LeetCode #284 — MEDIUM

Peeking Iterator

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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 pointer.

Note: Each language may have a different implementation of the constructor and Iterator, but they all support the int next() and boolean hasNext() functions.

Example 1:

Input
["PeekingIterator", "next", "peek", "next", "next", "hasNext"]
[[[1, 2, 3]], [], [], [], [], []]
Output
[null, 1, 2, 2, 3, false]

Explanation
PeekingIterator peekingIterator = new PeekingIterator([1, 2, 3]); // [1,2,3]
peekingIterator.next();    // return 1, the pointer moves to the next element [1,2,3].
peekingIterator.peek();    // return 2, the pointer does not move [1,2,3].
peekingIterator.next();    // return 2, the pointer moves to the next element [1,2,3]
peekingIterator.next();    // return 3, the pointer moves to the next element [1,2,3]
peekingIterator.hasNext(); // return False

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 1000
All the calls to next and peek are valid.
At most 1000 calls will be made to next, hasNext, and peek.

Follow up: How would you extend your design to be generic and work with all types, not just integer?

Step 01

Brute Force Baseline

Problem summary: 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 pointer. Note: Each language may have a different implementation of the constructor and Iterator, but they all support the int next() and boolean hasNext() functions.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Design

Example 1

["PeekingIterator","next","peek","next","next","hasNext"]
[[[1,2,3]],[],[],[],[],[]]

Core Insight

What unlocks the optimal approach

Think of "looking ahead". You want to cache the next element.
Is one variable sufficient? Why or why not?
Test your design with call order of <code>peek()</code> before <code>next()</code> vs <code>next()</code> before <code>peek()</code>.
For a clean implementation, check out <a href="https://github.com/google/guava/blob/703ef758b8621cfbab16814f01ddcc5324bdea33/guava-gwt/src-super/com/google/common/collect/super/com/google/common/collect/Iterators.java#L1125" target="_blank">Google's guava library source code</a>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #284: Peeking Iterator
// Java Iterator interface reference:
// https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html

class PeekingIterator implements Iterator<Integer> {
    private Iterator<Integer> iterator;
    private boolean hasPeeked;
    private Integer peekedElement;

    public PeekingIterator(Iterator<Integer> iterator) {
        // initialize any member here.
        this.iterator = iterator;
    }

    // Returns the next element in the iteration without advancing the iterator.
    public Integer peek() {
        if (!hasPeeked) {
            peekedElement = iterator.next();
            hasPeeked = true;
        }
        return peekedElement;
    }

    // hasNext() and next() should behave the same as in the Iterator interface.
    // Override them if needed.
    @Override
    public Integer next() {
        if (!hasPeeked) {
            return iterator.next();
        }
        Integer result = peekedElement;
        hasPeeked = false;
        peekedElement = null;
        return result;
    }

    @Override
    public boolean hasNext() {
        return hasPeeked || iterator.hasNext();
    }
}

// Accepted solution for LeetCode #284: Peeking Iterator
/*   Below is the interface for Iterator, which is already defined for you.
 *
 *   type Iterator struct {
 *
 *   }
 *
 *   func (this *Iterator) hasNext() bool {
 *		// Returns true if the iteration has more elements.
 *   }
 *
 *   func (this *Iterator) next() int {
 *		// Returns the next element in the iteration.
 *   }
 */

type PeekingIterator struct {
	iter          *Iterator
	hasPeeked     bool
	peekedElement int
}

func Constructor(iter *Iterator) *PeekingIterator {
	return &PeekingIterator{iter, iter.hasNext(), iter.next()}
}

func (this *PeekingIterator) hasNext() bool {
	return this.hasPeeked || this.iter.hasNext()
}

func (this *PeekingIterator) next() int {
	if !this.hasPeeked {
		return this.iter.next()
	}
	this.hasPeeked = false
	return this.peekedElement
}

func (this *PeekingIterator) peek() int {
	if !this.hasPeeked {
		this.peekedElement = this.iter.next()
		this.hasPeeked = true
	}
	return this.peekedElement
}

