Minimum Cost to Split an Array

Minimum Cost to Split an Array

1. What is this problem about?

The Minimum Cost to Split an Array problem involves partitioning an array into several non-empty subarrays. Each split has a fixed cost k. Additionally, each subarray has an "importance value," which is the sum of the number of elements that appear more than once in that subarray plus the fixed cost k. The goal is to minimize the total cost, which is the sum of the importance values of all subarrays. This is a optimization problem that balances the overhead of making many splits against the cost of large, diverse subarrays.

2. Why is this asked in interviews?

This "Hard" problem is asked by companies like Indeed to test a candidate's ability to optimize dynamic programming solutions. The Minimum Cost to Split an Array interview question requires identifying the state transitions and, more importantly, finding ways to calculate the "importance value" efficiently to avoid an $O(N^3)$ complexity. It assesses your proficiency with hash tables and frequency counting within a DP framework.

3. Algorithmic pattern used

The primary pattern is Dynamic Programming. Let $dp[i]$ be the minimum cost to split the prefix of the array ending at index $i-1$. To find $dp[i]$, we consider all possible last subarrays ending at $i-1$, starting from some $j < i$. The recurrence is $dp[i] = \min_{0 \leq j < i} (dp[j] + k + \text{trimmedLength}(j, i-1))$, where trimmedLength is the number of non-unique elements. This "Array, Hash Table, Counting, Dynamic Programming interview pattern" often requires an inner loop that updates the count of elements as it expands, keeping the complexity at $O(N^2)$.

4. Example explanation

Suppose nums = [1, 2, 1, 2, 1] and $k = 2$.

• Split into [1, 2], [1, 2, 1]:
  • [1, 2]: No repeats. Cost = $2 + 0 = 2$.
  • [1, 2, 1]: '1' repeats twice. Cost = $2 + 2 = 4$.
  • Total = 6.
• Split into [1, 2, 1, 2, 1]:
  • [1, 2, 1, 2, 1]: '1' repeats 3 times, '2' repeats 2 times.
  • Trimmed length = 3 + 2 = 5.
  • Total = $2 + 5 = 7$. The minimum cost might be achieved by other splits, like [1], [2], [1], [2], [1] (total 10) or [1, 2, 1, 2], [1] (total 6).

5. Common mistakes candidates make

A common mistake in the Minimum Cost to Split an Array coding problem is incorrectly calculating the "importance value." Many forget that an element contributes its entire count to the trimmed length once it appears twice (e.g., if '1' appears three times, all three are counted, not just the two "extra" ones). Another mistake is re-calculating the counts from scratch for every subarray, which leads to a $O(N^3)$ solution that will time out.

6. Interview preparation tip

When building a DP solution where the transition involves a range property (like trimmed length), try to update that property incrementally as you move the inner loop pointer. This turns an $O(N^3)$ problem into $O(N^2)$. Mastering this "incremental update" technique is vital for "Hard Dynamic Programming interview patterns" involving subarrays or segments.