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();
}