Problem Description

You are given n piles of stones arranged in a row, where the i-th pile contains stones[i] stones.

In each move, you must merge exactly k consecutive piles into a single pile. The cost of this merge operation equals the total number of stones in those k piles being merged.

Your goal is to find the minimum total cost to merge all piles into one single pile. If it's impossible to merge all piles into one, return -1.

For example, if you have stones [3, 2, 4, 1] and k = 2:

• You could merge piles at positions 1 and 2 (stones 3 and 2) with cost 5, resulting in [5, 4, 1]
• Then merge positions 1 and 2 (stones 5 and 4) with cost 9, resulting in [9, 1]
• Finally merge the last two piles with cost 10, resulting in [10]
• Total cost would be 5 + 9 + 10 = 24

The key constraint is that you can only merge exactly k consecutive piles at a time - no more, no less. This means that not all configurations can be reduced to a single pile. For instance, if you have 4 piles and k = 3, you can merge 3 piles to get 2 piles, but then you cannot merge 2 piles when k = 3, making it impossible to reach a single pile.

Intuition

Let's think about when it's possible to merge all stones into one pile. Each merge operation reduces the number of piles by K - 1 (we take K piles and produce 1 pile). Starting with n piles and ending with 1 pile, we need to reduce by n - 1 piles total. This means we need (n - 1) / (K - 1) merge operations. For this to be a whole number, (n - 1) % (K - 1) must equal 0. If not, it's impossible to merge all piles into one.

Now for the minimum cost calculation. The key insight is that merging stones is a range-based problem - we're always working with consecutive piles. This suggests using dynamic programming where we consider all possible ways to merge subarrays of stones.

Let's define f[i][j][m] as the minimum cost to merge stones from pile i to pile j into exactly m piles. Why do we need the third dimension m? Because intermediate states matter - sometimes we need to first merge subsections into a specific number of piles before we can perform the final merge.

The recurrence relation works as follows:

• To merge stones [i, j] into m piles, we can split the range at some point h and merge [i, h] into 1 pile and [h+1, j] into m-1 piles
• To merge stones [i, j] into 1 pile, we first need to merge them into K piles (which costs f[i][j][K]), then perform one final merge of those K piles (which costs the sum of all stones in range [i, j])

The base case is f[i][i][1] = 0 (a single pile is already 1 pile with no cost).

We use prefix sums to quickly calculate the total stones in any range [i, j], which represents the cost of the final merge operation when combining K piles into 1.

The answer will be f[1][n][1] - the minimum cost to merge all stones from position 1 to n into exactly 1 pile.

Learn more about Dynamic Programming and Prefix Sum patterns.

Solution Approach

The implementation follows the dynamic programming approach with these key components:

1. Impossibility Check:

if (n - 1) % (K - 1):
    return -1

First, we verify if merging is possible. If (n - 1) is not divisible by (K - 1), we cannot merge all piles into one.

2. Prefix Sum Array:

s = list(accumulate(stones, initial=0))

We create a prefix sum array to quickly calculate the sum of stones in any range [i, j] using s[j] - s[i - 1]. The initial=0 ensures our prefix sum is 1-indexed for easier calculation.

3. DP Table Initialization:

f = [[[inf] * (K + 1) for _ in range(n + 1)] for _ in range(n + 1)]
for i in range(1, n + 1):
    f[i][i][1] = 0

We create a 3D DP table f[i][j][k] where:

• i is the starting position (1-indexed)
• j is the ending position (1-indexed)
• k is the number of piles to merge into

Initialize all values to infinity except f[i][i][1] = 0 (single pile requires no merging).

4. Main DP Loop:

for l in range(2, n + 1):  # length of subarray
    for i in range(1, n - l + 2):  # starting position
        j = i + l - 1  # ending position

We iterate through all possible subarray lengths from 2 to n, and for each length, we consider all valid starting positions.

