Solutions

Prerequisites

Before attempting this problem, you should be comfortable with:

Iterator Design Pattern - Understanding how to implement next() and hasNext() methods to traverse elements sequentially
Two-Pointer Technique - Using multiple pointers to track positions across different data structures
Queue Data Structure - Using FIFO queues to manage and cycle through elements efficiently
Modular Arithmetic - Using the modulo operator to cycle through indices in a round-robin fashion

1. Two-Pointers

Intuition

To iterate through multiple vectors in zigzag order, we need to alternate between vectors while tracking our position in each. The key insight is to maintain two pointers: one for the current vector and one for the current element index. We cycle through vectors in round-robin fashion, and when we complete a full round across all vectors, we advance the element index. This ensures we visit elements in the correct zigzag order while handling vectors of different lengths.

Algorithm

Store references to both input vectors and initialize two pointers: p_vec for the current vector index and p_elem for the current element index.
Track the total number of elements and how many have been output for the hasNext() method.
In next():
- Iterate through vectors starting from the current vector pointer.
- If the current vector has an element at p_elem, capture it as the return value.
- Advance p_vec to the next vector (wrapping around using modulo).
- When p_vec wraps back to 0, increment p_elem to move to the next round.
- Return the captured element and increment the output count.
In hasNext(), return true if the output count is less than the total number of elements.

class ZigzagIterator:
    def __init__(self, v1: List[int], v2: List[int]):
        self.vectors = [v1, v2]

        self.p_elem = 0   # pointer to the index of element
        self.p_vec = 0    # pointer to the vector

        # variables for hasNext() function
        self.total_num = len(v1) + len(v2)
        self.output_count = 0

    def next(self) -> int:
        iter_num = 0
        ret = None

        # Iterate over the vectors
        while iter_num < len(self.vectors):
            curr_vec = self.vectors[self.p_vec]
            if self.p_elem < len(curr_vec):
                ret = curr_vec[self.p_elem]

            iter_num += 1
            self.p_vec = (self.p_vec + 1) % len(self.vectors)
            # increment the element pointer once iterating all vectors
            if self.p_vec == 0:
                self.p_elem += 1

            if ret is not None:
                self.output_count += 1
                return ret

        # no more element to output
        raise Exception


    def hasNext(self) -> bool:
        return self.output_count < self.total_num

class ZigzagIterator {

    private List<List<Integer>> vectors = new ArrayList<>();

    // pointer to vector, and pointer to element
    private Integer pVec = 0, pElem = 0;
    private Integer totalNum = 0, outputCount = 0;

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        this.vectors.add(v1);
        this.vectors.add(v2);
        for (List<Integer> vec : this.vectors) {
            this.totalNum += vec.size();
        }
    }

    public int next() {
        Integer iterNum = 0, ret = null;
        while (iterNum < this.vectors.size()) {
            List<Integer> currVec = this.vectors.get(this.pVec);
            if (this.pElem < currVec.size()) {
                ret = currVec.get(this.pElem);
                this.outputCount += 1;
            }

            iterNum += 1;
            this.pVec = (this.pVec + 1) % this.vectors.size();
            // increment the element pointer once iterating all vectors
            if (this.pVec == 0)
                this.pElem += 1;

            if (ret != null)
                return ret;
        }
        // one should raise an exception here.
        return 0;
    }

    public boolean hasNext() {
        return this.outputCount < this.totalNum;
    }
}

class ZigzagIterator {
private:
    vector<vector<int>> vectors;
    // pointer to vector, and pointer to element
    int pVec = 0, pElem = 0;
    int totalNum = 0, outputCount = 0;

public:
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        vectors.push_back(v1);
        vectors.push_back(v2);
        
        for (const auto& vec : vectors) {
            totalNum += vec.size();
        }
    }

    int next() {
        int iterNum = 0;
        int ret = -1;
        bool found = false;
        
        while (iterNum < vectors.size()) {
            vector<int>& currVec = vectors[pVec];
            
            if (pElem < currVec.size()) {
                ret = currVec[pElem];
                outputCount += 1;
                found = true;
            }
            
            iterNum += 1;
            pVec = (pVec + 1) % vectors.size();
            
