Lesson 9. Maximum slice problem - MaxDoubleSliceSum #1

1. 문제

A non-empty array A consisting of N integers is given.
A triplet (X, Y, Z), such that 0 ≤ X < Y < Z < N, is called a double slice.
The sum of double slice (X, Y, Z) is the total of A[X + 1] + A[X + 2] + ... + A[Y − 1] + A[Y + 1] + A[Y + 2] + ... + A[Z − 1].

For example, array A such that:
A[0] = 3
A[1] = 2
A[2] = 6
A[3] = -1
A[4] = 4
A[5] = 5
A[6] = -1
A[7] = 2
contains the following example double slices:
double slice (0, 3, 6), sum is 2 + 6 + 4 + 5 = 17,
double slice (0, 3, 7), sum is 2 + 6 + 4 + 5 − 1 = 16,
double slice (3, 4, 5), sum is 0.

The goal is to find the maximal sum of any double slice.

Write a function:
  int solution(vector<int> &A);

that, given a non-empty array A consisting of N integers, returns the maximal sum of any double slice.

For example, given:
A[0] = 3
A[1] = 2
A[2] = 6
A[3] = -1
A[4] = 4
A[5] = 5
A[6] = -1
A[7] = 2
the function should return 17, because no double slice of array A has a sum of greater than 17.

Write an efficient algorithm for the following assumptions:
  - N is an integer within the range [3..100,000];
  - each element of array A is an integer within the range [−10,000..10,000].

 

부분 집합 2개(X, Y, Z)를 더했을 때, 최대 합을 구하는 문제. \( (0 \leq X < Y < Z < N) \)

\( (X, Y, Z) = A[X + 1] + A[X + 2] + ... + A[Y - 1] + A[Y + 1]  + A[Y + 2] + A[Z - 1] \)

 

예시) return 17

A[0]	A[1]	A[2]	A[3]	A[4]	A[5]	A[6]	A[7]
3	2	6	-1	4	5	-1	2

 

(0, 3, 6) = 2 + 6 + 4 + 5 = 17

(0, 3, 7) = 2 + 6 + 4 + 5 - 1 = 16

(3, 4, 5) = 0

 

2. 매개변수 제한

• Vector A 크기: 3 ~ 100,000
• Vector A 요소 값: -10,000 ~ 10,000

 

3. 문제분석

• 맨 처음 요소와 맨 끝 요소는 부분 집합에 포함되지 않는다.

 

4. 코드 (C++)

전체 탐색 코드

#include <limits>

//부분 집합 합 구하기
int getSum(vector<int>& A, size_t x, size_t y, size_t z)
{
    int sum = 0;
    for(size_t i = x + 1; i < y; i++)
    {
        sum += A[i];
    }

    for(size_t i = y + 1; i < z; i++)
    {
        sum += A[i];
    }

    return sum;
}

int solution(vector<int> &A) {

    int max = numeric_limits<int>::min();
    for(size_t i = 0; i < A.size(); i++)
    {
        for(size_t j = i + 1; j < A.size(); j++)
        {
            for(size_t k = j + 1; k < A.size(); k++)
            {
                int sum = getSum(A, i, j, k);
                if(sum > max) max = sum;
            }
        }
    }

    return max;
}

 

 

5. 결과