#1000

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimum Cost to Merge Stones

Minimize the cost to merge stones with k consecutive piles using dynamic programming to achieve the optimal solution.

Problem Statement

You are given n piles of stones, where the ith pile contains stones[i] stones. A move consists of merging exactly k consecutive piles into one pile, and the cost of this move is equal to the sum of the number of stones in these k piles. The goal is to return the minimum cost required to merge all piles into a single pile. If it's impossible to merge the stones using the given k, return -1.

For example, consider stones = [3, 2, 4, 1] and k = 2. You can merge the first two piles [3, 2], resulting in a cost of 5. Then merge the remaining piles and continue until you have one pile left. The task requires carefully choosing merge points using dynamic programming to achieve the lowest cost. If merging with the given k is impossible, such as when only two piles are left and k > 2, return -1.

Examples

Example 1

Input: stones = [3,2,4,1], k = 2

Output: 20

We start with [3, 2, 4, 1]. We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1]. We merge [4, 1] for a cost of 5, and we are left with [5, 5]. We merge [5, 5] for a cost of 10, and we are left with [10]. The total cost was 20, and this is the minimum possible.

Example 2

Input: stones = [3,2,4,1], k = 3

Output: -1

After any merge operation, there are 2 piles left, and we can't merge anymore. So the task is impossible.

Example 3

Input: stones = [3,5,1,2,6], k = 3

Output: 25

We start with [3, 5, 1, 2, 6]. We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6]. We merge [3, 8, 6] for a cost of 17, and we are left with [17]. The total cost was 25, and this is the minimum possible.

Constraints

n == stones.length
1 <= n <= 30
1 <= stones[i] <= 100
2 <= k <= 30

Solution Approach

State Transition Dynamic Programming

The core of this problem is applying dynamic programming (DP) to track possible merges. We define a DP state to represent the minimum cost of merging stones[i...j]. For each possible merge length, we compute the cost of merging the subarray recursively, and use previous solutions to build up the minimum cost efficiently.

Prefix Sum Optimization

To efficiently calculate the cost of merging consecutive stones, we use a prefix sum array. This allows us to quickly find the sum of any subarray, which is necessary for calculating the cost of each merge operation without recalculating sums repeatedly.

Edge Case Handling

The solution must also handle edge cases where merging isn't possible. For example, if k is too large relative to the number of piles, or if the number of piles doesn't allow for exactly k piles to be merged at each step, we should immediately return -1. The DP approach must ensure these constraints are respected.

Complexity Analysis

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

The time complexity of this solution depends on the approach used for dynamic programming. With proper optimization using prefix sums and DP states, the time complexity can be reduced to O(n^3) where n is the number of piles. Space complexity also follows O(n^3) due to the DP table used to store intermediate results.

What Interviewers Usually Probe

Look for understanding of dynamic programming and optimization techniques like prefix sums.
Evaluate the candidate's ability to handle edge cases where merging is impossible.
Assess the candidate’s familiarity with state transitions and how it applies to minimizing cost.

Common Pitfalls or Variants

Common pitfalls

Failing to handle the case where merging isn't possible, leading to incorrect answers.
Not optimizing the prefix sum calculations, leading to excessive time complexity.
Overcomplicating the DP state transitions or missing a straightforward recursive relation.

Follow-up variants

Allowing a different number of consecutive piles (e.g., changing k).
Solving the problem under stricter time constraints for larger inputs.
Extending the problem to handle multiple groups of k piles simultaneously.

FAQ

What is the main pattern used to solve the Minimum Cost to Merge Stones problem?

The main pattern is state transition dynamic programming, where we track the minimum cost of merging stones using recursive relations and optimized calculations.

How can dynamic programming optimize the solution for merging stones?

Dynamic programming allows us to reuse previously computed solutions, which reduces redundant calculations and helps achieve an optimal merge sequence.

What is the importance of prefix sums in this problem?

Prefix sums help optimize the cost calculation by providing a fast way to compute the sum of any subarray of stones, avoiding repetitive calculations.

How do we handle cases where it is impossible to merge stones?

In such cases, the solution should immediately return -1. We detect these situations through checks in the DP approach or via basic logic, such as when there are not enough piles to merge with the given k.

Can the problem be solved efficiently for larger values of n?

Yes, with dynamic programming and prefix sum optimizations, the problem can be solved efficiently even for the maximum constraint (n = 30).

terminal

Solution

Solution 1

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class Solution:
    def mergeStones(self, stones: List[int], K: int) -> int:
        n = len(stones)
        if (n - 1) % (K - 1):
            return -1
        s = list(accumulate(stones, initial=0))
        f = [[[inf] * (K + 1) for _ in range(n + 1)] for _ in range(n + 1)]
        for i in range(1, n + 1):
            f[i][i][1] = 0
        for l in range(2, n + 1):
            for i in range(1, n - l + 2):
                j = i + l - 1
                for k in range(1, K + 1):
                    for h in range(i, j):
                        f[i][j][k] = min(f[i][j][k], f[i][h][1] + f[h + 1][j][k - 1])
                f[i][j][1] = f[i][j][K] + s[j] - s[i - 1]
        return f[1][n][1]

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