            // increment the element pointer once iterating all vectors
            if (pVec == 0)
                pElem += 1;
            
            if (found)
                return ret;
        }
        
        // one should raise an exception here.
        return 0;
    }

    bool hasNext() {
        return outputCount < totalNum;
    }
};

class ZigzagIterator {
    /**
     * @param {Integer[]} v1
     * @param {Integer[]} v2
     */
    constructor(v1, v2) {
        this.vectors = [v1, v2];
        this.p_elem = 0;   // pointer to the index of element
        this.p_vec = 0;    // pointer to the vector

        // variables for hasNext() function
        this.total_num = v1.length + v2.length;
        this.output_count = 0;
    }

    /**
     * @returns {boolean}
     */
    hasNext() {
        return this.output_count < this.total_num;
    }

    /**
     * @returns {integer}
     */
    next() {
        let iter_num = 0;
        let ret = null;

        // Iterate over the vectors
        while (iter_num < this.vectors.length) {
            let curr_vec = this.vectors[this.p_vec];

            if (this.p_elem < curr_vec.length) {
                ret = curr_vec[this.p_elem];
            }

            iter_num++;
            this.p_vec = (this.p_vec + 1) % this.vectors.length;

            // increment the element pointer once iterating all vectors
            if (this.p_vec === 0) {
                this.p_elem++;
            }

            if (ret !== null) {
                this.output_count++;
                return ret;
            }
        }

        // no more element to output
        throw new Error("No more elements");
    }
}

public class ZigzagIterator {
    private List<IList<int>> vectors = new List<IList<int>>();
    private int pVec = 0, pElem = 0;
    private int totalNum = 0, outputCount = 0;

    public ZigzagIterator(IList<int> v1, IList<int> v2) {
        vectors.Add(v1);
        vectors.Add(v2);
        foreach (var vec in vectors) {
            totalNum += vec.Count;
        }
    }

    public int Next() {
        int iterNum = 0;
        int? ret = null;
        while (iterNum < vectors.Count) {
            var currVec = vectors[pVec];
            if (pElem < currVec.Count) {
                ret = currVec[pElem];
                outputCount++;
            }

            iterNum++;
            pVec = (pVec + 1) % vectors.Count;
            if (pVec == 0)
                pElem++;

            if (ret != null)
                return ret.Value;
        }
        return 0;
    }

    public bool HasNext() {
        return outputCount < totalNum;
    }
}

type ZigzagIterator struct {
    vectors     [][]int
    pVec        int
    pElem       int
    totalNum    int
    outputCount int
}

func Constructor(v1, v2 []int) *ZigzagIterator {
    return &ZigzagIterator{
        vectors:     [][]int{v1, v2},
        pVec:        0,
        pElem:       0,
        totalNum:    len(v1) + len(v2),
        outputCount: 0,
    }
}

func (this *ZigzagIterator) next() int {
    iterNum := 0
    ret := -1
    found := false

    for iterNum < len(this.vectors) {
        currVec := this.vectors[this.pVec]
        if this.pElem < len(currVec) {
            ret = currVec[this.pElem]
            this.outputCount++
            found = true
        }

        iterNum++
        this.pVec = (this.pVec + 1) % len(this.vectors)
        if this.pVec == 0 {
            this.pElem++
        }

        if found {
            return ret
        }
    }
    return 0
}

func (this *ZigzagIterator) hasNext() bool {
    return this.outputCount < this.totalNum
}

class ZigzagIterator(v1: List<Int>, v2: List<Int>) {
    private val vectors = listOf(v1, v2)
    private var pVec = 0
    private var pElem = 0
    private val totalNum = v1.size + v2.size
    private var outputCount = 0

    fun next(): Int {
        var iterNum = 0
        var ret: Int? = null

        while (iterNum < vectors.size) {
            val currVec = vectors[pVec]
            if (pElem < currVec.size) {
                ret = currVec[pElem]
                outputCount++
            }

            iterNum++
            pVec = (pVec + 1) % vectors.size
            if (pVec == 0) {
                pElem++
            }

            if (ret != null) {
                return ret
            }
        }
        return 0
    }

