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Minimum Total Space Wasted With K Resizing Operations

Difficulty: Medium

Problem Description

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, size_t, 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 size_t - 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.

Key Insights

  • Each resizing operation allows you to adjust the size of the array to optimize for space.
  • The initial size of the array can be set without counting as a resize.
  • The goal is to minimize wasted space across all time indices, given a constraint on the number of resizing operations.
  • Dynamic programming (DP) can be employed to explore optimal resizing strategies without exceeding the allowed number of resizes.

Space and Time Complexity

Time Complexity: O(n^2 * k)
Space Complexity: O(n * k)

Solution

The solution uses dynamic programming to keep track of the minimum wasted space for each possible time index and each possible number of resizing operations used. We define a DP array where dp[i][j] represents the minimum wasted space considering the first i elements with j resizing operations.

  1. Initialize the DP table with infinite values.
  2. For each time index i, calculate the total space wasted if we choose to resize the array at various previous indices.
  3. Use nested loops to evaluate all possible segments of the array to determine the best place to resize, updating the DP table accordingly.
  4. The final answer will be the minimum value in the last row of the DP table after considering all resizing options.

Code Solutions

def minSpaceWastedKResizing(nums, k):
    n = len(nums)
    dp = [[float('inf')] * (k + 1) for _ in range(n + 1)]
    
    for j in range(k + 1):
        dp[0][j] = 0  # No waste with zero elements

    for i in range(1, n + 1):
        total = 0
        max_size = 0
        for j in range(i, 0, -1):
            total += nums[j - 1]
            max_size = max(max_size, nums[j - 1])
            dp[i][0] = min(dp[i][0], max_size * (i - j + 1) - total)
        
        for j in range(1, k + 1):
            for p in range(i):
                total = 0
                max_size = 0
                for q in range(p + 1, i + 1):
                    total += nums[q - 1]
                    max_size = max(max_size, nums[q - 1])
                    dp[i][j] = min(dp[i][j], dp[p][j - 1] + max_size * (q - p) - total)

    return dp[n][k]
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