Minimum Cost to Split an Array

Problem Description

You are given an integer array nums and an integer k. Split the array into some number of non-empty subarrays. The cost of a split is the sum of the importance value of each subarray in the split. The importance value of a subarray is defined as k + trimmed(subarray).length, where trimmed(subarray) is the version of the subarray where all numbers that appear only once are removed. Return the minimum possible cost of a split of nums.

Key Insights

• The importance value consists of a constant k plus the length of the trimmed subarray.
• A trimmed subarray retains only elements that appear more than once.
• Dynamic programming can be utilized to minimize the cost by evaluating possible splits.
• A hash table can be used to count occurrences of elements in the subarrays efficiently.

Space and Time Complexity

Time Complexity: O(n^2), where n is the length of the array nums, due to the nested loops for evaluating subarrays.
Space Complexity: O(n), for storing frequency counts of elements in the subarrays.

Solution

To solve this problem, we can use dynamic programming. We will maintain an array dp where dp[i] represents the minimum cost to split the subarray nums[0:i]. For each possible endpoint of a subarray, we will compute the trimmed length and update the cost accordingly. A hash table (or dictionary) will be used to keep track of the frequency of elements, allowing us to compute the trimmed length efficiently.

1. Initialize a dp array of size n + 1 with infinity, and set dp[0] to 0 (no cost for an empty subarray).
2. For each endpoint i from 1 to n, iterate over all possible starting points j.
3. Use a hash table to count occurrences of elements between j and i.
4. Calculate the trimmed length and update dp[i] with the minimum cost found using dp[j].

Code Solutions