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Minimum Total Space Wasted With K Resizing Operations

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

Solution Explanation

This problem asks to find the minimum total wasted space when resizing an array at most k times. The solution uses dynamic programming to efficiently explore all possible resizing combinations.

Understanding the Problem:

We are given an array nums where nums[i] represents the number of elements at time i. We can resize the array at most k times. Resizing means changing the array's size to any value greater than or equal to the current number of elements. The wasted space at time t is the difference between the array's size and nums[t]. The goal is to minimize the total wasted space across all times.

Dynamic Programming Approach:

The core idea is to use a 3D DP table (although a 2D table is sufficient due to the nested loops) to store the minimum wasted space.

  • g[i][j] : This 2D array pre-calculates the minimum wasted space for a subarray from index i to j (inclusive). It iterates through each subarray, finds the maximum element (mx) within that subarray, and calculates the wasted space if we use mx as the size for the entire subarray. The formula mx * (j - i + 1) - s represents this calculation, where s is the sum of elements in the subarray.

  • f[i][j] : This 2D array represents the minimum total wasted space considering the first i elements of nums and performing j resizing operations.

The DP relation is as follows:

f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1]) for 0 <= h < i

This means the minimum wasted space for the first i elements with j resizings is the minimum of all possible wasted spaces obtained by:

  1. Performing j-1 resizings on the first h elements (f[h][j - 1]).
  2. Resizing the array once at time h to accommodate the elements from h to i-1. The minimum wasted space for this is g[h][i - 1].

The final answer is f[n][k+1], where n is the length of nums. We add 1 to k because the initial size of the array is not counted as a resize.

Time and Space Complexity:

  • Time Complexity: O(n³k), where n is the length of nums and k is the maximum number of resizings. This comes from the three nested loops in the dynamic programming solution.

  • Space Complexity: O(n² + nk), which is dominated by the g and f arrays.

Code Explanation (Python):

The Python code efficiently implements the dynamic programming solution. The g array pre-computes wasted space for subarrays, and the f array stores the minimum wasted space for different numbers of resizings. The nested loops iterate through all possible subarray combinations and resizing operations to find the optimal solution. The inf variable handles cases where certain combinations are not possible.

Other Languages:

The provided code also includes Java, C++, and Go implementations. The logic remains the same across all languages; only the syntax and standard library functions differ. The efficiency and complexity analysis are consistent across all the provided implementations.