What is the time complexity of Peeking Iterator?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Peeking Iterator?

This problem uses Array, Design, Iterator. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Peeking Iterator?

Insight 1: The core idea is to cache the next element internally, so `peek()` can return it without changing the iterator's state. `next()` returns the cached element and updates the cache if there's another element available. Insight 2: We need to maintain a boolean flag to track if there's a 'peeked' element available or if the iterator is exhausted. This helps optimize the `hasNext()` method.

Which companies ask LeetCode 284. Peeking Iterator in interviews?

This problem is frequently asked at Google.

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Peeking Iterator - LeetCode 284 Solution

Peeking Iterator - Complete Solution Guide

Peeking Iterator is LeetCode problem 284, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Design an iterator that supports the peek operation on an existing iterator in addition to the hasNext and the next operations. Implement the PeekingIterator class: PeekingIterator(Iterator<int> nums) Initializes the object with the given integer iterator iterator . int next() Returns the next element in the array and moves the pointer to the next element. boolean hasNext() Returns true if there are still elements in the array. int peek() Returns the next element in the array without moving the

Detailed Explanation

The problem asks us to design an iterator that extends the functionality of a standard iterator by adding a `peek()` operation. The `peek()` operation allows us to look at the next element without advancing the iterator's position. We need to implement a `PeekingIterator` class that wraps an existing iterator and provides `next()`, `hasNext()`, and `peek()` methods. The constraints state that the input array's length is at most 1000 and its elements are between 1 and 1000 inclusive, calls to `next` and `peek` are always valid, and we can make at most 1000 calls to any of the methods.

Solution Approach

The solution uses a combination of caching and a boolean flag. The constructor initializes the PeekingIterator by taking an existing Iterator. It then tries to fetch the first element using `iterator.next()` and stores it in a `_peeked_element` variable. It also maintains a boolean flag `_has_next` to indicate if there's a next element available. The `peek()` method simply returns the `_peeked_element`. The `next()` method returns the current `_peeked_element`, and then updates the `_peeked_element` by advancing the underlying iterator, if there is one. If there are no more elements after calling `next()`, `_has_next` is set to `false` and the _peeked_element is set to null. The `hasNext()` method returns the value of the `_has_next` flag.

Step-by-Step Algorithm

Step 1: Constructor - Initialize the PeekingIterator with an existing Iterator. Cache the first element (if it exists) in `_peeked_element` and set the `_has_next` flag accordingly.
Step 2: peek() - Return the value of the cached `_peeked_element`. This doesn't advance the iterator.
Step 3: next() - Return the current value of `_peeked_element`. Advance the underlying iterator and update `_peeked_element` and `_has_next` if there are more elements. Handle the case where there are no more elements.
Step 4: hasNext() - Return the value of the `_has_next` flag, indicating if there are more elements in the iterator.

Key Insights

Insight 1: The core idea is to cache the next element internally, so `peek()` can return it without changing the iterator's state. `next()` returns the cached element and updates the cache if there's another element available.
Insight 2: We need to maintain a boolean flag to track if there's a 'peeked' element available or if the iterator is exhausted. This helps optimize the `hasNext()` method.
Insight 3: Careful handling of edge cases, such as when the iterator is initially empty, is crucial for correctness.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Array, Design, Iterator.

Companies

Asked at: Google.

Peeking Iterator | LeetCode 284

Medium

Array Design Iterator

Time: O(1)Space: O(1)

Solution Code

Available in:

// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html

import java.util.Iterator;

class PeekingIterator implements Iterator<Integer> {
    private Iterator<Integer> iterator;
    private Integer _peeked_element;
    private boolean _has_next;

	public PeekingIterator(Iterator<Integer> iterator) {
	    // initialize any member here.
	    this.iterator = iterator;
        this._peeked_element = null;
        this._has_next = this.iterator.hasNext();
        if (this._has_next) {
            this._peeked_element = this.iterator.next();
        }
	}
	
