Peeking Iterator LeetCode Solution | Easy Approach

Peeking Iterator 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.

Peeking Iterator Solutions

✅Time: O(1)
✅Space: O(n)

C++

 class PeekingIterator : public Iterator {
 public:
  PeekingIterator(const vector<int>& nums) : Iterator(nums) {}

  // Returns the next element in the iteration without advancing the iterator.
  int peek() {
    // Iterator(*this) makes a copy of current iterator, then call next on the
    // copied iterator to get the next value without affecting current iterator
    return Iterator(*this).next();
  }

  // hasNext() and next() should behave the same as in the Iterator interface.
  // Override them if needed.
  int next() {
    return Iterator::next();
  }

  bool hasNext() const {
    return Iterator::hasNext();
  }
};

Java

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

class PeekingIterator implements Iterator<Integer> {
  public PeekingIterator(Iterator<Integer> iterator) {
    this.iterator = iterator;
    buffer = iterator.hasNext() ? iterator.next() : null;
  }

  // Returns the next element in the iteration without advancing the iterator.
  public Integer peek() {
    return buffer;
  }

  // hasNext() and next() should behave the same as in the Iterator interface.
  // Override them if needed.
  @Override
  public Integer next() {
    Integer next = buffer;
    buffer = iterator.hasNext() ? iterator.next() : null;
    return next;
  }

  @Override
  public boolean hasNext() {
    return buffer != null;
  }

  private Iterator<Integer> iterator;
  private Integer buffer;
}

Python

 class PeekingIterator:
  def __init__(self, iterator: Iterator):
    self.iterator = iterator
    self.buffer = self.iterator.next() if self.iterator.hasNext() else None

  def peek(self) -> int:
    """
    Returns the next element in the iteration without advancing the iterator.
    """
    return self.buffer

  def next(self) -> int:
    next = self.buffer
    self.buffer = self.iterator.next() if self.iterator.hasNext() else None
    return next

  def hasNext(self) -> bool:
    return self.buffer is not None

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