Minimum Total Space Wasted With K Resizing Operations
MedPrompt
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 <= 2001 <= nums[i] <= 1060 <= k <= nums.length - 1
Approaches
3 approaches with complexity analysis and trade-offs.
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 withi(i.e.,opt[i] <= opt[i+1]). This property holds for this problem's cost function. - Algorithm:
- For each number of resizes
kifrom 1 tok, we compute thedp[...][ki]array fromdp[...][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]foriin[i_start, i_end], given that their optimal split pointsjare in[j_start, j_end]. - Inside
compute:- Find the midpoint
i_mid = (i_start + i_end) / 2. - Find the optimal split point
opt_jfori_midby iteratingjfromj_starttomin(i_mid - 1, j_end). - Set
dp[i_mid][ki]to the minimum value found. - Recursively call
computefor the left part:compute(i_start, i_mid - 1, j_start, opt_j). - Recursively call
computefor the right part:compute(i_mid + 1, i_end, opt_j, j_end).
- Find the midpoint
- For each number of resizes
- 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).
Walkthrough
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
Time
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
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.
Trade-offs
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.
This approach involves exploring all possible ways to partition the array nums. Since we can perform at most k resizes, this is equivalent to partitioning the array into at most k+1 contiguous segments. A brute-force recursive method can be designed to try every valid partition, calculate the total wasted space for each, and return the minimum among them.
Algorithm
- Define a recursive function, let's say
solve(i, k_rem), which calculates the minimum wasted space for the subarraynums[i:]withk_remresizing operations remaining. - Base Cases:
- If
ireaches the end of the array (i == n), it means we have successfully partitioned the entire array, so the waste is 0. - If
k_rembecomes negative, it means we have used more thankresizes, which is an invalid state. Return a value representing infinity.
- If
- Recursive Step:
- To find the minimum waste for
nums[i:], we can try all possible endpointsjfor the current partition, which starts ati. The partition would benums[i...j]. - For each
jfromiton-1:- Calculate the cost for this partition. The optimal size for this segment is
max(nums[i...j]). The wasted space ismax(nums[i...j]) * (j - i + 1) - sum(nums[i...j]). - Recursively call the function for the rest of the array:
solve(j + 1, k_rem - 1). Note that one resize operation is consumed. - The total waste for this choice of
jiscost + solve(j + 1, k_rem - 1).
- Calculate the cost for this partition. The optimal size for this segment is
- The function returns the minimum total waste found among all possible choices for
j.
- To find the minimum waste for
- The initial call to the function would be
solve(0, k).
Walkthrough
We can define a recursive function solve(i, k_rem) that computes the minimum wasted space for the subarray starting at index i with k_rem resizes available.
In this function, we iterate through all possible split points j from i to n-1. Each j defines a segment nums[i...j]. For this segment, we set the array size to be the maximum element within it to minimize waste for that segment. The waste is max(nums[i...j]) * length - sum(nums[i...j]). After this segment, we recursively call the function for the remaining part of the array nums[j+1:] with k_rem - 1 resizes.
The base cases for the recursion are when we've processed the whole array (i == n), in which case the waste is 0, or when we've run out of resizes (k_rem < 0), which is an invalid path.
This method explores the entire search space of partitions, but its overlapping subproblems cause a massive number of redundant calculations, leading to an exponential time complexity.
Complexity
Time
Exponential, roughly O(n * C(n-1, k)). The function explores all ways to choose `k` split points from `n-1` possible locations, leading to a combinatorial explosion of calls.
Space
O(n), for the depth of the recursion stack in the worst case.
Trade-offs
Pros
Conceptually simple and directly follows the problem's definition.
Cons
Extremely inefficient due to recomputing the same subproblems multiple times.
Will result in a 'Time Limit Exceeded' error for the given constraints.
The problem exhibits optimal substructure and overlapping subproblems, which is a clear indicator for dynamic programming. We can solve this by building a table dp[i][k] that stores the minimum wasted space for the prefix of the array nums[0...i] using k resizes. By iterating through all possible endpoints of the last segment, we can build up the solution.
Algorithm
- Let
dp[i][k]be the minimum total wasted space for the prefixnums[0...i]using exactlykresizing operations. - The size of our DP table will be
n x (k+1). - Base Case (k=0):
- If we use 0 resizes, the entire prefix
nums[0...i]must be one segment. dp[i][0] = max(nums[0...i]) * (i + 1) - sum(nums[0...i]). This can be calculated for allifrom 0 ton-1.
