Package g3101_3200.s3196_maximize_total_cost_of_alternating_subarrays

Class Solution

java.lang.Object
g3101_3200.s3196_maximize_total_cost_of_alternating_subarrays.Solution
public class Solution extends Object
3196 - Maximize Total Cost of Alternating Subarrays.

Medium

You are given an integer array nums with length n.

The cost of a subarray nums[l..r], where 0 <= l <= r < n, is defined as:

cost(l, r) = nums[l] - nums[l + 1] + … + nums[r] * (\u22121)r \u2212 l

Your task is to split nums into subarrays such that the total cost of the subarrays is maximized , ensuring each element belongs to exactly one subarray.

Formally, if nums is split into k subarrays, where k > 1, at indices i1, i2, …, ik \u2212 1, where 0 <= i1 < i2 < … < ik - 1 < n - 1, then the total cost will be:

cost(0, i1) + cost(i1 + 1, i2) + … + cost(ik \u2212 1 + 1, n \u2212 1)

Return an integer denoting the maximum total cost of the subarrays after splitting the array optimally.

Note: If nums is not split into subarrays, i.e. k = 1, the total cost is simply cost(0, n - 1).

Example 1:

Input: nums = [1,-2,3,4]

Output: 10

Explanation:

One way to maximize the total cost is by splitting [1, -2, 3, 4] into subarrays [1, -2, 3] and [4]. The total cost will be (1 + 2 + 3) + 4 = 10.

Example 2:

Input: nums = [1,-1,1,-1]

Output: 4

Explanation:

One way to maximize the total cost is by splitting [1, -1, 1, -1] into subarrays [1, -1] and [1, -1]. The total cost will be (1 + 1) + (1 + 1) = 4.

Example 3:

Input: nums = [0]

Output: 0

Explanation:

We cannot split the array further, so the answer is 0.

Example 4:

Input: nums = [1,-1]

Output: 2

Explanation:

Selecting the whole array gives a total cost of 1 + 1 = 2, which is the maximum.

Constraints:

  • 1 <= nums.length <= 105
  • -109 <= nums[i] <= 109

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • maximumTotalCost

      public long maximumTotalCost(int[] nums)