MaxDoubleSliceSum Solution
Task
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] = 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].
Solution
The key to this is noticing that given a value of Y, the optimum values of X and Z are independent.
So the solution is to iterate over the Y values, finding the optimum max slice for the interval (Y,Z), as in the max slice sum task, and using a precomputed solution for the optimum slice (X,Y).
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;
}Test Function
This code tests the solution
void testSoln(vector<int> A, int exp) { assert(solution(A) == exp); }
int main() {
testSoln(vector<int>{3, 2, 6, -1, 4, 5, -1, 2}, 17);
testSoln(vector<int>{-9000, -9000, -9000}, 0);
testSoln(vector<int>(100000, 0), 0);
testSoln(vector<int>(100000, 1), 99997);
testSoln(vector<int>{1, 2, 3}, 0);
testSoln(vector<int>{1, 1, 1, 1}, 1);
testSoln(vector<int>{1, 10, 1, 1}, 10);
testSoln(vector<int>{1, -10, -10, 7, 1}, 7);
testSoln(vector<int>{-1, -10, -10, -7, -1}, 0);
testSoln(vector<int>{-2 – 3, 4, -1, -2, 1, 5, -3}, 9);
testSoln(vector<int>{1, 1, 0, 10, -100, 10, 0}, 21);
testSoln(vector<int>{0, 10, -5, -2, 0}, 10);
}Results
Codility Results
| Correctness | 100% |
| Performance | 100% |
| Time Complexity | O(N) |
Other Solutions
These are some other solutions to this problem on the web. These all use some variation on the above.