Codility - Lesson 9 Maximum Slice Problem - 1. MaxDoubleSliceSum

Source Link:
https://codility.com/programmers/lessons/9-maximum_slice_problem/max_double_slice_sum/

Question:

A non-empty zero-indexed 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 zero-indexed 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. Assume that:

• 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].

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

My Solution:
Notes:
1. (X, Y, Z)
// 1st slice: A[X+1] + ... + A[Y-1]  
// 2nd slice: A[Y+1] + ... + A[Z-1]
// Key Point:
// The array will be split at "Y"    
2. main idea: if the middle point is "Y", find "maxLeft" and "maxRight" 
int maxLeft[] = new int[A.length];
int maxRight[] = new int[A.length]; 
3. find "maxLeft"
// maxLeft[i] is the maximum sum "contiguous subsequence" ending at index i  
// note: because it is "contiguous", we only need the ending index (important)
for(int i=1; i< A.length ;i++){ // be careful: from i=1 (because of maxLeft[i-1])
maxLeft[i] = Math.max(0, maxLeft[i-1]+A[i] ); //golden slice algorithm: Math.max(0, maxLeft[i-1]+A[i] )
}    
4. find "maxRight"
// maxRight[i] is the maximum sum "contiguous subsequence" starting at index i  
// note: because it is "contiguous", we only need the starting index (important)
for(int i=A.length-2; i >=0; i--){ // be careful: from i=A.length-2 (because of maxLeft[i+1])
maxRight[i] = Math.max(0, maxRight[i+1]+A[i] ); //golden slice algorithm: Math.max(0, maxRight[i+1]+A[i] )
}    
5. find the maximum of "maxLeft + maxRight"
int maxDoubleSlice =0;
for(int i=1; i < A.length-1; i++){ // where "i" means "Y" in this problem
if(maxLeft[i-1] + maxRight[i+1] > maxDoubleSlice) // be careful: left end at "i-1" and right begins at "i+1"
maxDoubleSlice = maxLeft[i-1] + maxRight[i+1]; // be careful: "not" maxLeft[i] + maxRight[i]
}   
return maxDoubleSlice;