# Accepted solution for LeetCode #284: Peeking Iterator
# Below is the interface for Iterator, which is already defined for you.
#
# class Iterator:
#     def __init__(self, nums):
#         """
#         Initializes an iterator object to the beginning of a list.
#         :type nums: List[int]
#         """
#
#     def hasNext(self):
#         """
#         Returns true if the iteration has more elements.
#         :rtype: bool
#         """
#
#     def next(self):
#         """
#         Returns the next element in the iteration.
#         :rtype: int
#         """


class PeekingIterator:
    def __init__(self, iterator):
        """
        Initialize your data structure here.
        :type iterator: Iterator
        """
        self.iterator = iterator
        self.has_peeked = False
        self.peeked_element = None

    def peek(self):
        """
        Returns the next element in the iteration without advancing the iterator.
        :rtype: int
        """
        if not self.has_peeked:
            self.peeked_element = self.iterator.next()
            self.has_peeked = True
        return self.peeked_element

    def next(self):
        """
        :rtype: int
        """
        if not self.has_peeked:
            return self.iterator.next()
        result = self.peeked_element
        self.has_peeked = False
        self.peeked_element = None
        return result

    def hasNext(self):
        """
        :rtype: bool
        """
        return self.has_peeked or self.iterator.hasNext()


# Your PeekingIterator object will be instantiated and called as such:
# iter = PeekingIterator(Iterator(nums))
# while iter.hasNext():
#     val = iter.peek()   # Get the next element but not advance the iterator.
#     iter.next()         # Should return the same value as [val].

// Accepted solution for LeetCode #284: Peeking Iterator
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #284: Peeking Iterator
// // Java Iterator interface reference:
// // https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
// 
// class PeekingIterator implements Iterator<Integer> {
//     private Iterator<Integer> iterator;
//     private boolean hasPeeked;
//     private Integer peekedElement;
// 
//     public PeekingIterator(Iterator<Integer> iterator) {
//         // initialize any member here.
//         this.iterator = iterator;
//     }
// 
//     // Returns the next element in the iteration without advancing the iterator.
//     public Integer peek() {
//         if (!hasPeeked) {
//             peekedElement = iterator.next();
//             hasPeeked = true;
//         }
//         return peekedElement;
//     }
// 
//     // hasNext() and next() should behave the same as in the Iterator interface.
//     // Override them if needed.
//     @Override
//     public Integer next() {
//         if (!hasPeeked) {
//             return iterator.next();
//         }
//         Integer result = peekedElement;
//         hasPeeked = false;
//         peekedElement = null;
//         return result;
//     }
// 
//     @Override
//     public boolean hasNext() {
//         return hasPeeked || iterator.hasNext();
//     }
// }

// Accepted solution for LeetCode #284: Peeking Iterator
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #284: Peeking Iterator
// // Java Iterator interface reference:
// // https://docs.oracle.com/javase/8/docs/api/java/util/Iterator.html
// 
// class PeekingIterator implements Iterator<Integer> {
//     private Iterator<Integer> iterator;
//     private boolean hasPeeked;
//     private Integer peekedElement;
// 
//     public PeekingIterator(Iterator<Integer> iterator) {
//         // initialize any member here.
//         this.iterator = iterator;
//     }
// 
//     // Returns the next element in the iteration without advancing the iterator.
//     public Integer peek() {
//         if (!hasPeeked) {
//             peekedElement = iterator.next();
//             hasPeeked = true;
//         }
//         return peekedElement;
//     }
// 
//     // hasNext() and next() should behave the same as in the Iterator interface.
//     // Override them if needed.
//     @Override
//     public Integer next() {
//         if (!hasPeeked) {
//             return iterator.next();
//         }
//         Integer result = peekedElement;
//         hasPeeked = false;
//         peekedElement = null;
//         return result;
//     }
// 
//     @Override
//     public boolean hasNext() {
//         return hasPeeked || iterator.hasNext();
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(1) per op

Space

O(n)

Approach Breakdown

NAIVE

O(n) per op time

O(n) space

Use a simple list or array for storage. Each operation (get, put, remove) requires a linear scan to find the target element — O(n) per operation. Space is O(n) to store the data. The linear search makes this impractical for frequent operations.

OPTIMIZED DESIGN

O(1) per op time

O(n) space

Design problems target O(1) amortized per operation by combining data structures (hash map + doubly-linked list for LRU, stack + min-tracking for MinStack). Space is always at least O(n) to store the data. The challenge is achieving constant-time operations through clever structure composition.

Shortcut: Combine two data structures to get O(1) for each operation type. Space is always O(n).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.