    fun hasNext(): Boolean {
        return outputCount < totalNum
    }
}

class ZigzagIterator {
    private var vectors: [[Int]]
    private var pVec: Int = 0
    private var pElem: Int = 0
    private var totalNum: Int
    private var outputCount: Int = 0

    init(_ v1: [Int], _ v2: [Int]) {
        vectors = [v1, v2]
        totalNum = v1.count + v2.count
    }

    func next() -> Int {
        var iterNum = 0
        var ret: Int? = nil

        while iterNum < vectors.count {
            let currVec = vectors[pVec]
            if pElem < currVec.count {
                ret = currVec[pElem]
                outputCount += 1
            }

            iterNum += 1
            pVec = (pVec + 1) % vectors.count
            if pVec == 0 {
                pElem += 1
            }

            if let result = ret {
                return result
            }
        }
        return 0
    }

    func hasNext() -> Bool {
        return outputCount < totalNum
    }
}

Time & Space Complexity

Time Complexity:
- For the next() function, at most it will take us $K$ iterations to find a valid element to output. Hence, its time complexity is $O(K)$ .
- For the hasNext() function, its time complexity is $O(1)$ .
Space Complexity:
- For the next() function, we keep the references to all the input vectors in the variable self.vectors.
- As a result, we would need $O(K)$ space for $K$ vectors.
- In addition, we used some constant-space variables such as the pointers to the vector and the element.
- Hence, the overall space complexity for this function is $O(K)$ .
- Note: we did not copy the input vectors, but simply keep references to them.

Where $K$ is the number of input vectors. Although it is always two in the setting of this problem, this variable becomes relevant once the input becomes $K$ vectors.

2. Queue of Pointers

Intuition

Using a queue to manage pointers provides a cleaner and more efficient approach. Instead of scanning through all vectors on each call, we maintain a queue of (vector index, element index) pairs representing which elements are ready to be returned. After returning an element, we add the pointer for the next element in that vector (if any) to the back of the queue. This naturally handles vectors of different lengths and ensures O(1) time for each next() call.

Algorithm

Initialize a queue with pointers to the first element of each non-empty vector, stored as (vector index, element index) pairs.
In next():
- Dequeue the front pointer to get the current vector and element indices.
- Retrieve the element to return from the specified position.
- If there are more elements in that vector, enqueue a pointer to the next element.
- Return the retrieved element.
In hasNext(), simply check if the queue is non-empty.

class ZigzagIterator:
    def __init__(self, v1: List[int], v2: List[int]):
        self.vectors = [v1, v2]
        self.queue = deque()
        for index, vector in enumerate(self.vectors):
            # <index_of_vector, index_of_element_to_output>
            if len(vector) > 0:
                self.queue.append((index, 0))

    def next(self) -> int:

        if self.queue:
            vec_index, elem_index = self.queue.popleft()
            next_elem_index = elem_index + 1
            if next_elem_index < len(self.vectors[vec_index]):
                # append the pointer for the next round
                # if there are some elements left
                self.queue.append((vec_index, next_elem_index))

            return self.vectors[vec_index][elem_index]

        # no more element to output
        raise Exception

    def hasNext(self) -> bool:
        return len(self.queue) > 0

class ZigzagIterator {

    private List<List<Integer>> vectors = new ArrayList<>();
    private LinkedList<Pair<Integer, Integer>> queue = new LinkedList<>();

    public ZigzagIterator(List<Integer> v1, List<Integer> v2) {
        this.vectors.add(v1);
        this.vectors.add(v2);
        int index = 0;
        for (List<Integer> vec : this.vectors) {
            if (vec.size() > 0)
                // <index_to_vec, index_to_element_within_vec>
                this.queue.add(new Pair<Integer, Integer>(index, 0));
            index++;
        }
    }

    public int next() {
        // <index_to_vec, index_to_element_within_vec>
        Pair<Integer, Integer> pointer = this.queue.removeFirst();
        Integer vecIndex = pointer.getKey();
        Integer elemIndex = pointer.getValue();
        Integer nextElemIndex = elemIndex + 1;
        // append the pointer for the next round
        // if there are some elements left.
        if (nextElemIndex < this.vectors.get(vecIndex).size())
            this.queue.addLast(new Pair<>(vecIndex, nextElemIndex));

