In this guide, you will get 284. Peeking Iterator LeetCode Solution with the best time and space complexity. The solution to Peeking Iterator problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
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 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.
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
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?
Complexity Analysis
Time Complexity: O(1)
Space Complexity: O(n)
284. Peeking Iterator LeetCode Solution in 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();
}
};
/* code provided by PROGIEZ */
284. Peeking Iterator LeetCode Solution in 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;
}
// code provided by PROGIEZ
284. Peeking Iterator LeetCode Solution in 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
# code by PROGIEZ