5. Calculate Minimum Cost for k Piles:

for k in range(1, K + 1):
    for h in range(i, j):
        f[i][j][k] = min(f[i][j][k], f[i][h][1] + f[h + 1][j][k - 1])

For merging range [i, j] into k piles, we try all possible split points h. We merge [i, h] into 1 pile and [h+1, j] into k-1 piles, taking the minimum cost among all splits.

6. Final Merge to One Pile:

f[i][j][1] = f[i][j][K] + s[j] - s[i - 1]

To merge range [i, j] into 1 pile, we first merge it into K piles (cost: f[i][j][K]), then perform the final merge of those K piles into 1. The cost of this final merge equals the total stones in the range, calculated as s[j] - s[i - 1].

7. Return Result:

return f[1][n][1]

The answer is the minimum cost to merge all stones (from position 1 to n) into exactly 1 pile.

The time complexity is O(n³ * K) and space complexity is O(n² * K), where n is the number of stone piles.

Example Walkthrough

Let's walk through a concrete example with stones = [3, 2, 4, 1] and K = 2.

Step 1: Check if merging is possible

• We have n = 4 piles
• Check: (n - 1) % (K - 1) = (4 - 1) % (2 - 1) = 3 % 1 = 0 ✓
• Since the remainder is 0, we can merge all piles into one

Step 2: Build prefix sum array

• s = [0, 3, 5, 9, 10]
• This helps us quickly calculate range sums. For example, sum of stones[2:4] = s[4] - s[1] = 10 - 3 = 7

Step 3: Initialize DP table

• Create 3D array f[i][j][m] initialized to infinity
• Set base cases: f[1][1][1] = 0, f[2][2][1] = 0, f[3][3][1] = 0, f[4][4][1] = 0
• (Each single pile is already 1 pile with zero cost)

Step 4: Fill DP table for increasing lengths

Length 2 subarrays:

• For [3, 2] (i=1, j=2):
  • To get 2 piles: f[1][2][2] = f[1][1][1] + f[2][2][1] = 0 + 0 = 0
  • To get 1 pile: f[1][2][1] = f[1][2][2] + (s[2] - s[0]) = 0 + 5 = 5
• For [2, 4] (i=2, j=3):
  • To get 2 piles: f[2][3][2] = 0
  • To get 1 pile: f[2][3][1] = 0 + 6 = 6
• For [4, 1] (i=3, j=4):
  • To get 2 piles: f[3][4][2] = 0
  • To get 1 pile: f[3][4][1] = 0 + 5 = 5

Length 3 subarrays:

• For [3, 2, 4] (i=1, j=3):
  • To get 2 piles (split at h=1): f[1][3][2] = f[1][1][1] + f[2][3][1] = 0 + 6 = 6
  • To get 2 piles (split at h=2): f[1][3][2] = min(6, f[1][2][1] + f[3][3][1]) = min(6, 5 + 0) = 5
  • To get 1 pile: f[1][3][1] = f[1][3][2] + (s[3] - s[0]) = 5 + 9 = 14
• For [2, 4, 1] (i=2, j=4):
  • To get 2 piles: f[2][4][2] = min(f[2][2][1] + f[3][4][1], f[2][3][1] + f[4][4][1]) = min(0 + 5, 6 + 0) = 5
  • To get 1 pile: f[2][4][1] = 5 + 7 = 12

Length 4 (entire array):

• For [3, 2, 4, 1] (i=1, j=4):
  • To get 2 piles:
    • Split at h=1: f[1][1][1] + f[2][4][1] = 0 + 12 = 12
    • Split at h=2: f[1][2][1] + f[3][4][1] = 5 + 5 = 10
    • Split at h=3: f[1][3][1] + f[4][4][1] = 14 + 0 = 14
    • Minimum: f[1][4][2] = 10
  • To get 1 pile: f[1][4][1] = f[1][4][2] + (s[4] - s[0]) = 10 + 10 = 20

Step 5: Return result The minimum cost to merge all stones into one pile is f[1][4][1] = 20.

