Lesson9
Maximum slice problem
Medium
MaxDoubleSliceSum
Find the maximal sum of any double slice.
Task description
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:
class Solution { public int solution(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]. ***
Code Walkthrough
class Solution {
public int solution(int[] A) {
int[] prevSubSum = new int[A.length];
int[] postSubSum = new int[A.length];
for (int i = 1; i < A.length - 1; i++) {
prevSubSum[i] = Math.max(0, prevSubSum[i - 1] + A[i]);
}
for (int i = A.length - 2; i > 0; i--) {
postSubSum[i] = Math.max(0, postSubSum[i + 1] + A[i]);
}
int globalMaxSum = 0;
for (int i = 1; i < A.length - 1; i++) {
globalMaxSum = Math.max(prevSubSum[i - 1] + postSubSum[i + 1], globalMaxSum);
}
return globalMaxSum;
}
}
Conclusion
- Detected time complexity: O(N)
- Detected space complexity: O(N)