Watch 10 video solutions for Minimum Cost to Merge Stones, a hard level problem involving Array, Dynamic Programming, Prefix Sum. This walkthrough by NeetCode has 427,699 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
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.length1 <= n <= 301 <= stones[i] <= 1002 <= k <= 30