[Lesson 3] TapeEquilibrium

 

문제

A non-empty array A consisting of N integers is given. Array A represents numbers on a tape.
Any integer P, such that 0 < P < N, splits this tape into two non-empty parts: A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].
The difference between the two parts is the value of: |(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|
In other words, it is the absolute difference between the sum of the first part and the sum of the second part.
For example, consider array A such that:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3

We can split this tape in four places:
P = 1, difference = |3 − 10| = 7
P = 2, difference = |4 − 9| = 5
P = 3, difference = |6 − 7| = 1
P = 4, difference = |10 − 3| = 7
Write a function:
int solution(vector<int> &A);
that, given a non-empty array A of N integers, returns the minimal difference that can be achieved.
For example, given:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 4 A[4] = 3

the function should return 1, as explained above.
Write an efficient algorithm for the following assumptions:
N is an integer within the range [2..100,000];each element of array A is an integer within the range [−1,000..1,000].

 

풀이

처음에 누적합(Prefix Sum)을 생각했는데, 범위에 음수가 포함되어있어서, 한번 풀이과정을 길게 늘려봤다.

첫번째 요소와 그 뒤의 합을 기준으로 A[idx] 값이 더하고, 빼지는 규칙을 기준으로 코드를 작성할 수 있다.

// you can use includes, for example:
// #include <algorithm>

// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;

int solution(vector<int> &A) {
    // Implement your solution here
    // 2 <= A.size()
    //Sum if all element
    int Sum1 = A[0];
    int Sum2 = 0;
    int res = 9876; //Minimal result
    for(int i = 1; i < A.size(); i++)
    {
        Sum2 += A[i];
    }

    for(int i = 1; i < A.size(); i++)
    {
        res = min(res, abs(Sum1 - Sum2));

        Sum1 += A[i];
        Sum2 -= A[i];
    }



    return res;
}