Package g3001_3100.s3013_divide_an_array_into_subarrays_with_minimum_cost_ii

Class Solution

  • All Implemented Interfaces:
    
public final class Solution

    3013 - Divide an Array Into Subarrays With Minimum Cost II\.

    Hard

    You are given a 0-indexed array of integers nums of length n, and two positive integers k and dist.

    The cost of an array is the value of its first element. For example, the cost of [1,2,3] is 1 while the cost of [3,4,1] is 3.

    You need to divide nums into k disjoint contiguous subarrays, such that the difference between the starting index of the second subarray and the starting index of the kth subarray should be less than or equal to dist. In other words, if you divide nums into the subarrays <code>nums0..(i<sub>1</sub> - 1), numsi<sub>1</sub>..(i<sub>2</sub> - 1), ..., numsi<sub>k-1</sub>..(n - 1)</code>, then <code>i<sub>k-1</sub> - i<sub>1</sub><= dist</code>.

    Return the minimum possible sum of the cost of these subarrays.

    Example 1:

    Input: nums = 1,3,2,6,4,2, k = 3, dist = 3

    Output: 5

    Explanation: The best possible way to divide nums into 3 subarrays is: 1,3, 2,6,4, and 2. This choice is valid because i<sub>k-1</sub> - i<sub>1</sub> is 5 - 2 = 3 which is equal to dist.

    The total cost is nums0 + nums2 + nums5 which is 1 + 2 + 2 = 5.

    It can be shown that there is no possible way to divide nums into 3 subarrays at a cost lower than 5.

    Example 2:

    Input: nums = 10,1,2,2,2,1, k = 4, dist = 3

    Output: 15

    Explanation: The best possible way to divide nums into 4 subarrays is: 10, 1, 2, and 2,2,1. This choice is valid because i<sub>k-1</sub> - i<sub>1</sub> is 3 - 1 = 2 which is less than dist.

    The total cost is nums0 + nums1 + nums2 + nums3 which is 10 + 1 + 2 + 2 = 15.

    The division 10, 1, 2,2,2, and 1 is not valid, because the difference between i<sub>k-1</sub> and i<sub>1</sub> is 5 - 1 = 4, which is greater than dist.

    It can be shown that there is no possible way to divide nums into 4 subarrays at a cost lower than 15.

    Example 3:

    Input: nums = 10,8,18,9, k = 3, dist = 1

    Output: 36

    Explanation: The best possible way to divide nums into 4 subarrays is: 10, 8, and 18,9. This choice is valid because i<sub>k-1</sub> - i<sub>1</sub> is 2 - 1 = 1 which is equal to dist.

    The total cost is nums0 + nums1 + nums2 which is 10 + 8 + 18 = 36.

    The division 10, 8,18, and 9 is not valid, because the difference between i<sub>k-1</sub> and i<sub>1</sub> is 3 - 1 = 2, which is greater than dist.

    It can be shown that there is no possible way to divide nums into 3 subarrays at a cost lower than 36.

    Constraints:

    • <code>3 <= n <= 10<sup>5</sup></code>

    • <code>1 <= numsi<= 10<sup>9</sup></code>

    • 3 &lt;= k &lt;= n

    • k - 2 &lt;= dist &lt;= n - 2

    • Nested Class Summary

      Nested Classes 
      Modifier and Type Class Description

    • Field Summary

      Fields 
      Modifier and Type Field Description

    • Constructor Summary

      Constructors 
      Constructor Description
      Solution()

    • Enum Constant Summary

      Enum Constants 
      Enum Constant Description

    • Method Summary

      Modifier and Type Method Description
      final Long minimumCost(IntArray nums, Integer k, Integer dist)

      • Methods inherited from class java.lang.Object

        clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait