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Formatted question description: https://leetcode.ca/all/1000.html

# 1000. Minimum Cost to Merge Stones (Hard)

There are N piles of stones arranged in a row.  The i-th 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.

Find 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 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 .
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 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 .
The total cost was 25, and this is the minimum possible.


Note:

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

Related Topics:
Dynamic Programming

Similar Questions:

## Solution 1. DP

// OJ: https://leetcode.com/problems/minimum-cost-to-merge-stones/
// Time: O(N^3 / K)
// Space: O(N^2)
// Ref: https://leetcode.com/problems/minimum-cost-to-merge-stones/discuss/247567/JavaC%2B%2BPython-DP
class Solution {
public:
int mergeStones(vector<int>& stones, int K) {
int N = stones.size();
if ((N - 1) % (K - 1)) return -1;
vector<int> prefix(N + 1);
partial_sum(stones.begin(), stones.end(), prefix.begin() + 1);
vector<vector<int>> dp(N, vector<int>(N));
for (int m = K; m <= N; ++m) {
for (int i = 0; i + m <= N; ++i) {
int j = i + m - 1;
dp[i][j] = INT_MAX;
for (int mid = i; mid < j; mid += K - 1) {
dp[i][j] = min(dp[i][j], dp[i][mid] + dp[mid + 1][j]);
}
if ((j - i) % (K - 1) == 0) dp[i][j] += prefix[j + 1] - prefix[i];
}
}
return dp[N - 1];
}
};

• class Solution {
public int mergeStones(int[] stones, int K) {
int length = stones.length;
if (length == 1)
return 0;
if ((length - 1) % (K - 1) != 0)
return -1;
int[] prefixSum = new int[length];
prefixSum = stones;
for (int i = 1; i < length; i++)
prefixSum[i] = prefixSum[i - 1] + stones[i];
int[][][] dp = new int[length][length][K];
for (int i = 0; i < length; i++) {
for (int j = 0; j < length; j++) {
for (int k = 0; k < K; k++)
dp[i][j][k] = Integer.MAX_VALUE;
}
}
for (int i = 0; i < length; i++)
dp[i][i] = 0;
for (int curLength = 2; curLength <= length; curLength++) {
int maxStart = length - curLength;
for (int i = 0; i <= maxStart; i++) {
int end = i + curLength - 1;
for (int j = 1; j < K; j++) {
for (int k = i; k < end; k += K - 1) {
dp[i][end][j] = Math.min(dp[i][end][j], dp[i][k] + dp[k + 1][end][j - 1]);
dp[i][end] = sum(prefixSum, i, end) + dp[i][end][j];
}
}
}
}
return dp[length - 1];
}

public int sum(int[] prefixSum, int start, int end) {
if (start == 0)
return prefixSum[end];
else
return prefixSum[end] - prefixSum[start - 1];
}
}

############

class Solution {
public int mergeStones(int[] stones, int K) {
int n = stones.length;
if ((n - 1) % (K - 1) != 0) {
return -1;
}
int[] s = new int[n + 1];
for (int i = 1; i <= n; ++i) {
s[i] = s[i - 1] + stones[i - 1];
}
int[][][] f = new int[n + 1][n + 1][K + 1];
final int inf = 1 << 20;
for (int[][] g : f) {
for(int[] e : g) {
Arrays.fill(e, inf);
}
}
for (int i = 1; i <= n; ++i) {
f[i][i] = 0;
}
for (int l = 2; l <= n; ++l) {
for (int i = 1; i + l - 1 <= n; ++i) {
int j = i + l - 1;
for (int k = 1; k <= K; ++k) {
for (int h = i; h < j; ++h) {
f[i][j][k] = Math.min(f[i][j][k], f[i][h] + f[h + 1][j][k - 1]);
}
}
f[i][j] = f[i][j][K] + s[j] - s[i - 1];
}
}
return f[n];
}
}

• // OJ: https://leetcode.com/problems/minimum-cost-to-merge-stones/
// Time: O(N^3 / K)
// Space: O(N^2)
// Ref: https://leetcode.com/problems/minimum-cost-to-merge-stones/discuss/247567/JavaC%2B%2BPython-DP
class Solution {
public:
int mergeStones(vector<int>& stones, int K) {
int N = stones.size();
if ((N - 1) % (K - 1)) return -1;
vector<int> prefix(N + 1);
partial_sum(stones.begin(), stones.end(), prefix.begin() + 1);
vector<vector<int>> dp(N, vector<int>(N));
for (int m = K; m <= N; ++m) {
for (int i = 0; i + m <= N; ++i) {
int j = i + m - 1;
dp[i][j] = INT_MAX;
for (int mid = i; mid < j; mid += K - 1) {
dp[i][j] = min(dp[i][j], dp[i][mid] + dp[mid + 1][j]);
}
if ((j - i) % (K - 1) == 0) dp[i][j] += prefix[j + 1] - prefix[i];
}
}
return dp[N - 1];
}
};

• # 1000. Minimum Cost to Merge Stones
# https://leetcode.com/problems/minimum-cost-to-merge-stones/

class Solution:
def mergeStones(self, stones: List[int], k: int) -> int:
n = len(stones)
if (n - 1) % (k - 1): return -1
prefix =  + list(accumulate(stones))

@cache
def go(i, j):
if (j - i + 1) < k: return 0

res = min(go(i, mid) + go(mid + 1, j) for mid in range(i, j, k - 1))

if (j - i) % (k - 1) == 0:
res += prefix[j + 1] - prefix[i]

return res

return go(0, n - 1)


• func mergeStones(stones []int, K int) int {
n := len(stones)
if (n-1)%(K-1) != 0 {
return -1
}
s := make([]int, n+1)
for i, x := range stones {
s[i+1] = s[i] + x
}
f := make([][][]int, n+1)
for i := range f {
f[i] = make([][]int, n+1)
for j := range f[i] {
f[i][j] = make([]int, K+1)
for k := range f[i][j] {
f[i][j][k] = 1 << 20
}
}
}
for i := 1; i <= n; i++ {
f[i][i] = 0
}
for l := 2; l <= n; l++ {
for i := 1; i <= n-l+1; i++ {
j := i + l - 1
for k := 2; k <= K; k++ {
for h := i; h < j; h++ {
f[i][j][k] = min(f[i][j][k], f[i][h][k-1]+f[h+1][j])
}
}
f[i][j] = f[i][j][K] + s[j] - s[i-1]
}
}
return f[n]
}

func min(a, b int) int {
if a < b {
return a
}
return b
}