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Peeking Iterator

Difficulty: Medium

Problem Description

Design an iterator that supports the peek operation on an existing iterator in addition to the hasNext and the next operations.

Key Insights

  • The PeekingIterator must maintain the state of the underlying iterator to provide peek functionality.
  • We need to handle the case where the iterator has elements left and where it has reached the end.
  • The peek method should return the next element without advancing the iterator.

Space and Time Complexity

Time Complexity: O(1) for next(), hasNext(), and peek() operations.
Space Complexity: O(1) for storing the next element to be returned by peek().

Solution

We can create a class PeekingIterator that wraps an existing iterator. The class will have an internal variable to store the next element retrieved from the iterator. Upon initialization, we will load the first element from the iterator into this variable.

The next() method will return the stored next element and then update it to the subsequent element from the iterator. The peek() method will return the stored next element without modifying the iterator's state. The hasNext() method checks if there are more elements in the underlying iterator.

Code Solutions

class PeekingIterator:
    def __init__(self, iterator):
        self.iterator = iterator  # Store the original iterator
        self.next_element = None   # Variable to store the next element
        self.has_next = self.iterator.hasNext()  # Check if there is a next element
        if self.has_next:
            self.next_element = self.iterator.next()  # Load the first element

    def peek(self):
        return self.next_element  # Return the next element without advancing

    def next(self):
        current = self.next_element  # Store the current next element
        self.has_next = self.iterator.hasNext()  # Check if there is a next element
        if self.has_next:
            self.next_element = self.iterator.next()  # Advance to the next element
        else:
            self.next_element = None  # No more elements
        return current  # Return the current element

    def hasNext(self):
        return self.has_next  # Return whether there are more elements
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