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

// Dynamic Programming
// DP state transition:
// dp[i][j] = min(dp[i][j], dp[i][t] + dp[t + 1][j], i <= t < j)
// dp[i][j] += sums[j + 1] - sums[i] if (j - i) % (K - 1) == 0
int mergeStones(vector<int>& stones, int K) { // time: O(n^3 / K); space: O(n^2)
    int n = stones.size();
    if ((n - 1) % (K - 1) != 0) return -1;
    vector<int> sums(n + 1, 0);
    for (int i = 1; i <= n; ++i) sums[i] = sums[i - 1] + stones[i - 1];
    vector<vector<int> > dp(n, vector<int>(n, 0));
    for (int len = K; len <= n; ++len) {
        for (int i = 0; i <= n - len; ++i) {
            int j = i + len - 1;
            dp[i][j] = INT_MAX;
            for (int t = i; t < j; t += K - 1) {
                dp[i][j] = min(dp[i][j], dp[i][t] + dp[t + 1][j]);
            }
            if ((j - i) % (K - 1) == 0) {
                dp[i][j] += sums[j + 1] - sums[i];
            }
        }
    }
    return dp.front().back();
}