Lesson 9 - maxDoubleSliceSum

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:

class Solution { public int solution(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].

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못 풀어서 구글링했다. 답을 찾아보니 30분 고민한걸로는 영원히 못풀 문제더라.

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베스트 답안

무려 하루를 고민하셨다고 한다. 설명은 여기를 참조하자.

function solution(A) {
    const N = A.length;

    if(N === 3) return 0;

    let headSum = A.map(i => 0);
    let tailSum = A.map(i => 0);

    for(let i = 1; i < N-1; i++) {
        headSum[i] = Math.max(0, headSum[i-1] + A[i]);
    }

    for(let i = N-2; i >= 1; i--) {
        tailSum[i] = Math.max(0, tailSum[i+1] + A[i]);
    }

    let maxSum = 0;

    for(let i = 1; i < N-1; i++) {
        maxSum = Math.max(maxSum, headSum[i-1] + tailSum[i+1]);    
    } 

    return maxSum;
}

여기도 영어로 해당 문제에 대해 잘 설명해주고 있다.

public int solution(int[] A) {
  int N = A.length;
  int[] K1 = new int[N];
  int[] K2 = new int[N];

  for(int i = 1; i < N-1; i++){
    K1[i] = Math.max(K1[i-1] + A[i], 0);
  }
  for(int i = N-2; i > 0; i--){
    K2[i] = Math.max(K2[i+1]+A[i], 0);
  }

  int max = 0;

  for(int i = 1; i < N-1; i++){
    max = Math.max(max, K1[i-1]+K2[i+1]);
  }

  return max;
}