#1959

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Total Space Wasted With K Resizing Operations

Find the minimum total space wasted in a dynamic array with at most k resizing operations.

Problem Statement

You are designing a dynamic array where nums[i] represents the number of elements at time i. You are allowed up to k resizing operations on the array, where each resize ensures the array is large enough to hold all elements up to that point. The array size must be at least nums[i] at time i, with wasted space calculated as the difference between the size and the actual number of elements at that time.

Return the minimum total space wasted by resizing the array at most k times, considering the waste caused by selecting different array sizes at each point. Your goal is to find the optimal way to resize the array to minimize the wasted space.

Examples

Example 1

Input: nums = [10,20], k = 0

Output: 10

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

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

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 Approach

State Transition Dynamic Programming

Use dynamic programming to evaluate the total wasted space with different array sizes, transitioning between possible sizes as you process each element. For each resize operation, calculate the space wasted and transition between states to minimize the total space wasted.

Precomputing Total Waste for Different Ranges

Precompute the total wasted space for different ranges within the array. This allows for quick calculation of the total waste in any subarray when a resize operation is made. This approach speeds up the dynamic programming solution.

Handling the Resize Operations Efficiently

When applying the resize operations, ensure that the array size is adjusted optimally at each step. This involves making decisions based on past calculations of space wasted and possible future transitions between sizes.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the specific dynamic programming solution used, but it is generally O(n * k), where n is the length of the array and k is the number of allowed resizes. The space complexity is typically O(n * k) as well, due to the storage of intermediate results in the dynamic programming table.

What Interviewers Usually Probe

Assess the candidate's understanding of dynamic programming and its application in array resizing problems.
Evaluate the candidate's ability to break down the problem into subproblems and use state transitions to minimize wasted space.
Check if the candidate can optimize their solution by precomputing waste and leveraging past results effectively.

Common Pitfalls or Variants

Common pitfalls

Forgetting to account for the minimal wasted space when no resizes are allowed (k = 0).
Incorrectly handling the dynamic programming state transitions, leading to excessive or suboptimal resizing operations.
Neglecting to precompute the total waste for ranges, resulting in slower solutions that don't take advantage of overlapping subproblems.

Follow-up variants

Modify the problem to allow resizing only once and observe how the solution changes.
Change the array size and constraints to test the scalability of the solution for larger inputs.
Introduce additional constraints such as limiting the maximum array size to test how the solution adapts.

FAQ

What is the minimum total space wasted with k resizing operations?

The goal is to minimize the total wasted space by selecting the optimal array size and performing the fewest resizing operations, which is best achieved using dynamic programming.

How can dynamic programming help solve the problem of resizing an array with minimum wasted space?

Dynamic programming allows you to break the problem into subproblems where you calculate the minimum wasted space for each possible resize configuration and use those results to find the global optimum.

How does the number of resizing operations k impact the solution?

The number of resizing operations k limits the number of transitions between different array sizes, and the optimal solution must account for how to distribute these operations across the array to minimize waste.

What happens if k equals zero in the problem?

If k equals zero, no resizing operations can be made, and the array size must be chosen initially such that it minimizes the waste across all times in the array.

What is the best approach to optimize space complexity in this problem?

To optimize space complexity, precompute the total waste for different ranges and use dynamic programming to store intermediate results, ensuring you avoid recalculating the same waste multiple times.

terminal

Solution

Solution 1: Dynamic Programming

The problem is equivalent to dividing the array $\textit{nums}$ into $k + 1$ segments. The wasted space for each segment is the maximum value of that segment multiplied by the length of the segment minus the sum of the elements in that segment. By summing the wasted space of each segment, we get the total wasted space. By adding 1 to $k$, we are effectively dividing the array into $k$ segments.

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class Solution:
    def minSpaceWastedKResizing(self, nums: List[int], k: int) -> int:
        k += 1
        n = len(nums)
        g = [[0] * n for _ in range(n)]
        for i in range(n):
            s = mx = 0
            for j in range(i, n):
                s += nums[j]
                mx = max(mx, nums[j])
                g[i][j] = mx * (j - i + 1) - s
        f = [[inf] * (k + 1) for _ in range(n + 1)]
        f[0][0] = 0
        for i in range(1, n + 1):
            for j in range(1, k + 1):
                for h in range(i):
                    f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1])
        return f[-1][-1]

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