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Home/ Questions/Q 6318951
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T15:43:07+00:00 2026-05-24T15:43:07+00:00

I have some code and an array where each iteration I delete the first

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I have some code and an array where each iteration I delete the first element, add an element at the end, and then sum the contents of the array. Naturally, the array stays the same size. I have tried using both vector and list, but both seem pretty slow.

int length = 400;

vector <int> v_int(length, z);
list <int>   l_int(length, z);

for(int q=0; q < x; q++)
{
    int sum =0;

    if(y)                       //if using vector
    {      
        v_int.erase(v_int.begin());    //takes `length` amount of time to shift memory
        v_int.push_back(z);   
        for(int w=0; w < v_int.size(); w++)
            sum += v_int[w];
    }      
    else                        //if using list
    {
        l_int.pop_front();             //constant time
        l_int.push_back(z);
        list<int>::iterator it;
        for ( it=l_int.begin() ; it != l_int.end(); it++ ) //seems to take much  
            sum += *it;                                    //longer than vector does
    }
}

The problem is that erasing the first element of the vector requires that each other element be shifted down, multiplying, by the size of the vector, the amount of time taken each iteration. Using a linked list avoids this (constant time removal of elements), and should not sacrifice any time summing the array (linear time traversal of the array), except that in my program it seems to be taking way longer to sum the contents than the vector does (at least 1 order of magnitude longer).

Is there a better container to use here? or a different way to approach the problem?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-24T15:43:09+00:00Added an answer on May 24, 2026 at 3:43 pm

    Efficient additions and deletions of end elements in a container is what the deque was made for.

    If you are just inserting at one end and deleting at the other then you can use a queue

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