        return this.vectors.get(vecIndex).get(elemIndex);
    }

    public boolean hasNext() {
        return this.queue.size() > 0;
    }
}

class ZigzagIterator {
private:
    vector<vector<int>> vectors;
    queue<pair<int, int>> q;
    
public:
    ZigzagIterator(vector<int>& v1, vector<int>& v2) {
        vectors.push_back(v1);
        vectors.push_back(v2);
        
        for (int index = 0; index < vectors.size(); index++) {
            if (vectors[index].size() > 0) {
                // <index_to_vec, index_to_element_within_vec>
                q.push({index, 0});
            }
        }
    }
    
    int next() {
        // <index_to_vec, index_to_element_within_vec>
        pair<int, int> pointer = q.front();
        q.pop();
        
        int vecIndex = pointer.first;
        int elemIndex = pointer.second;
        int nextElemIndex = elemIndex + 1;
        
        // append the pointer for the next round
        // if there are some elements left.
        if (nextElemIndex < vectors[vecIndex].size()) {
            q.push({vecIndex, nextElemIndex});
        }
        
        return vectors[vecIndex][elemIndex];
    }
    
    bool hasNext() {
        return q.size() > 0;
    }
};

class ZigzagIterator {
    /**
     * @param {Integer[]} v1
     * @param {Integer[]} v2
     */
    constructor(v1, v2) {
        this.vectors = [v1, v2];
        this.queue = new Deque(); // Using @datastructures-js/deque

        for (let index = 0; index < this.vectors.length; index++) {
            const vector = this.vectors[index];
            // <index_of_vector, index_of_element_to_output>
            if (vector.length > 0) {
                this.queue.pushBack([index, 0]);
            }
        }
    }

    /**
     * @returns {boolean}
     */
    hasNext() {
        return !this.queue.isEmpty();
    }

    /**
     * @returns {integer}
     */
    next() {
        if (!this.queue.isEmpty()) {
            const [vecIndex, elemIndex] = this.queue.popFront();
            const nextElemIndex = elemIndex + 1;

            if (nextElemIndex < this.vectors[vecIndex].length) {
                // append the pointer for the next round
                // if there are some elements left
                this.queue.pushBack([vecIndex, nextElemIndex]);
            }

            return this.vectors[vecIndex][elemIndex];
        }

        // no more element to output
        throw new Error("No more elements");
    }
}

public class ZigzagIterator {
    private List<IList<int>> vectors = new List<IList<int>>();
    private Queue<(int, int)> queue = new Queue<(int, int)>();

    public ZigzagIterator(IList<int> v1, IList<int> v2) {
        vectors.Add(v1);
        vectors.Add(v2);
        for (int index = 0; index < vectors.Count; index++) {
            if (vectors[index].Count > 0) {
                // <index_to_vec, index_to_element_within_vec>
                queue.Enqueue((index, 0));
            }
        }
    }

    public int Next() {
        // <index_to_vec, index_to_element_within_vec>
        var (vecIndex, elemIndex) = queue.Dequeue();
        int nextElemIndex = elemIndex + 1;
        // append the pointer for the next round
        // if there are some elements left.
        if (nextElemIndex < vectors[vecIndex].Count) {
            queue.Enqueue((vecIndex, nextElemIndex));
        }

        return vectors[vecIndex][elemIndex];
    }

    public bool HasNext() {
        return queue.Count > 0;
    }
}

type ZigzagIterator struct {
    vectors [][]int
    queue   [][2]int
}

func Constructor(v1, v2 []int) *ZigzagIterator {
    zi := &ZigzagIterator{
        vectors: [][]int{v1, v2},
        queue:   [][2]int{},
    }
    for index, vec := range zi.vectors {
        if len(vec) > 0 {
            // [index_to_vec, index_to_element_within_vec]
            zi.queue = append(zi.queue, [2]int{index, 0})
        }
    }
    return zi
}

func (this *ZigzagIterator) next() int {
    // [index_to_vec, index_to_element_within_vec]
    pointer := this.queue[0]
    this.queue = this.queue[1:]

    vecIndex := pointer[0]
    elemIndex := pointer[1]
    nextElemIndex := elemIndex + 1

