- Difficulty: Medium
- Tags: LeetCode, Medium, Array, Dynamic Programming, leetcode-1959, O(k * n^2), O(k * n)
Problem
You are currently designing a dynamic array. You are given a 0-indexed integer array nums
, where nums[i]
is the number of elements that will be in the array at time i
. In addition, you are given an integer k
, the maximum number of times you can resize the array (to any size).
The size of the array at time t
, sizet
, must be at least nums[t]
because there needs to be enough space in the array to hold all the elements. The space wasted at time t
is defined as sizet - nums[t]
, and the total space wasted is the sum of the space wasted across every time t
where 0 <= t < nums.length
.
Return the minimum total space wasted if you can resize the array at most k
times.
Note: The array can have any size at the start and does not count towards the number of resizing operations.
Example 1:
Input: nums = [10,20], k = 0 Output: 10 Explanation: size = [20,20]. We can set the initial size to be 20. The total wasted space is (20 - 10) + (20 - 20) = 10.
Example 2:
Input: nums = [10,20,30], k = 1 Output: 10 Explanation: size = [20,20,30]. We can set the initial size to be 20 and resize to 30 at time 2. The total wasted space is (20 - 10) + (20 - 20) + (30 - 30) = 10.
Example 3:
Input: nums = [10,20,15,30,20], k = 2 Output: 15 Explanation: size = [10,20,20,30,30]. We can set the initial size to 10, resize to 20 at time 1, and resize to 30 at time 3. The total wasted space is (10 - 10) + (20 - 20) + (20 - 15) + (30 - 30) + (30 - 20) = 15.
Constraints:
1 <= nums.length <= 200
1 <= nums[i] <= 106
0 <= k <= nums.length - 1