- If we use 0 resizes, the entire prefix
- State Transition (k > 0):
- To compute
dp[i][k], we consider all possible split pointsjfor the last segment. The last segment isnums[j...i](forj > 0). - This implies that a resize happened at index
j, and the prefixnums[0...j-1]was handled withk-1resizes. - The recurrence relation is:
dp[i][k] = min_{1 <= j <= i} (dp[j-1][k-1] + cost(j, i)), wherecost(j, i)is the waste for the segmentnums[j...i].
- To compute
- Calculation:
- We can implement this bottom-up. Iterate
kifrom 1 tok. For eachki, iterateifrom 1 ton-1. For eachi, iteratejfromidown to 1 to find the minimum. - The
cost(j, i)can be computed efficiently inside thejloop by maintaining a running sum and maximum.
- We can implement this bottom-up. Iterate
- Final Answer: The result is
dp[n-1][k]. - Space Optimization: Notice that
dp[...][k]only depends ondp[...][k-1]. We can optimize space toO(n)by using two 1D arrays, one for the previouskstate and one for the current.
Walkthrough
We define dp[i][k] as the minimum wasted space for the subarray nums[0...i] with k resizes. Our goal is to find dp[n-1][k].
The base case is for k=0 resizes. This means the entire prefix nums[0...i] is treated as a single block. The array size must be max(nums[0...i]) for this whole duration. The waste is dp[i][0] = max(nums[0...i]) * (i+1) - sum(nums[0...i]).
For k > 0, we build upon previous results. To calculate dp[i][k], we assume the last resize happened at some index j (1 <= j <= i). This creates a final segment nums[j...i]. The total waste is the sum of the waste from the prefix nums[0...j-1] using k-1 resizes (which is dp[j-1][k-1]) and the waste from the segment nums[j...i]. We try all possible split points j and take the minimum.
The recurrence is dp[i][k] = min_{1 <= j <= i} (dp[j-1][k-1] + cost(j, i)). We can implement this with three nested loops. The space complexity can be optimized from O(n*k) to O(n) since computing the k-th column only requires the (k-1)-th column.
class Solution { public int minSpaceWastedKResizing(int[] nums, int k) { int n = nums.length; int[] dp = new int[n]; Arrays.fill(dp, Integer.MAX_VALUE); int sum = 0; int maxVal = 0; // Base case: k = 0 resizes for (int i = 0; i < n; i++) { sum += nums[i]; maxVal = Math.max(maxVal, nums[i]); dp[i] = maxVal * (i + 1) - sum; } // Iterate for k = 1 to K resizes for (int ki = 1; ki <= k; ki++) { int[] next_dp = new int[n]; Arrays.fill(next_dp, Integer.MAX_VALUE); for (int i = 1; i < n; i++) { int currentSum = 0; int currentMax = 0; // j is the start of the last segment for (int j = i; j >= 1; j--) { currentSum += nums[j]; currentMax = Math.max(currentMax, nums[j]); int cost = currentMax * (i - j + 1) - currentSum; if (dp[j - 1] != Integer.MAX_VALUE) { next_dp[i] = Math.min(next_dp[i], dp[j - 1] + cost); } } } dp = next_dp; } return dp[n - 1]; }}Complexity
Time
O(n^2 * k). There are three nested loops: over `k` (number of resizes), `i` (end of prefix), and `j` (start of the last segment).
Space
O(n * k) for the DP table. This can be optimized to O(n) by only storing the DP states for the previous and current number of resizes.
Trade-offs
Pros
Guaranteed to find the optimal solution.
Efficient enough to pass the given constraints.
Can be space-optimized from O(n*k) to O(n).
Cons
The O(n^2 * k) complexity might be too slow if constraints were slightly larger.
Solutions
Solution
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]; }}class Solution {public: int minSpaceWastedKResizing(vector<int> &nums, int k) { ++k; int n = nums.size(); vector<vector<int>> g(n, vector<int>(n)); for (int i = 0; i < n; ++i) { int s = 0, mx = 0; for (int j = i; j < n; ++j) { mx = max(mx, nums[j]); s += nums[j]; g[i][j] = mx * (j - i + 1) - s; } } int inf = 0x3f3f3f3f; vector<vector<int>> f(n + 1, vector<int>(k + 1, 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] = min(f[i][j], f[h][j - 1] + g[h][i - 1]); } } } return f[n][k]; }};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]Video walkthrough
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