Zigzag Iterator

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ArraysTwo Pointers

Given two vectors of integers v1 and v2, implement an iterator to return their elements alternately.

Implement the ZigzagIterator class:

ZigzagIterator(List<Integer> v1, List<Integer> v2) initializes the object with the two vectors v1 and v2.
Integer next() returns the next element from the two vectors and moves the iterator one step forward.
boolean hasNext() returns true if there are still elements in both vectors or at least one of the vectors.

Example 1:

Input: v1 = [1,2], v2 = [3,4,5,6]
Output: [1,3,2,4,5,6]
Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,3,2,4,5,6].

Example 2:

Input: v1 = [ ], v2 = [1,2,3]
Output: [1,2,3]

Constraints:

0 <= v1.size, v2.size <= 1000
1 <= v1.size + v2.size <= 2000
-1000 <= v1[i], v2[i] <= 1000

Follow up: What if you are given k vectors? How well can your code be extended?

Clarification for the Follow-up Question: The "Zigzag" order here is not defined very well. Generally, it is expected to be in the order of the first element of v1, the first element of v2, the first element of v3, and so on until the first element of vk, then the second element of v1, the second element of v2, and so on until the second element of vk, and so on.

Zigzag Iterator

Medium

Asked by:

10 views

Topics:

ArraysTwo Pointers

Given two vectors of integers v1 and v2, implement an iterator to return their elements alternately.

Implement the ZigzagIterator class:

ZigzagIterator(List<Integer> v1, List<Integer> v2) initializes the object with the two vectors v1 and v2.
Integer next() returns the next element from the two vectors and moves the iterator one step forward.
boolean hasNext() returns true if there are still elements in both vectors or at least one of the vectors.

Example 1:

Input: v1 = [1,2], v2 = [3,4,5,6]
Output: [1,3,2,4,5,6]
Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,3,2,4,5,6].

Example 2:

Input: v1 = [ ], v2 = [1,2,3]
Output: [1,2,3]

Constraints:

0 <= v1.size, v2.size <= 1000
1 <= v1.size + v2.size <= 2000
-1000 <= v1[i], v2[i] <= 1000

Follow up: What if you are given k vectors? How well can your code be extended?

Clarification for the Follow-up Question: The "Zigzag" order here is not defined very well. Generally, it is expected to be in the order of the first element of v1, the first element of v2, the first element of v3, and so on until the first element of vk, then the second element of v1, the second element of v2, and so on until the second element of vk, and so on.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can the input lists be of different sizes, and if so, how should the iterator handle the case where one list is exhausted before the other(s)?
What data types will the elements within the lists be? Are null/None values permitted within the lists, and how should those be handled?
If one or more of the input lists are empty, what should the iterator do? Should it return an empty iterator, or simply skip those lists?
Is the order of iteration among the lists determined by the order they are given as input, or is there a different priority/ordering?
Are there any memory constraints or practical limitations on the number or total size of the input lists?

Brute Force Solution

Approach

The brute force approach to combining lists takes every possible combination of elements from the lists, one at a time. It checks if a combination is valid, and repeats until all valid interleavings are explored. It's like exhaustively trying every possible path to see what works.

Here's how the algorithm would work step-by-step:

Start by picking the first element from the first list.
Then, pick the first element from the second list.
Keep a record of this selection to form one possible output.
Now, go back and pick the second element from the first list, while still keeping the first element from the second list.
Continue exploring all combinations by advancing through both lists, one element at a time, and recording each unique sequence of elements.
Repeat this process, ensuring all possible orderings from both lists are tried and recorded.
Finally, return the recorded sequences as the result, representing all possible interleavings.