    // append the pointer for the next round
    // if there are some elements left.
    if nextElemIndex < len(this.vectors[vecIndex]) {
        this.queue = append(this.queue, [2]int{vecIndex, nextElemIndex})
    }

    return this.vectors[vecIndex][elemIndex]
}

func (this *ZigzagIterator) hasNext() bool {
    return len(this.queue) > 0
}

class ZigzagIterator(v1: List<Int>, v2: List<Int>) {
    private val vectors = listOf(v1, v2)
    private val queue = ArrayDeque<Pair<Int, Int>>()

    init {
        for ((index, vec) in vectors.withIndex()) {
            if (vec.isNotEmpty()) {
                // <index_to_vec, index_to_element_within_vec>
                queue.add(Pair(index, 0))
            }
        }
    }

    fun next(): Int {
        // <index_to_vec, index_to_element_within_vec>
        val (vecIndex, elemIndex) = queue.removeFirst()
        val nextElemIndex = elemIndex + 1

        // append the pointer for the next round
        // if there are some elements left.
        if (nextElemIndex < vectors[vecIndex].size) {
            queue.add(Pair(vecIndex, nextElemIndex))
        }

        return vectors[vecIndex][elemIndex]
    }

    fun hasNext(): Boolean {
        return queue.isNotEmpty()
    }
}

class ZigzagIterator {
    private var vectors: [[Int]]
    private var queue: [(Int, Int)] = []

    init(_ v1: [Int], _ v2: [Int]) {
        vectors = [v1, v2]
        for (index, vec) in vectors.enumerated() {
            if !vec.isEmpty {
                // (index_to_vec, index_to_element_within_vec)
                queue.append((index, 0))
            }
        }
    }

    func next() -> Int {
        // (index_to_vec, index_to_element_within_vec)
        let (vecIndex, elemIndex) = queue.removeFirst()
        let nextElemIndex = elemIndex + 1

        // append the pointer for the next round
        // if there are some elements left.
        if nextElemIndex < vectors[vecIndex].count {
            queue.append((vecIndex, nextElemIndex))
        }

        return vectors[vecIndex][elemIndex]
    }

    func hasNext() -> Bool {
        return !queue.isEmpty
    }
}

Time & Space Complexity

Time Complexity: $O(1)$
- For both the next() function and the hasNext() function, we have a constant time complexity, as we discussed before.
Space Complexity: $O(K)$
- We use a queue to keep track of the pointers to the input vectors in the variable self.vectors.
- As a result, we would need $O(K)$ space for $K$ vectors.
- Although the size of the queue will reduce over time once we exhaust some shorter vectors, the space complexity for both functions is still $O(K)$ .

Where $K$ is the number of input vectors. Although it is always two in the setting of this problem, this variable becomes relevant once the input becomes $K$ vectors.

Common Pitfalls

Skipping Empty Vectors in Initialization

When one or both input vectors are empty, failing to handle them causes index errors or incorrect iteration. Empty vectors should either be skipped entirely or handled gracefully in the next() logic.

# Wrong: Adding empty vectors to queue causes issues
for index, vector in enumerate(self.vectors):
    self.queue.append((index, 0))  # Should check: if len(vector) > 0

Incrementing Element Pointer at Wrong Time

In the two-pointer approach, the element index should only increment after completing a full round through all vectors (when p_vec wraps back to 0). Incrementing it every step breaks the zigzag order.

# Wrong: Incrementing p_elem on every iteration
self.p_vec = (self.p_vec + 1) % len(self.vectors)
self.p_elem += 1  # Should only increment when p_vec == 0

Not Returning After Finding a Valid Element

In the two-pointer approach, after finding a valid element in a vector, you must return immediately. Continuing to iterate through other vectors in the same call corrupts the zigzag ordering and may return wrong elements.

# Wrong: Missing early return
if self.p_elem < len(curr_vec):
    ret = curr_vec[self.p_elem]
    # Missing: return ret after incrementing output_count!
self.p_vec = (self.p_vec + 1) % len(self.vectors)