LeetCode 1000. Minimum Cost to Merge Stones

原题链接在这里：https://leetcode.com/problems/minimum-cost-to-merge-stones/

题目：

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

题解：

Each merge step, piles number decreased by K-1. Eventually there is only 1 pile. n - mergeTimes * (K-1) == 1. megeTimes = (n-1)/(K-1). If it is not divisable, then it could not merge into one pile, thus return -1.

Let dp[i][j] denotes minimum cost to merge [i, j] inclusively.

m = i, i+1, ... j-1. Let i to m be one pile, and m+1 to j to certain piles. dp[i][j] = min(dp[i][m] + dp[m+1][j]).

In order to make i to m as one pile, [i,m] inclusive length is multiple of K. m moves K-1 each step.

If [i, j] is multiple of K, then dp[i][j] could be merged into one pile. dp[i][j] += preSum[j+1] - preSum[i].

return dp[0][n-1], minimum cost to merge [0, n-1] inclusively.

Time Complexity: O(n^3/K).

Space: O(n^2).

AC Java:

 class Solution {
     public int mergeStones(int[] stones, int K) {
         int n = stones.length;
         if((n-1)%(K-1) != 0){
             return -1;
         }

         int [] preSum = new int[n+1];
         for(int i = 1; i<=n; i++){
             preSum[i] = preSum[i-1] + stones[i-1];
         }

         int [][] dp = new int[n][n];
         for(int size = 2; size<=n; size++){
             for(int i = 0; i<=n-size; i++){
                 int j = i+size-1;
                 dp[i][j] = Integer.MAX_VALUE;

                 for(int m = i; m<j; m += K-1){
                     dp[i][j] = Math.min(dp[i][j], dp[i][m]+dp[m+1][j]);
                 }

                 if((size-1) % (K-1) == 0){
                     dp[i][j] += preSum[j+1] - preSum[i];
                 }
             }
         }

         return dp[0][n-1];
     }
 }

类似Burst Balloons.