Package g0901_1000.s1000_minimum_cost_to_merge_stones

Class Solution

  • public class Solution
extends Object
    1000 - Minimum Cost to Merge Stones.

    Hard

    There are n piles of stones arranged in a row. The ith 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.

    Return 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.

    Constraints:

    • n == stones.length
    • 1 <= n <= 30
    • 1 <= stones[i] <= 100
    • 2 <= k <= 30

    • Constructor Detail

      • Solution

        public Solution()

    • Method Detail

      • mergeStones

        public int mergeStones​(int[] stones,
                       int k)