How to flatten a Vector of Vectors or 2D Vector in C++

Given a Vector of Vectors (2D vector), the task is to flatten this 2d vector. Examples:

Input: vector = [[1, 2, 3, 4], [5, 6], [7, 8]] Output: 1 2 3 4 5 6 7 8 Input: vector = [[1, 2], [3], [4, 5, 6, 8]] Output: 1 2 3 4 5 6 8

Algorithm:

1. 2D Vector can be flattened using iterators.
2. Store starting and ending iterator of every vector in two arrays, iStart & iEnd respectively.
3. Create a hasNext() method to check if it has the vector has next element or not.
4. Print the current element, if hasNext() yields true

Below is the implementation of the above approach: 

CPP

// C++ program to flatten a
// Vector of Vectors or 2D Vector

#include <bits/stdc++.h>
using namespace std;

// Class to flatten the 2d vector
class FlattenVector {

public:
    int n;

    vector<vector<int>::iterator> iStart;
    vector<vector<int>::iterator> iEnd;
    int currIndex;

    // Store ending and starting iterators.
    FlattenVector(vector<vector<int> >& v)
    {

        // Get the number
        // of rows in 2d vector
        n = v.size();
        currIndex = 0;
        iStart.resize(n);
        iEnd.resize(n);

        for (int i = 0; i < n; i++) {
            iStart[i] = v[i].begin();
            iEnd[i] = v[i].end();
        }
    }

    // Returns true if any element is left.
    bool hasNext()
    {
        for (int i = 0; i < n; i++) {
            if (iStart[i] != iEnd[i])
                return true;
        }
        return false;
    }

    int next()
    {
        // Vector at currIndex is printed,
        // increment currIndex.
        if (iStart[currIndex]
            == iEnd[currIndex]) {
            currIndex++;
            return next();
        }

        // Increment iterator
        // and return the value.
        else
            return *iStart[currIndex]++;
    }
};

// Driver code
int main()
{
    vector<vector<int> >
        v{ { 1, 2 },
        { 3 },
        { 4, 5, 6 },
        { 7, 8, 9, 10 } };
    FlattenVector iter(v);

    while (iter.hasNext())
        cout << iter.next() << " ";

    return 0;
}

Output:

1 2 3 4 5 6 7 8 9 10

Time Complexity: O(N)

Auxiliary Space: O(N)