[Codility] Lesson9 - Maximum slice problem: MaxDoubleSliceSum

Problem

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].

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Solution(c++)

// you can use includes, for example:
#include <bits/stdc++.h>

using namespace std;

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

int solution(vector<int> &A) {
    // write your code in C++14 (g++ 6.2.0)
    
    int len = A.size();
    vector<int> sum_left(len);
    vector<int> sum_right(len);
    int ret = 0;
    if(A.size()<=3){
        return 0;
    }

    for(int i=1; i<len-1; i++){
        sum_left[i] = max(sum_left[i-1]+A[i],0);
    }

    for(int i=len-1; i>1; i--){
        sum_right[i-1] = max(sum_right[i]+A[i-1],0);
    }

    for(int i=1; i<len-1; i++){
        ret = max(sum_left[i-1] + sum_right[i+1], ret);
    }


    return ret;
}

Test Result

app.codility.com/demo/results/trainingPP7UR2-D8T/

Test results - Codility