Minimum Cost to Merge Stones — C++

// Time:  O(n^3 / k)
// Space: O(n^2)

class Solution {
public:
    int mergeStones(vector<int>& stones, int K) {
        if ((stones.size() - 1) % (K - 1)) {
            return -1;
        }
        vector<int> prefix(stones.size() + 1, 0);
        partial_sum(cbegin(stones), cend(stones), next(begin(prefix)), plus<int>());

        vector<vector<int> > dp(stones.size(), vector<int>(stones.size()));
        for (int l = K - 1; l < stones.size(); ++l) {
            for (int i = 0; i + l < stones.size(); ++i) {
                dp[i][i + l] = numeric_limits<int>::max();
                for (int j = i; j + 1 <= i + l; j += K - 1) {
                    dp[i][i + l] = min(dp[i][i + l], dp[i][j] + dp[j + 1][i + l]);
                }
                if (l % (K - 1) == 0) {
                    dp[i][i + l] += prefix[i + l + 1] - prefix[i];
                }
            }
        }
        return dp[0][stones.size() - 1];
    }
};