#2547

Hard

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Cost to Split an Array

Minimize the cost of splitting an array into k subarrays by calculating importance values and applying dynamic programming.

array hash table dynamic programming counting

Problem Statement

You are given an integer array nums and an integer k. Your task is to split the array into some number of non-empty subarrays. The cost of a split is the sum of the importance values of the subarrays. The importance value of a subarray is calculated by removing any number that appears only once, then summing the remaining values. The goal is to find the minimum cost of splitting the array into k subarrays.

The problem can be solved using dynamic programming. You are asked to minimize the total cost of partitioning the array by considering the importance values of each possible split. Additionally, you need to efficiently calculate these values using techniques like array scanning and hash table lookups.

Examples

Example 1

Input: nums = [1,2,1,2,1,3,3], k = 2

Output: 8

We split nums to have two subarrays: [1,2], [1,2,1,3,3]. The importance value of [1,2] is 2 + (0) = 2. The importance value of [1,2,1,3,3] is 2 + (2 + 2) = 6. The cost of the split is 2 + 6 = 8. It can be shown that this is the minimum possible cost among all the possible splits.

Example 2

Input: nums = [1,2,1,2,1], k = 2

Output: 6

We split nums to have two subarrays: [1,2], [1,2,1]. The importance value of [1,2] is 2 + (0) = 2. The importance value of [1,2,1] is 2 + (2) = 4. The cost of the split is 2 + 4 = 6. It can be shown that this is the minimum possible cost among all the possible splits.

Example 3

Input: nums = [1,2,1,2,1], k = 5

Output: 10

We split nums to have one subarray: [1,2,1,2,1]. The importance value of [1,2,1,2,1] is 5 + (3 + 2) = 10. The cost of the split is 10. It can be shown that this is the minimum possible cost among all the possible splits.

Constraints

1 <= nums.length <= 1000
0 <= nums[i] < nums.length
1 <= k <= 109

Solution Approach

Dynamic Programming Approach

Use dynamic programming to track the minimum cost of splitting the array up to each element. Define dp[r] as the minimum cost to partition the first r elements of nums. This allows for efficient transitions when considering new partitions, and helps minimize the overall cost.

Array Scanning with Hash Lookup

During the partitioning process, scan through the array and use a hash table to store the importance values of elements. This allows for fast lookups and efficient computation of the importance values for each subarray, reducing the time complexity of recalculating these values.

Minimizing Split Cost

To minimize the split cost, explore various ways to partition the array, computing the cost of each partition and keeping track of the minimum cost encountered. The dynamic programming array should update with the lowest cost partitioning as it progresses through the array.

Complexity Analysis

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

The time and space complexity of the solution depend on the approach chosen. Dynamic programming with array scanning and hash table lookups can reduce the overall complexity, but a brute force approach might have higher time complexity due to repeated calculations.

What Interviewers Usually Probe

The candidate should be able to identify the role of dynamic programming in solving the problem efficiently.
Look for understanding of how hash table lookups help reduce the cost calculation time.
Assess the candidate’s ability to balance the trade-off between array scanning and space/time complexity when partitioning.

Common Pitfalls or Variants

Common pitfalls

Incorrectly computing the importance values due to failing to remove unique elements first.
Misunderstanding dynamic programming transitions, leading to inefficient solutions.
Not utilizing hash tables properly, leading to slower lookups and higher time complexity.

Follow-up variants

Extend the problem to support multiple ways of splitting the array with different cost constraints.
Consider optimizing the approach for arrays with very large lengths or extreme k values.
Modify the problem to use different subarray importance calculation methods, such as weighted sums.

FAQ

How does dynamic programming help in solving the Minimum Cost to Split an Array problem?

Dynamic programming allows you to track the minimum cost of splitting the array up to each element, providing an optimal solution through efficient state transitions.

What role do hash tables play in this problem?

Hash tables store the importance values of subarrays, allowing for quick lookups and reducing the computational overhead of recalculating values repeatedly.

How can I minimize the cost of the split in this problem?

To minimize the split cost, you need to explore all possible partitions and choose the one with the lowest calculated cost using dynamic programming and efficient importance value calculations.

What is the time complexity of solving this problem?

The time complexity depends on the approach used, with dynamic programming providing a more efficient solution compared to brute-force methods.

Can the problem be solved without dynamic programming?

While it can be solved with simpler methods, dynamic programming is crucial for efficiently minimizing the split cost, especially for larger arrays.

terminal

Solution

Solution 1: Memoization Search

We design a function $dfs(i)$, which represents the minimum cost of splitting from index $i$. So the answer is $dfs(0)$.

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class Solution:
    def minCost(self, nums: List[int], k: int) -> int:
        @cache
        def dfs(i):
            if i >= n:
                return 0
            cnt = Counter()
            one = 0
            ans = inf
            for j in range(i, n):
                cnt[nums[j]] += 1
                if cnt[nums[j]] == 1:
                    one += 1
                elif cnt[nums[j]] == 2:
                    one -= 1
                ans = min(ans, k + j - i + 1 - one + dfs(j + 1))
            return ans

        n = len(nums)
        return dfs(0)

Continue Practicing

#1994 the number of good subsets #3186 maximum total damage with spell casting #2581 count number of possible root nodes