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# 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

class Solution {

public int mergeStones(int[] stones, int K) {

int n = stones.length;

if ((n - 1) % (K - 1) != 0) return -1;

int[] sum = new int[n + 1];

for (int i = 0; i < n; i++) {

sum[i + 1] = sum[i] + stones[i];

}

int[][] dp = new int[n][n];

for (int len = K; len <= n; len++) {

for (int i = 0; i + len <= n; i++) {

int j = i + len - 1;

dp[i][j] = Integer.MAX_VALUE;

for (int mid = i; mid < j; mid += K - 1) {

dp[i][j] = Math.min(dp[i][j], dp[i][mid] + dp[mid + 1][j]);

}

if ((j - i) % (K - 1) == 0) {

dp[i][j] += sum[j + 1] - sum[i];

}

}

}

return dp[0][n - 1];

}

}

Last modified 2yr ago