{ "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 <= 1000cost.length == nums.length1 <= nums[i], cost[i] <= 10001 <= 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 <= 1000cost.length == nums.length1 <= nums[i], cost[i] <= 10001 <= k <= 1000dp[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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