Peak Index in a Mountain Array

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.

• 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.

Leetcode link

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.

Code

Python Code

class Solution:
    def __init__(self, iterator):
        self.it = iterator
        self.cache = None

    def peek(self):
        if self.cache:
            return self.cache
        self.cache = self.it.next()
        return self.cache

    def next(self):
        if self.cache:
            temp = self.cache
            self.cache = None
            return temp
        return self.it.next()

    def hasNext(self):
        if self.cache:
            return True
        return self.it.hasNext()