Efficiently iterate over vectors consistently c++ [closed] - c++

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I have n vectors and I need to iterate over all of them consistently.
I mean that first I need to iterate over all elements of first vector then to iterate over all elements of second vector and so on.
I have task to compare all elements consistently including the last element of k-1'th vector and first element of k'th vector.
I know the solution where I first concatanete all this vectors in one big final vector and then iterate over final vector.
But it seems to me that this solution wastes a lot of memory and time.
Please help me to find more optimized solution.
If it exists.
For example, with complexities: O(1) additional memory and O(n) time .

Answer 1

A simple approach would be to use two nested loops.
In C++11 and later, simply do
std::vector<std::vector<int> > vec;
// some code that sets up the vectors in vec
for (auto &v : vec) // iterate over the vectors in vec
{
for (auto &e : v) // iterate over the elements of v
{
// do whatever is needed to element e
}
}
If need be, v and e may be const qualified (to indicate not changing the vectors or their elements). The total number of iterations will be the number of elements of the contained vectors. The additional memory usage will be (at any point in time in the inner loop) two sets of references - one set for controlling the inner loop and one for the outer loop (i.e. no copies of the vectors or elements themselves).
Before C++11, the above loops can be rewritten less verbosely using for loops that work with iterators rather than ranges;
for (std::vector<std::vector<int> >::iterator v(vec.begin()), vend(vec.end());
v != vend; ++v)
{
for (std::vector<int>::iterator e(v->begin()), eend(v->end());
e != eend; ++e)
{
// do whatever is needed to element *e
}
}
Note the need to dereference iterators in this form.

Answer 2

If you only need something that can be iterated over, you probably want something like ranges::view::concat
Given N source ranges, produce a result range that is the concatenation of all of them.
This doesn't copy any values, so is O(1) storage and O(1) time.
If the thing you need is exactly std::vector<MyClass>, then you are out of luck, you will have to populate a vector with all the values.