{ "data": { "question": { "questionId": "3765", "questionFrontendId": "3500", "categoryTitle": "Algorithms", "boundTopicId": 3632308, "title": "Minimum Cost to Divide Array Into Subarrays", "titleSlug": "minimum-cost-to-divide-array-into-subarrays", "content": "
You are given two integer arrays, nums
and cost
, of the same size, and an integer k
.
You can divide nums
into subarrays. The cost of the ith
subarray consisting of elements nums[l..r]
is:
(nums[0] + nums[1] + ... + nums[r] + k * i) * (cost[l] + cost[l + 1] + ... + cost[r])
.Note that i
represents the order of the subarray: 1 for the first subarray, 2 for the second, and so on.
Return the minimum total cost possible from any valid division.
\n\n\n
Example 1:
\n\nInput: nums = [3,1,4], cost = [4,6,6], k = 1
\n\nOutput: 110
\n\nExplanation:
\nThe minimum total cost possible can be achieved by dividingnums
into subarrays [3, 1]
and [4]
.\n\n[3,1]
is (3 + 1 + 1 * 1) * (4 + 6) = 50
.[4]
is (3 + 1 + 4 + 1 * 2) * 6 = 60
.Example 2:
\n\nInput: nums = [4,8,5,1,14,2,2,12,1], cost = [7,2,8,4,2,2,1,1,2], k = 7
\n\nOutput: 985
\n\nExplanation:
\nThe minimum total cost possible can be achieved by dividingnums
into subarrays [4, 8, 5, 1]
, [14, 2, 2]
, and [12, 1]
.\n\n[4, 8, 5, 1]
is (4 + 8 + 5 + 1 + 7 * 1) * (7 + 2 + 8 + 4) = 525
.[14, 2, 2]
is (4 + 8 + 5 + 1 + 14 + 2 + 2 + 7 * 2) * (2 + 2 + 1) = 250
.[12, 1]
is (4 + 8 + 5 + 1 + 14 + 2 + 2 + 12 + 1 + 7 * 3) * (1 + 2) = 210
.\n
Constraints:
\n\n1 <= nums.length <= 1000
cost.length == nums.length
1 <= nums[i], cost[i] <= 1000
1 <= k <= 1000
给你两个长度相等的整数数组 nums
和 cost
,和一个整数 k
。
你可以将 nums
分割成多个子数组。第 i
个子数组由元素 nums[l..r]
组成,其代价为:
(nums[0] + nums[1] + ... + nums[r] + k * i) * (cost[l] + cost[l + 1] + ... + cost[r])
。注意,i
表示子数组的顺序:第一个子数组为 1,第二个为 2,依此类推。
返回通过任何有效划分得到的 最小 总代价。
\n\n子数组 是一个连续的 非空 元素序列。
\n\n\n\n
示例 1:
\n\n输入: nums = [3,1,4], cost = [4,6,6], k = 1
\n\n输出: 110
\n\n解释:
\n将nums
分割为子数组 [3, 1]
和 [4]
,得到最小总代价。\n\n[3,1]
的代价是 (3 + 1 + 1 * 1) * (4 + 6) = 50
。[4]
的代价是 (3 + 1 + 4 + 1 * 2) * 6 = 60
。示例 2:
\n\n输入: nums = [4,8,5,1,14,2,2,12,1], cost = [7,2,8,4,2,2,1,1,2], k = 7
\n\n输出: 985
\n\n解释:
\n将nums
分割为子数组 [4, 8, 5, 1]
,[14, 2, 2]
和 [12, 1]
,得到最小总代价。\n\n[4, 8, 5, 1]
的代价是 (4 + 8 + 5 + 1 + 7 * 1) * (7 + 2 + 8 + 4) = 525
。[14, 2, 2]
的代价是 (4 + 8 + 5 + 1 + 14 + 2 + 2 + 7 * 2) * (2 + 2 + 1) = 250
。[12, 1]
的代价是 (4 + 8 + 5 + 1 + 14 + 2 + 2 + 12 + 1 + 7 * 3) * (1 + 2) = 210
。\n\n
提示:
\n\n1 <= nums.length <= 1000
cost.length == nums.length
1 <= nums[i], cost[i] <= 1000
1 <= k <= 1000
dp[i]
is the minimum cost to split the array suffix starting at i
.",
"Observe that no matter how many subarrays we have, if we have the first subarray on the left, the total cost of the previous subarrays increases by k * total_cost_of_the_subarray
. This is because when we increase i
to (i + 1)
, the cost increase is just the suffix sum of the cost array."
],
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