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Kotlin-based LeetCode algorithm problem solutions, regularly updated
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`