Formatted question description: https://leetcode.ca/all/1000.html

1000. Minimum Cost to Merge Stones (Hard)

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

Related Topics:
Dynamic Programming

Similar Questions:

Solution 1. DP

// OJ: https://leetcode.com/problems/minimum-cost-to-merge-stones/

// Time: O(N^3 / K)
// Space: O(N^2)
// Ref: https://leetcode.com/problems/minimum-cost-to-merge-stones/discuss/247567/JavaC%2B%2BPython-DP
class Solution {
public:
    int mergeStones(vector<int>& stones, int K) {
        int N = stones.size();
        if ((N - 1) % (K - 1)) return -1;
        vector<int> prefix(N + 1);
        partial_sum(stones.begin(), stones.end(), prefix.begin() + 1);
        vector<vector<int>> dp(N, vector<int>(N));
        for (int m = K; m <= N; ++m) {
            for (int i = 0; i + m <= N; ++i) {
                int j = i + m - 1;
                dp[i][j] = INT_MAX;
                for (int mid = i; mid < j; mid += K - 1) {
                    dp[i][j] = min(dp[i][j], dp[i][mid] + dp[mid + 1][j]);
                }
                if ((j - i) % (K - 1) == 0) dp[i][j] += prefix[j + 1] - prefix[i];
            }
        }
        return dp[0][N - 1];
    }
};

Java

class Solution {
    public int mergeStones(int[] stones, int K) {
        int length = stones.length;
        if (length == 1)
            return 0;
        if ((length - 1) % (K - 1) != 0)
            return -1;
        int[] prefixSum = new int[length];
        prefixSum[0] = stones[0];
        for (int i = 1; i < length; i++)
            prefixSum[i] = prefixSum[i - 1] + stones[i];
        int[][][] dp = new int[length][length][K];
        for (int i = 0; i < length; i++) {
            for (int j = 0; j < length; j++) {
                for (int k = 0; k < K; k++)
                    dp[i][j][k] = Integer.MAX_VALUE;
            }
        }
        for (int i = 0; i < length; i++)
            dp[i][i][0] = 0;
        for (int curLength = 2; curLength <= length; curLength++) {
            int maxStart = length - curLength;
            for (int i = 0; i <= maxStart; i++) {
                int end = i + curLength - 1;
                for (int j = 1; j < K; j++) {
                    for (int k = i; k < end; k += K - 1) {
                        dp[i][end][j] = Math.min(dp[i][end][j], dp[i][k][0] + dp[k + 1][end][j - 1]);
                        dp[i][end][0] = sum(prefixSum, i, end) + dp[i][end][j];
                    }
                }
            }
        }
        return dp[0][length - 1][0];
    }

    public int sum(int[] prefixSum, int start, int end) {
        if (start == 0)
            return prefixSum[end];
        else
            return prefixSum[end] - prefixSum[start - 1];
    }
}

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