Description
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] * (−1)r − 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 − 1
, where 0 <= i1 < i2 < ... < ik - 1 < n - 1
, then the total cost will be:
cost(0, i1) + cost(i1 + 1, i2) + ... + cost(ik − 1 + 1, n − 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
Solution
Python3
class Solution:
def maximumTotalCost(self, nums: List[int]) -> int:
N = len(nums)
# if flag -> 1, minus, otherwise add
@cache
def go(index, flag):
if index == N: return 0
# start over
res = go(index + 1, 1) + nums[index]
# flip based on the flag
sign = 1 if flag == 0 else -1
res = max(res, go(index + 1, 1 - flag) + nums[index] * sign)
return res
return go(0, 0)