Find the maximal sum of any double slice.
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:
func Solution(A []int) int
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]. Copyright 2009–2021 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.
A[X+1]到A[Y-1] + A[Y+1]到A[Z-1] 最大的和
正向尋過array, 獲得到達每個index可以得到的最大值序列, 然后反向尋過array獲得到達每個index可以得到的最大值序列， 反向的的最大值序列需要倒轉.然後間隔一個位置， 最後尋遍array起兩者相加最大值
- https://app.codility.com/programmers/lessons/9-maximum_slice_problem/max_double_slice_sum/
- https://github.com/Anfany/Codility-Lessons-By-Python3/blob/master/L9_Maximum%20Slice%20Problem/9.3%20%20MaxDoubleSliceSum.md
package MaxDoubleSliceSum
import (
"math"
)
func Solution(A []int) int {
if len(A) < 4 {
return 0
}
N := len(A) - 2
forwardSum := make([]int, N)
reverseSum := make([]int, N)
// 0 ≤ X < Y < Z < N,
// A[X + 1] + A[X + 2] + ... + A[Y − 1] + A[Y + 1] + A[Y + 2] + ... + A[Z − 1].
// A : [ 3, 2, 6, -1, 4, 5, -1, 2]
// forwardSum : [ 0, 2, 8, 7, 11, 16]
// reverseSum : [14, 8, 9, 5, 0, 0]
for i := 0; i < N-1; i++ {
forwardVal := A[i+1]
reverseVal := A[N-i]
forwardSum[i+1] = int(math.Max(0, float64(forwardVal)+float64(forwardSum[i])))
reverseSum[N-i-2] = int(math.Max(0, float64(reverseVal)+float64(reverseSum[N-i-1])))
}
combineMax := math.MinInt64
for i := 0; i < N; i++ {
combineMax = int(math.Max(float64(combineMax), float64(forwardSum[i])+float64(reverseSum[i])))
}
return combineMax
}