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Minimum Total Space Wasted With K Resizing Operations

Number: 2081

Difficulty: Medium

Paid? No

Companies: Media.net

Problem Description

Given an array nums where nums[i] represents the number of elements in a dynamic array at time i, and an integer k representing the maximum number of times we can resize the array, determine the minimum total wasted space when you are allowed to resize at most k times. The wasted space at time t is defined as (current array size - nums[t]). You can choose the initial size freely (it does not count as a resize), and each contiguous segment (between resizes) must have a fixed size chosen as the maximum element encountered in that segment.

Key Insights

  • The problem can be segmented into contiguous subarrays where each segment is assigned a size equal to the maximum element within it.
  • The cost (waste) for a segment from index i to j is given by (j-i+1) * max(nums[i..j]) - sum(nums[i..j]).
  • We can solve this using dynamic programming by trying all possible segmentations using at most k+1 segments (since the initial size counts as one segment).
  • Precompute prefix sums and the wasted cost for every subarray to allow efficient DP transition.
  • Use a DP state dp[i][r] representing the minimum wasted space for the subarray starting at index i when r resizes are still allowed.

Space and Time Complexity

Time Complexity: O(n^2 * k) where n is the length of nums, due to precomputing the cost for each subarray (O(n^2)) and iterating over possible segmentations in the DP (O(n^2 * k)). Space Complexity: O(n^2) for storing subarray costs; however, the DP table requires O(n * k), so overall space is dominated by the subarray cost storage.

Solution

The solution uses dynamic programming with the following steps:

  1. Precompute the cost (waste) for every subarray nums[i..j] using prefix sums and by tracking the maximum element.
  2. Define a DP state dp[i][r] where i is the current index and r is the number of remaining resizes allowed. When no resizes are allowed (r = 0), the cost is the waste for the segment from index i to the end.
  3. For dp[i][r] with r > 0, iterate over possible break points j (from i to n-1) and combine the waste from the current segment (i to j) with the optimal solution for the subarray starting at j+1 with one fewer resize.
  4. The final answer is dp[0][k], since we start at index 0 with k resizes allowed.
  5. Use memoization or iterative DP to avoid redundant recalculations.

Code Solutions

def minWastedSpace(nums, k):
    n = len(nums)
    # Precompute prefix sums for quick subarray sum query.
    prefixSum = [0] * (n + 1)
    for i in range(n):
        prefixSum[i + 1] = prefixSum[i] + nums[i]
    
    # Precompute waste(i, j) for all subarrays.
    waste = [[0] * n for _ in range(n)]
    for i in range(n):
        currentMax = nums[i]
        for j in range(i, n):
            currentMax = max(currentMax, nums[j])
            segment_length = j - i + 1
            total = prefixSum[j + 1] - prefixSum[i]
            waste[i][j] = currentMax * segment_length - total

    # dp[i][r] will hold the minimal wasted space for subarray starting at index i with r resizes allowed.
    dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
    # Base case: when starting index is n, no waste.
    for r in range(k + 1):
        dp[n][r] = 0

    # Fill dp table from the end towards the beginning.
    for i in range(n - 1, -1, -1):
        for r in range(k + 1):
            # If no more resizes allowed, the segment covers from i to end.
            if r == 0:
                dp[i][r] = waste[i][n - 1]
            else:
                # Try placing a resize at every possible split point.
                for j in range(i, n):
                    # dp[j+1][r-1] is added if j+1 < n else 0.
                    cost = waste[i][j] + dp[j + 1][r - 1]
                    dp[i][r] = min(dp[i][r], cost)
    return dp[0][k]

# Example usage:
nums = [10,20,15,30,20]
k = 2
print(minWastedSpace(nums, k))  # Expected output: 15
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