1000. Minimum Cost to Merge Stones

Description

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

 

Solutions

Solution: Dynamic Programming

• Time complexity: O(n³/k)
• Space complexity: O(n²)

 

JavaScript

/**
 * @param {number[]} stones
 * @param {number} k
 * @return {number}
 */
const mergeStones = function (stones, k) {
  const n = stones.length;

  if ((n - 1) % (k - 1)) return -1;

  const prefixSum = new Array(n + 1).fill(0);

  for (let index = 0; index < n; index++) {
    prefixSum[index + 1] = prefixSum[index] + stones[index];
  }

  const dp = new Array(n)
    .fill('')
    .map(_ => new Array(n).fill(Number.MAX_SAFE_INTEGER));

  for (let index = 0; index < n; index++) {
    dp[index][index] = 0;
  }

  for (let length = 2; length <= n; length++) {
    for (let left = 0; left + length <= n; left++) {
      const right = left + length - 1;

      for (let index = left; index < right; index += k - 1) {
        dp[left][right] = Math.min(dp[left][index] + dp[index + 1][right], dp[left][right]);
      }

      if ((length - 1) % (k - 1) === 0) {
        dp[left][right] += prefixSum[right + 1] - prefixSum[left];
      }
    }
  }
  return dp[0][n - 1];
};