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] = 2contains 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] = 2the 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].
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
int solution(vector<int> &A) {
assert(A.size() >= 3);
vector<int> partials(A.size());
int sliceMax = 0;
for (size_t i = 1; i < A.size(); i++) {
int a = A[i];
// slice can be empty
sliceMax = max(sliceMax + a, 0);
partials[i] = sliceMax;
}
sliceMax = 0;
int res = 0; // 0 slways possible as a result
for (size_t y = A.size() - 2; y > 0; y--) {
int a = A[y + 1];
// min Z is N-1, so if Y= N-2 then upper slice is 0
if (y == A.size() - 2)
a = 0;
sliceMax = max(sliceMax + a, 0);
int dblemax = sliceMax + partials[y - 1];
res = max(res, dblemax);
}
return res;
}
func.cpp: In function 'int solution(std::vector<int>&)': func.cpp:8:23: error: 'assert' was not declared in this scope assert(A.size() >= 3); ^
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
#include <cassert>
int solution(vector<int> &A) {
assert(A.size() >= 3);
vector<int> partials(A.size());
int sliceMax = 0;
for (size_t i = 1; i < A.size(); i++) {
int a = A[i];
// slice can be empty
sliceMax = max(sliceMax + a, 0);
partials[i] = sliceMax;
}
sliceMax = 0;
int res = 0; // 0 slways possible as a result
for (size_t y = A.size() - 2; y > 0; y--) {
int a = A[y + 1];
// min Z is N-1, so if Y= N-2 then upper slice is 0
if (y == A.size() - 2)
a = 0;
sliceMax = max(sliceMax + a, 0);
int dblemax = sliceMax + partials[y - 1];
res = max(res, dblemax);
}
return res;
}
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
#include <cassert>
int solution(vector<int> &A) {
assert(A.size() >= 3);
vector<int> partials(A.size());
int sliceMax = 0;
for (size_t i = 1; i < A.size(); i++) {
int a = A[i];
// slice can be empty
sliceMax = max(sliceMax + a, 0);
partials[i] = sliceMax;
}
sliceMax = 0;
int res = 0; // 0 slways possible as a result
for (size_t y = A.size() - 2; y > 0; y--) {
int a = A[y + 1];
// min Z is N-1, so if Y= N-2 then upper slice is 0
if (y == A.size() - 2)
a = 0;
sliceMax = max(sliceMax + a, 0);
int dblemax = sliceMax + partials[y - 1];
res = max(res, dblemax);
}
return res;
}
// you can use includes, for example:
// #include <algorithm>
// you can write to stdout for debugging purposes, e.g.
// cout << "this is a debug message" << endl;
#include <cassert>
int solution(vector<int> &A) {
assert(A.size() >= 3);
vector<int> partials(A.size());
int sliceMax = 0;
for (size_t i = 1; i < A.size(); i++) {
int a = A[i];
// slice can be empty
sliceMax = max(sliceMax + a, 0);
partials[i] = sliceMax;
}
sliceMax = 0;
int res = 0; // 0 slways possible as a result
for (size_t y = A.size() - 2; y > 0; y--) {
int a = A[y + 1];
// min Z is N-1, so if Y= N-2 then upper slice is 0
if (y == A.size() - 2)
a = 0;
sliceMax = max(sliceMax + a, 0);
int dblemax = sliceMax + partials[y - 1];
res = max(res, dblemax);
}
return res;
}
The solution obtained perfect score.