1000. Minimum Cost to Merge Stones

There are N piles of stones arranged in a row. The i-th pile has 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 total number of stones in these K piles.

Find the minimum cost to merge all piles of stones into one pile. If it is impossible, return -1.

Example 1:

Input: stones = [3,2,4,1], K = 2
Output: 20
Explanation:
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
Explanation: 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
Explanation:
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.

Note:

• 1 <= stones.length <= 30
• 2 <= K <= 30
• 1 <= stones[i] <= 100

# @lc code=start
using LeetCode

function merge_stones(stones::Vector{Int}, K::Int)
    len = length(stones)
    (len - 1) % (K - 1) != 0 && return -1
    pre = fill(0, len + 1)
    cumsum!(@view(pre[2:len + 1]), stones)
    dp = fill(0, len, len)
    for m in K:len, i in 1:(len - m + 1)
        dp[i, i + m - 1] = minimum(dp[i, k] + dp[k + 1, i + m - 1]
                               for k in i:(K - 1):(i + m - 2)) +
                           ((m - 1) % (K - 1) == 0 ? pre[i + m] - pre[i] : 0)
    end
    return dp[1, len]
end
# @lc code=end

merge_stones (generic function with 1 method)

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