Minimum Total Space Wasted With K Resizing Operations

MEDIUM

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.

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

Approaches

Checkout 3 different approaches to solve Minimum Total Space Wasted With K Resizing Operations. Click on different approaches to view the approach and algorithm in detail.

DP with Divide and Conquer Optimization

The standard O(n^2 * k) DP can be further optimized. The calculation of each row of the DP table, dp[...][k], from the previous row dp[...][k-1] can be accelerated from O(n^2) to O(n log n). This is possible due to a property of the cost function which implies that the optimal choice of the last partition point is monotonic. This allows for a 'Divide and Conquer' optimization strategy.

Algorithm

The DP recurrence is dp[i][k] = min_{0 <= j < i} (dp[j][k-1] + cost(j, i-1)).
This structure is a candidate for Divide and Conquer Optimization if the optimal split point, opt[i] = argmin_{j} (dp[j][k-1] + cost(j, i-1)), is monotonic with i (i.e., opt[i] <= opt[i+1]). This property holds for this problem's cost function.
Algorithm:
- For each number of resizes ki from 1 to k, we compute the dp[...][ki] array from dp[...][ki-1].
- Instead of nested loops, we use a recursive helper function: compute(i_start, i_end, j_start, j_end).
- This function computes dp[i][ki] for i in [i_start, i_end], given that their optimal split points j are in [j_start, j_end].
- Inside compute:
  1. Find the midpoint i_mid = (i_start + i_end) / 2.
  2. Find the optimal split point opt_j for i_mid by iterating j from j_start to min(i_mid - 1, j_end).
  3. Set dp[i_mid][ki] to the minimum value found.
  4. Recursively call compute for the left part: compute(i_start, i_mid - 1, j_start, opt_j).
  5. Recursively call compute for the right part: compute(i_mid + 1, i_end, opt_j, j_end).
To make cost(j, i) calculation efficient (O(1)), we can precompute prefix sums for the sum part and use a Sparse Table or Segment Tree for Range Maximum Queries (RMQ).

This approach, also known as Knuth-Yao DP optimization (though more commonly applied via the Divide and Conquer trick), speeds up the computation of DP states. The key insight is that for a fixed number of resizes k, the optimal position j to make the (k-1)-th resize for computing dp[i][k] is non-decreasing as i increases.

Let opt[i][k] be the smallest j that minimizes dp[j-1][k-1] + cost(j, i). The property is opt[i][k] <= opt[i+1][k]. This allows us to constrain the search space for the optimal j.

We define a recursive function, say compute(i_start, i_end, j_start, j_end), which calculates dp[i][k] for all i in [i_start, i_end]. The j_start and j_end parameters provide a restricted range where the optimal split point j must lie. We first find the optimal split opt_j for the midpoint i_mid. Due to the monotonicity, the optimal split for any i < i_mid must be in [j_start, opt_j], and for any i > i_mid, it must be in [opt_j, j_end]. This allows us to solve the subproblems recursively with smaller search ranges.

For this to be efficient, calculating cost(j, i) must be fast. We can precompute prefix sums for O(1) range sum queries and build a Sparse Table for O(1) Range Maximum Queries. The Sparse Table construction takes O(n log n). The compute function for each k then takes O(n log n). The total time complexity becomes O(n * k * log n).

Complexity Analysis

Time Complexity: O(n * k * log n). Building the RMQ structure takes `O(n log n)`. Then, for each of the `k` resizes, the divide and conquer DP calculation takes `O(n log n)`.Space Complexity: O(n * k) or O(n * log n). The DP table takes `O(n*k)` (or `O(n)` if optimized), and the Sparse Table for RMQ requires `O(n log n)` space.

Pros and Cons

Pros:

Asymptotically faster than the standard DP approach.
Provides a significant performance improvement for larger n.

Cons:

Significantly more complex to understand and implement correctly.
The proof that the optimal split point is monotonic is non-trivial.
Requires auxiliary data structures like a Sparse Table for efficient RMQ.

Code Solutions

Checking out 3 solutions in different languages for Minimum Total Space Wasted With K Resizing Operations. Click on different languages to view the code.

class Solution {
public
  int minSpaceWastedKResizing(int[] nums, int k) {
    ++k;
    int n = nums.length;
    int[][] g = new int[n][n];
    for (int i = 0; i < n; ++i) {
      int s = 0, mx = 0;
      for (int j = i; j < n; ++j) {
        s += nums[j];
        mx = Math.max(mx, nums[j]);
        g[i][j] = mx * (j - i + 1) - s;
      }
    }
    int[][] f = new int[n + 1][k + 1];
    int inf = 0x3f3f3f3f;
    for (int i = 0; i < f.length; ++i) {
      Arrays.fill(f[i], inf);
    }
    f[0][0] = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= k; ++j) {
        for (int h = 0; h < i; ++h) {
          f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
        }
      }
    }
    return f[n][k];
  }
}

Video Solution

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Patterns:

Dynamic Programming

Data Structures:

Array

Companies:

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