{ "data": { "question": { "questionId": "3313", "questionFrontendId": "3077", "categoryTitle": "Algorithms", "boundTopicId": 2674662, "title": "Maximum Strength of K Disjoint Subarrays", "titleSlug": "maximum-strength-of-k-disjoint-subarrays", "content": "
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.
\n\nNote that the selected subarrays don't need to cover the entire array.
\n\n\n
Example 1:
\n\n\nInput: nums = [1,2,3,-1,2], k = 3\nOutput: 22\nExplanation: 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.\n\n\n
Example 2:
\n\n\nInput: nums = [12,-2,-2,-2,-2], k = 5\nOutput: 64\nExplanation: 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.\n\n\n
Example 3:
\n\n\nInput: nums = [-1,-2,-3], k = 1\nOutput: -1\nExplanation: The best possible way to select 1 subarray is: nums[0..0]. The strength is -1.\n\n\n
\n
Constraints:
\n\n1 <= n <= 104
-109 <= nums[i] <= 109
1 <= k <= n
1 <= n * k <= 106
k
is odd.给你一个长度为 n
下标从 0 开始的整数数组 nums
和一个 正奇数 整数 k
。
x
个子数组的能量值定义为 strength = sum[1] * x - sum[2] * (x - 1) + sum[3] * (x - 2) - sum[4] * (x - 3) + ... + sum[x] * 1
,其中 sum[i]
是第 i
个子数组的和。更正式的,能量值是满足 1 <= i <= x
的所有 i
对应的 (-1)i+1 * sum[i] * (x - i + 1)
之和。
你需要在 nums
中选择 k
个 不相交子数组 ,使得 能量值最大 。
请你返回可以得到的 最大能量值 。
\n\n注意,选出来的所有子数组 不 需要覆盖整个数组。
\n\n\n\n
示例 1:
\n\n\n输入:nums = [1,2,3,-1,2], k = 3\n输出:22\n解释:选择 3 个子数组的最好方式是选择:nums[0..2] ,nums[3..3] 和 nums[4..4] 。能量值为 (1 + 2 + 3) * 3 - (-1) * 2 + 2 * 1 = 22 。\n\n\n
示例 2:
\n\n\n输入:nums = [12,-2,-2,-2,-2], k = 5\n输出:64\n解释:唯一一种选 5 个不相交子数组的方案是:nums[0..0] ,nums[1..1] ,nums[2..2] ,nums[3..3] 和 nums[4..4] 。能量值为 12 * 5 - (-2) * 4 + (-2) * 3 - (-2) * 2 + (-2) * 1 = 64 。\n\n\n
示例 3:
\n\n\n输入:nums = [-1,-2,-3], k = 1\n输出:-1\n解释:选择 1 个子数组的最优方案是:nums[0..0] 。能量值为 -1 。\n\n\n
\n\n
提示:
\n\n1 <= n <= 104
-109 <= nums[i] <= 109
1 <= k <= n
1 <= n * k <= 106
k
是奇数。dp[i][j][x == 0/1]
be the maximum strength to select j
disjoint subarrays from the original array’s suffix (nums[i..(n - 1)]
), x denotes whether we select the element or not.",
"Initially dp[n][0][0] == 0
.",
"We have \r\ndp[i][j][1] = nums[i] * get(j) + max(dp[i + 1][j - 1][0], dp[i + 1][j][1])
where get(j) = j
if j
is odd, otherwise -j
.",
"We can select nums[i]
as a separate subarray or select at least nums[i]
and nums[i + 1]
as the first subarray.\r\ndp[i][j][0] = max(dp[i + 1][j][0], dp[i][j][1])
.",
"The answer is dp[0][k][0]
."
],
"solution": null,
"status": null,
"sampleTestCase": "[1,2,3,-1,2]\n3",
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