# Program to find minimum cost to merge stones in Python

Suppose we have N piles of stones arranged in a row. Here the i-th pile has stones[i] number of stones. A move consists of merging K consecutive piles into one pile, now the cost of this move is equal to the total number of stones in these K number of piles. We have to find the minimum cost to merge all piles of stones into one pile. If there is no such solution then, return -1.

So, if the input is like nums = [3,2,4,1], K = 2, then the output will be 20, because, initially have [3, 2, 4, 1]. Then merge [3, 2] with cost 5, and we have [5, 4, 1]. After that merge [4, 1] with cost 5, and we have [5, 5]. Then merge [5, 5] with cost 10, and we have [10]. So, the total cost was 20, and this is the minimum one.

To solve this, we will follow these steps −

• n := size of nums

• if (n-1) mod (K-1) is not 0, then

• return -1

• dp := one n x n array and fill with 0

• sums := n array of size (n+1) and fill with 0

• for i in range 1 to n, do

• sums[i] := sums[i-1]+nums[i-1]

• for length in range K to n, do

• for i in range 0 to n-length, do

• j := i+length-1

• dp[i, j] := infinity

• for t in range i to j-1, update in each step by K-1, do

• dp[i][j] = minimum of dp[i, j] and (dp[i, t] + dp[t+1, j])

• if (j-i) mod (K-1) is same as 0, then

• dp[i, j] := dp[i, j] + sums[j+1]-sums[i]

• return dp[0, n-1]

## Example

Let us see the following implementation to get better understanding

import heapq
def solve(nums, K):
n = len(nums)
if (n-1)%(K-1) != 0:
return -1
dp = [[0]*n for _ in range(n)]
sums = [0]*(n+1)
for i in range(1,n+1):
sums[i] = sums[i-1]+nums[i-1]
for length in range(K,n+1):
for i in range(n-length+1):
j = i+length-1
dp[i][j] = float('inf')
for t in range(i,j,K-1):
dp[i][j] = min(dp[i][j], dp[i][t]+dp[t+1][j])
if (j-i)%(K-1)==0:
dp[i][j] += sums[j+1]-sums[i]
return dp[0][n-1]

nums = [3,2,4,1]
K = 2
print(solve(nums, K))

## Input

[3,2,4,1], 2


## Output

20

Updated on: 07-Oct-2021

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