Code Implementation

def zigzag_brute_force(list1, list2):
    results = []
    def find_interleavings(current_interleaving, index1, index2):
        # Base case: if we have exhausted both lists
        if index1 == len(list1) and index2 == len(list2):
            results.append(current_interleaving.copy())
            return

        # Explore adding element from the first list
        if index1 < len(list1):
            current_interleaving.append(list1[index1])

            find_interleavings(current_interleaving, index1 + 1, index2)

            # Backtrack: remove the last added element to explore other paths
            current_interleaving.pop()

        # Explore adding element from the second list
        if index2 < len(list2):
            current_interleaving.append(list2[index2])

            find_interleavings(current_interleaving, index1, index2 + 1)

            # Backtrack: remove the last added element to explore other paths
            current_interleaving.pop()

    find_interleavings([], 0, 0)
    return results

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(2^(m+n)) – The brute force approach explores all possible interleavings of the two lists. In the worst case, where we have 'm' elements in the first list and 'n' elements in the second list, we potentially generate all possible combinations of picking elements from each list. This results in exploring 2^(m+n) possible interleavings as each element has two choices: to be included or not. Therefore, the time complexity grows exponentially with the sum of the lengths of the input lists.

Space Complexity

O(N!) – The algorithm records each unique sequence of elements, which represents interleavings of the input lists. In the worst-case scenario, where the algorithm explores all possible orderings, it stores all possible interleavings. The number of interleavings grows factorially with the total number of elements across all lists, which we can denote as N. Therefore, the algorithm needs to store a number of sequences proportional to N! in memory.

Optimal Solution

Approach

The Zigzag Iterator cleverly combines elements from different sources in an alternating fashion. The efficient approach uses a structure to keep track of the sources and a mechanism to switch between them, ensuring each source contributes its elements fairly.

Here's how the algorithm would work step-by-step:

Maintain a list of the sources you need to iterate through.
Start with the first source in the list.
Take the next element from the current source, if it has any elements left.
Move to the next source in the list. If you reach the end of the list, go back to the beginning.
If a source becomes empty (has no more elements), remove it from the list of sources.
Keep repeating steps 3-5 until all sources are empty.
The order in which you extracted the elements is the zigzag order.

Code Implementation

class ZigzagIterator:

    def __init__(self, vector_one, vector_two):
        self.vectors = []
        if vector_one:
            self.vectors.append(vector_one)
        if vector_two:
            self.vectors.append(vector_two)
        self.vector_index = 0

    def next(self):
        if not self.has_next():
            return None

        current_vector = self.vectors[self.vector_index]
        next_element = current_vector.pop(0)

        # Advance to the next vector, or cycle back to the beginning
        self.vector_index = (self.vector_index + 1) % len(self.vectors)

        return next_element

    def has_next(self):
        # Remove empty vectors to avoid infinite looping
        self.vectors = [vector for vector in self.vectors if vector]

        # Returns true if there are any vectors with elements remaining.
        return len(self.vectors) > 0

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n) – The primary operation is iterating through all the elements from all input lists, which we can denote as 'n' total elements. Within the next() method, we iterate through a list of iterators, which in the worst case could be the number of input lists. However, each iterator is only visited once for each element we extract. Removing empty iterators from the list of iterators is also amortized O(1) since it happens only when an iterator is exhausted, contributing at most a constant factor per element. Thus, the overall time complexity is directly proportional to the total number of elements 'n', leading to O(n).

Space Complexity

O(K) – The algorithm maintains a list of iterators (or the sources themselves) where K is the number of input lists. The space used by this list is proportional to the number of lists we are iterating through. As we remove empty lists from the list, the space usage can decrease but it never exceeds K. Therefore, the auxiliary space complexity is O(K).

Edge Cases

Case	How to Handle
Both input lists are empty	Return an empty list, as there are no elements to iterate through.
One list is empty, the other is not	Iterate through the non-empty list only.
Lists have drastically different lengths (e.g., one is empty, the other has 1000 elements)	The solution should gracefully handle this by iterating until both lists are exhausted, using appropriate index bounds checking.
Input lists contain null elements	The solution should either skip null elements or throw an IllegalArgumentException depending on the defined behavior.
Very large lists that may exceed memory limits	Consider an iterator-based approach instead of building the entire zigzag list in memory at once, to manage memory efficiently.
Integer overflow on index calculations for very long lists	Use appropriate data types (e.g., long) for index variables to prevent overflow errors.
Concurrent modification of input lists	If the input lists are mutable, copy the lists to prevent modification during iteration, or document that concurrent modification is not supported.
Input lists containing same element multiple times	The zigzag pattern should ensure that elements are selected sequentially from each list, regardless of duplicate values within each list.