3077. Maximum Strength of K Disjoint Subarrays
Description
You are given a 0-indexed array of integers nums
of length n
, and a positive odd integer k
.
The strength of x
subarrays is defined as strength = sum[1] * x - sum[2] * (x - 1) + sum[3] * (x - 2) - sum[4] * (x - 3) + ... + sum[x] * 1
where sum[i]
is the sum of the elements in the ith
subarray. Formally, strength is sum of (-1)i+1 * sum[i] * (x - i + 1)
over all i
's such that 1 <= i <= x
.
You need to select k
disjoint subarrays from nums
, such that their strength is maximum.
Return the maximum possible strength that can be obtained.
Note that the selected subarrays don't need to cover the entire array.
Example 1:
Input: nums = [1,2,3,-1,2], k = 3 Output: 22 Explanation: The best possible way to select 3 subarrays is: nums[0..2], nums[3..3], and nums[4..4]. The strength is (1 + 2 + 3) * 3 - (-1) * 2 + 2 * 1 = 22.
Example 2:
Input: nums = [12,-2,-2,-2,-2], k = 5 Output: 64 Explanation: The only possible way to select 5 disjoint subarrays is: nums[0..0], nums[1..1], nums[2..2], nums[3..3], and nums[4..4]. The strength is 12 * 5 - (-2) * 4 + (-2) * 3 - (-2) * 2 + (-2) * 1 = 64.
Example 3:
Input: nums = [-1,-2,-3], k = 1 Output: -1 Explanation: The best possible way to select 1 subarray is: nums[0..0]. The strength is -1.
Constraints:
1 <= n <= 104
-109 <= nums[i] <= 109
1 <= k <= n
1 <= n * k <= 106
k
is odd.
Solutions
Solution 1: Dynamic Programming
For the $i$th number $nums[i - 1]$, if it is selected and is in the $j$th subarray, then its contribution to the answer is $nums[i - 1] \times (k - j + 1) \times (-1)^{j+1}$. We denote $(-1)^{j+1}$ as $sign$, so its contribution to the answer is $sign \times nums[i - 1] \times (k - j + 1)$.
We define $f[i][j][0]$ as the maximum energy value when selecting $j$ subarrays from the first $i$ numbers, and the $i$th number is not selected. We define $f[i][j][1]$ as the maximum energy value when selecting $j$ subarrays from the first $i$ numbers, and the $i$th number is selected. Initially, $f[0][0][1] = 0$, and the rest of the values are $-\infty$.
When $i > 0$, we consider how $f[i][j]$ transitions.
If the $i$th number is not selected, then the $i-1$th number can either be selected or not selected, so $f[i][j][0] = \max(f[i-1][j][0], f[i-1][j][1])$.
If the $i$th number is selected, if the $i-1$th number and the $i$th number are in the same subarray, then $f[i][j][1] = \max(f[i][j][1], f[i-1][j][1] + sign \times nums[i-1] \times (k - j + 1))$, otherwise $f[i][j][1] = \max(f[i][j][1], \max(f[i-1][j-1][0], f[i-1][j-1][1]) + sign \times nums[i-1] \times (k - j + 1))$.
The final answer is $\max(f[n][k][0], f[n][k][1])$.
The time complexity is $O(n \times k)$, and the space complexity is $O(n \times k)$. Where $n$ is the length of the array $nums$.
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