**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 Constraints n ≤ 100,000 where n is the number of calls to peek, next and hasnext Example 1 Input methods = ["constructor", "peek", "next", "hasnext", "peek", "next", "hasnext"] arguments = [[[1, 2]], [], [], [], [], [], []]` Output [None, 1, 1, True, 2, 2, False] Explanation 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 {
public:
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();
}
private:
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;
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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