Peekable Iterator - Google Top Interview Questions

Problem Statement :

Implement an iterator of a list of integers nums where

peek() returns the next element, without moving the iterator

next() polls the next element in the iterator

hasnext() which returns whether the next element exists


n ≤ 100,000 where n is the number of calls to peek, next and hasnext

Example 1


methods = ["constructor", "peek", "next", "hasnext", "peek", "next", "hasnext"]

arguments = [[[1, 2]], [], [], [], [], [], []]`


[None, 1, 1, True, 2, 2, False]


First we create a PeekableIterator with values [1, 2]

We peek the next element which is 1

We poll the next element which is 1

We check if the next element exists, which it does since 2 is next in the iterator.

We peek the next element which is 2

We poll the next element which is 2

We check if the next element exists which it doesn't

Solution :


                        Solution in C++ :

class PeekableIterator {
    PeekableIterator(vector<int>& nums) {
        arr.insert(arr.end(), nums.begin(), nums.end());
        idx = 0;

    int peek() {
        if (hasnext()) {
            return arr[idx];

        return -1;

    int next() {
        int val = -1;
        if (hasnext()) {
            val = arr[idx++];

        return val;

    bool hasnext() {
        return idx < arr.size();

    int idx;
    vector<int> arr;

                        Solution in Java :

import java.util.*;

class PeekableIterator {
    int curr = 0;
    int[] arr;
    public PeekableIterator(int[] nums) {
        arr = nums;

    public int peek() {
        return arr[curr];

    public int next() {
        int idx = curr;
        return arr[idx];

    public boolean hasnext() {
        return curr < arr.length;

                        Solution in Python : 
class PeekableIterator:
    def __init__(self, nums):
        self.nums = deque(nums)

    def peek(self):
        return self.nums[0]

    def next(self):
        return self.nums.popleft()

    def hasnext(self):
        return True if self.nums else False

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