quick Check:

   Iterator pattern = Access collection without exposing structure [OK]

Hint: Iterator = access elements without showing internal details [OK]

Common Mistakes:

• Confusing Iterator with data storage or cloning patterns
• Thinking Iterator manages security or authentication
• Assuming Iterator modifies the collection

2.

Which of the following is the correct method signature for the next() method in an iterator interface?

easy

A. def next() -> void

B. def next(self, index) -> Element

C. def next(self) -> Element

D. def next(self, element) -> bool

Solution

1. Step 1: Recall the standard iterator method signature

   The next() method typically takes no parameters except the implicit self and returns the next element in the collection.

2. Step 2: Analyze each option

   def next(self) -> Element matches the standard signature: it takes self and returns an element. Options B and D incorrectly add parameters, and C returns void which is incorrect.

3. Final Answer:

   def next(self) -> Element -> Option C

4. Quick Check:

   next() takes no args, returns element [OK]

Hint: next() returns next element, no extra parameters [OK]

Common Mistakes:

• Adding parameters to next() method
• Returning void instead of element
• Confusing next() with hasNext() method

3.

Consider the following Python code implementing a simple iterator:

class MyIterator:
    def __init__(self, data):
        self.data = data
        self.index = 0
    def __iter__(self):
        return self
    def __next__(self):
        if self.index < len(self.data):
            result = self.data[self.index]
            self.index += 1
            return result
        else:
            raise StopIteration

it = MyIterator([10, 20, 30])
print(next(it))
print(next(it))

What will be the output?

medium

A. 20\n30

B. 10\n20

C. 10\n30

D. Error at runtime

Solution

1. Step 1: Trace the iterator's next calls

   First call to next(it) returns data[0] = 10 and increments index to 1. Second call returns data[1] = 20 and increments index to 2.

2. Step 2: Confirm no errors occur

   Since index is less than length during both calls, no StopIteration is raised.

3. Final Answer:

   10 20 -> Option B

4. Quick Check:

   First two elements printed: 10 and 20 [OK]

Hint: next() returns elements in order, increments index [OK]

Common Mistakes:

• Assuming next() skips elements
• Expecting error before StopIteration
• Mixing up index increments

4.

Given this iterator implementation in Python, identify the bug:

class BuggyIterator:
    def __init__(self, data):
        self.data = data
        self.index = 0
    def __iter__(self):
        return self
    def __next__(self):
        if self.index <= len(self.data):
            result = self.data[self.index]
            self.index += 1
            return result
        else:
            raise StopIteration

What is the cause of the error when iterating?

medium

A. IndexError due to accessing out-of-range element

B. StopIteration raised too early

C. Infinite loop because index never increments

D. Syntax error in method definitions

Solution

1. Step 1: Analyze the condition in __next__

   The condition uses <= len(self.data), which allows index to equal length, causing out-of-range access.

2. Step 2: Understand the error caused

   Accessing self.data[self.index] when index == len(self.data) causes IndexError because list indices go from 0 to len-1.

3. Final Answer:

   IndexError due to accessing out-of-range element -> Option A

4. Quick Check:

   Condition allows index == length causing IndexError [OK]

Hint: Use < not <= to avoid out-of-range errors [OK]

Common Mistakes:

• Using <= instead of < in boundary check
• Assuming StopIteration triggers before error
• Ignoring index increment effects

5.

You need to design an iterator for a complex data structure that contains nested lists of integers. Which approach best follows the Iterator pattern principles to allow clients to iterate over all integers seamlessly?

1. Flatten the nested lists into a single list before iteration.
2. Implement a recursive iterator that yields integers from nested lists on demand.
3. Expose the internal nested list structure and let clients handle iteration.
4. Provide separate iterators for each nested list and require clients to manage them.

hard

A. Implement a recursive iterator that yields integers from nested lists on demand

B. Flatten the nested lists into a single list before iteration

C. Expose the internal nested list structure and let clients handle iteration

D. Provide separate iterators for each nested list and require clients to manage them

Solution

1. Step 1: Understand Iterator pattern goal

   The pattern aims to hide internal structure and provide a simple way to access elements sequentially.

2. Step 2: Evaluate each approach

   Flatten the nested lists into a single list before iteration flattens data upfront, which may be inefficient and breaks lazy access. Implement a recursive iterator that yields integers from nested lists on demand uses a recursive iterator to yield elements on demand, hiding complexity and supporting lazy iteration. Options C and D expose internal structure or complexity to clients, violating encapsulation.

3. Final Answer:

   Implement a recursive iterator that yields integers from nested lists on demand -> Option A

4. Quick Check:

   Recursive iterator hides structure, yields elements lazily [OK]

Hint: Use recursive iterator to hide nested structure [OK]

Common Mistakes:

• Flattening data upfront losing lazy iteration benefits
• Exposing internal structure breaking encapsulation
• Forcing clients to manage multiple iterators