Minimum Cost to Merge Stones

Minimum Cost to Merge Stones

1. What is this problem about?

Minimum Cost to Merge Stones is a classic optimization problem involving a row of stones. You are given an array where each element represents the weight of a pile of stones, and an integer $K$. In one move, you can merge $K$ consecutive piles into one single pile. The cost of this merge is the sum of the weights of the piles being merged. The goal is to continue merging until only one pile remains. If it's impossible to reach a single pile, you return -1. Otherwise, you find the minimum total cost of all merge operations combined.

2. Why is this asked in interviews?

This problem is frequently encountered in interviews at companies like Google and Amazon because it is a premier example of Range Dynamic Programming. The Minimum Cost to Merge Stones interview question tests a candidate's ability to identify overlapping subproblems and optimal substructure in a non-trivial setting. It also requires a solid grasp of mathematical conditions for feasibility—determining whether $N$ piles can actually be reduced to 1 using $K$-pile merges.

3. Algorithmic pattern used

The core algorithmic pattern is Range Dynamic Programming (DP) combined with Prefix Sums for fast cost calculation. The DP state $dp[i][j][m]$ typically represents the minimum cost to merge the range of piles from index $i$ to $j$ into $m$ piles. We use a prefix sum array to quickly find the total weight of any range $[i, j]$, which is the cost incurred when merging those piles. This "Array, Dynamic Programming, Prefix Sum interview pattern" is essential for solving problems where we make decisions about contiguous subarrays.

4. Example explanation

Suppose stones = [3, 2, 4, 1] and $K = 2$.

• Step 1: Merge [3, 2] -> [5, 4, 1]. Cost = 5.
• Step 2: Merge [4, 1] -> [5, 5]. Cost = 5.
• Step 3: Merge [5, 5] -> [10]. Cost = 10.
• Total cost = 5 + 5 + 10 = 20.

Alternatively:

• Step 1: Merge [2, 4] -> [3, 6, 1]. Cost = 6.
• Step 2: Merge [3, 6] -> [9, 1]. Cost = 9.
• Step 3: Merge [9, 1] -> [10]. Cost = 10.
• Total cost = 6 + 9 + 10 = 25. The minimum cost is 20. Note that with $K=2$, we are merging two piles at a time, which is similar to the matrix chain multiplication problem.

5. Common mistakes candidates make

A major pitfall in the Minimum Cost to Merge Stones coding problem is failing to check the feasibility condition: $(N-1) \pmod{K-1} == 0$. If this isn't true, you can never reach exactly one pile. Another mistake is using an inefficient DP state or forgetting to use prefix sums, leading to $O(N^4)$ or worse complexity. Some candidates also struggle with the transitions, specifically how to split the range $[i, j]$ into a smaller range that can be merged into 1 pile and a remaining range that merges into $m-1$ piles.

6. Interview preparation tip

Range DP is a frequent topic for "Hard" questions. Practice problems like "Matrix Chain Multiplication" or "Optimal Binary Search Tree" to get comfortable with the $dp[i][j]$ structure. For this specific problem, understanding how $K$ affects the reduction of the number of piles is crucial. This "Dynamic Programming interview pattern" is all about breaking down a range into smaller, manageable chunks and combining their results optimally.