    // Returns the next element in the iteration without advancing the iterator.
	public Integer peek() {
        return this._peeked_element;
	}
	
	// hasNext() and next() should behave the same as in the Iterator interface.
	// Override them if needed.
	@Override
	public Integer next() {
        Integer result = this._peeked_element;
        
        if (this.iterator.hasNext()) {
            this._peeked_element = this.iterator.next();
        } else {
            this._has_next = false;
            this._peeked_element = null;
        }
        
        return result;
	}
	
	@Override
	public boolean hasNext() {
        return this._has_next;
	}
}

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Problem Info

Description

Problem 284: Peeking Iterator - Detailed explanation and solution approaches.

Topics

Array Design Iterator

Companies

Google

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:284

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

4 steps

1
Step 1: Constructor - Initialize the PeekingIterator with an existing Iterator. Cache the first element (if it exists) in `_peeked_element` and set the `_has_next` flag accordingly.
2
Step 2: peek() - Return the value of the cached `_peeked_element`. This doesn't advance the iterator.
3
Step 3: next() - Return the current value of `_peeked_element`. Advance the underlying iterator and update `_peeked_element` and `_has_next` if there are more elements. Handle the case where there are no more elements.
4
Step 4: hasNext() - Return the value of the `_has_next` flag, indicating if there are more elements in the iterator.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

Since all operations only require a constant number of operations (reading cached values, or advancing the underlying iterator by one position) and the storage is only for two variables, the complexity stays at O(1) for both time and space.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Storing all elements in a list

We could store all the elements from the given iterator in an internal list during the constructor, and then maintain an index pointing to the current position. This way, `peek` and `next` can be easily implemented by accessing the list at the appropriate index.

Time: O(N) for the constructor to copy the elements, O(1) for `next`, `peek` and `hasNext`.Space: O(N) to store the list of elements.

Trade-offs: This approach uses O(N) space, which is less efficient than the O(1) space of the presented solution. However, it could be simpler to implement if modifying the original iterator is prohibited. The initial overhead in the constructor is a significant disadvantage if the methods (especially `peek` or `next`) are called only a few times.

✨

Key Insights

1Insight 1: The core idea is to cache the next element internally, so `peek()` can return it without changing the iterator's state. `next()` returns the cached element and updates the cache if there's another element available.
2Insight 2: We need to maintain a boolean flag to track if there's a 'peeked' element available or if the iterator is exhausted. This helps optimize the `hasNext()` method.
3Insight 3: Careful handling of edge cases, such as when the iterator is initially empty, is crucial for correctness.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: The input iterator is initially empty. The constructor should handle this case correctly by setting `_has_next` to `false` and `_peeked_element` to `null` (or a similar equivalent in the respective language). This prevents issues when `peek()` or `next()` are called on an empty iterator.
Edge case 2: Calling `next()` repeatedly until the iterator is exhausted. The code should correctly update `_has_next` to `false` and `_peeked_element` to `null` when the last element has been returned, so `hasNext()` returns `false` and `next()` doesn't throw errors.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle the case where the input iterator is initially empty. This leads to null pointer exceptions or incorrect behavior when `peek()` or `next()` are called.

✗Common mistake 2: Forgetting to update the `_has_next` flag correctly after calling `next()`. This can lead to `hasNext()` returning the wrong result and `next()` being called on an empty iterator.

✗Common mistake 3: Modifying the internal state of the iterator within the peek() method. The peek() method must not move the iterator cursor.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Implementing a custom data stream processor where you need to look ahead at the next element in the stream without consuming it. This is useful for conditional processing or determining the type of the next element before deciding how to handle it.

🏦Real-world scenario 2: Implementing a compiler or parser where you need to peek at the next token in the input stream to determine the next parsing action. For example, you might need to peek at the next token to decide whether to start a new statement or continue the current one.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 13, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Peeking Iterator"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether design techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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