# L3-TapeEquilibrium
###### tags: `Codility_lessons`
## Question
https://app.codility.com/programmers/lessons/3-time_complexity/tape_equilibrium/
## Key
直接用一個Loop迭代不同的p(位置)，去算difference，然後取最小的就是答案，但如何避免用太多loop算difference才不會time exceed算是重點
先把總和算出來，因為loop是由小到大迭代，所以左邊的部分就直接累加，右邊的部分則在迭代過程中從總和扣除，這樣只需要一個loop
## Reference
## Solution

#include <iostream>
#include <algorithm>
#include <vector>
#include <numeric>


using namespace std;

int solution(vector<int> &A)
{

    long diff = 0;
    int suml = A[0];
    int sumr = accumulate(A.begin()+1, A.end(), 0);
    long tape = abs(suml - sumr);

    for(unsigned int i=1; i<A.size()-1; i++)
    {
        suml += A[i];
        sumr -= A[i];
        diff = abs(suml - sumr);
        if(diff<=tape)
        {
            tape = diff;
        }
    }
    return tape;
}