c++程序的时间和空间复杂度

Question

#include<iostream>
#include<vector>

using namespace std;

vector <int> removeFirstOrder(const vector<int>& orders)
{
    return vector<int>(++orders.begin() , orders.end());
}

bool isFirstComeFirstServed(const vector<int>& takeOutOrders,
                            const vector<int>& dineInOrders,
                            const vector<int>& servedOrders)
{
    //base case
    if(servedOrders.empty())
    {
        return true;
    }

    if(!takeOutOrders.empty() && takeOutOrders[0]==servedOrders[0])
    {
        return isFirstComeFirstServed(removeFirstOrder(takeOutOrders),
                                      dineInOrders,removeFirstOrder(servedOrders));

    }

    else if(!dineInOrders.empty() && dineInOrders[0]==servedOrders[0])
    {
        return isFirstComeFirstServed(takeOutOrders, removeFirstOrder(takeOutOrders),
                                        removeFirstOrder(servedOrders));

    }

    else
    {
        return false;
    }
}


int main()
{
    vector<int> takeOutOrders{17,8,4};

    vector<int> dineInOrders{12,19,2};
    vector<int> servedOrders{17,8,12,19,24,2};
    isFirstComeFirstServed(takeOutOrders,dineInOrders,servedOrders);



    return 0;
}

My doubt is that here Author of this program says that it has O(n^2) time complexity and O(n^2) space complexity.

I agree with time complexity of this program because isFirstComeFirstServed function will be called n times which is size of servedOrders Vector Right? and removeFirstOrder will be call n times in first function call of isFirstComeFirstServed and n-1 times in second function call of isFirstComeFirstServed and so on till there are no element left in servedOrder Vector Right?

But my doubt is that how it can O(n^2) space complexity? can someone help me to visualize it?

Answer 1

Each time removeFirstOrder is called the returned vector is smaller by 1.

n-1 + n-2 + n-3 + ... + 1

From arithmetic progression rules, the sum is (n+1)*n / 2 which is order n^2.

Tail call optimization could make it O(n) space behind the scene, but it's not guaranteed be performed at all.