This represents the optimal merging sequence:

1. Merge [3, 2] → [5] with cost 5
2. Merge [4, 1] → [5] with cost 5
3. Merge [5, 5] → [10] with cost 10 Total cost: 5 + 5 + 10 = 20

Solution Implementation

Time and Space Complexity

Time Complexity: O(n³ * K)

The algorithm uses dynamic programming with four nested loops:

• The outermost loop iterates through lengths from 2 to n: O(n) iterations
• The second loop iterates through starting positions i: O(n) iterations
• The third loop iterates through k from 1 to K: O(K) iterations
• The innermost loop iterates through split positions h from i to j-1: O(n) iterations in the worst case

This gives us a total time complexity of O(n * n * K * n) = O(n³ * K).

Additionally, the prefix sum calculation takes O(n) time, but this doesn't affect the overall complexity.

Space Complexity: O(n² * K)

The algorithm uses a 3D DP array f with dimensions (n+1) × (n+1) × (K+1), which requires O(n² * K) space.

The prefix sum array s uses O(n) additional space, but this is dominated by the DP array.

Therefore, the total space complexity is O(n² * K).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Split Point Iteration

Pitfall: When trying to merge a range [start, end] into num_piles piles, developers often iterate through all possible split points from start to end-1. However, this leads to redundant calculations and incorrect results because not all split points are valid when merging into exactly K piles at a time.

Why it's wrong:

# INCORRECT approach:
for split in range(start, end):  # Checking every possible split
    dp[start][end][num_piles] = min(
        dp[start][end][num_piles],
        dp[start][split][1] + dp[split + 1][end][num_piles - 1]
    )

This checks unnecessary splits that don't align with the constraint of merging exactly K piles at a time.

Correct approach:

# CORRECT approach:
for split in range(start, end, K - 1):  # Step by K-1
    dp[start][end][num_piles] = min(
        dp[start][end][num_piles],
        dp[start][split][1] + dp[split + 1][end][num_piles - 1]
    )

The split points should increment by K-1 because when we merge K piles into 1, we reduce the total count by K-1. This ensures we only consider valid merging patterns.

2. Misunderstanding the Impossibility Condition

Pitfall: Developers might check n % K == 0 or other incorrect conditions to determine if merging is possible.

Why it's wrong:

# INCORRECT checks:
if n % K != 0:  # Wrong!
    return -1

if n % K == 1:  # Also wrong!
    return -1

Correct approach:

# CORRECT check:
if (n - 1) % (K - 1) != 0:
    return -1

Explanation: Each merge operation takes K piles and produces 1 pile, reducing the total by K-1. To go from n piles to 1 pile, we need exactly (n-1)/(K-1) merge operations. This is only possible if (n-1) is divisible by (K-1).

3. Off-by-One Errors in Prefix Sum Calculation

Pitfall: Using 0-indexed prefix sums with 1-indexed DP table leads to index confusion.

Why it's wrong:

# INCORRECT: Mixing 0-indexed and 1-indexed
prefix_sum = list(accumulate(stones))  # 0-indexed
# Later in code:
cost = prefix_sum[end] - prefix_sum[start - 1]  # Index error when start = 1

Correct approach:

# CORRECT: Consistent 1-indexed approach
prefix_sum = [0]
for stone in stones:
    prefix_sum.append(prefix_sum[-1] + stone)
# Or using accumulate:
prefix_sum = list(accumulate(stones, initial=0))

# Now safe to use:
cost = prefix_sum[end] - prefix_sum[start - 1]

4. Forgetting the Final Merge Cost

Pitfall: Only calculating the cost to arrange stones into K piles without adding the final merge cost.

Why it's wrong:

# INCORRECT: Missing the final merge cost
dp[start][end][1] = dp[start][end][K]  # Forgot to add merge cost!

Correct approach:

# CORRECT: Include the cost of the final merge
dp[start][end][1] = dp[start][end][K] + prefix_sum[end] - prefix_sum[start - 1]

The cost to merge K piles into 1 includes both:

1. The cost to arrange into K piles (dp[start][end][K])
2. The cost of merging those K piles (sum of all stones in the range)