[Dynamic Programming] Minimum Cost to Merge Stones (Simplified) - Medium

The Minimum Cost to Merge Stones problem is about combining stones with the least cost. We need to merge a set of stones into one. The cost to merge is the sum of the stones we combine. We can solve this problem well using dynamic programming. This method helps us break the problem into smaller parts, save their solutions, and build up to the final answer efficiently.

In this article, we will look closely at the Minimum Cost to Merge Stones problem. First, we will explain the problem statement. Then, we will present a clear dynamic programming approach that shows the solution. We will also give examples in Java, Python, and C++ to help us understand better. Additionally, we will talk about ways to improve our solution, check the time and space cost, point out common mistakes, and answer some frequent questions to make the topic clearer.

Dynamic Programming Solution for Minimum Cost to Merge Stones Simplified Medium Problem
Understanding the Problem Statement for Minimum Cost to Merge Stones
Dynamic Programming Approach for Minimum Cost to Merge Stones
Java Implementation of Minimum Cost to Merge Stones
Python Implementation of Minimum Cost to Merge Stones
C++ Implementation of Minimum Cost to Merge Stones
Optimization Techniques for Minimum Cost to Merge Stones
Analyzing Time and Space Complexity for Minimum Cost to Merge Stones
Common Pitfalls in Minimum Cost to Merge Stones
Frequently Asked Questions

If we want to learn more about dynamic programming, we can read articles like Dynamic Programming: Fibonacci Number and Dynamic Programming: Minimum Cost Climbing Stairs. They give good insights and examples.

Understanding the Problem Statement for Minimum Cost to Merge Stones

The “Minimum Cost to Merge Stones” problem is a challenge in dynamic programming. It involves merging stones into one stone for the lowest cost. The cost comes from the sum of the weights of the stones that we merge.

Problem Definition

We get an array stones of length n. Here, stones[i] shows the weight of the i-th stone.
Our goal is to merge all stones into one. We can merge k stones that are next to each other. The cost of this merge is the total weight of those stones.
We can only merge stones in groups of k until one stone is left.
If the total number of stones is not a multiple of k, we cannot merge the leftover stones until they can form a complete group.

Constraints

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

Example

For stones = [3, 2, 4, 1] and k = 2:

First, we merge the first two stones (3 and 2). The cost is 5, and we have [5, 4, 1].
Then, we merge the next two stones (5 and 4). The cost is 9, and now we have [9, 1].
Finally, we merge the last two stones (9 and 1). The cost is 10.

So the total cost is 5 + 9 + 10 = 24.

Input/Output Format

Input: A list of integers stones and an integer k.
Output: The minimum cost to merge all stones into one.

We can solve this problem well with dynamic programming. We need to manage the states carefully and calculate the costs. This way, we can explore all merge combinations while keeping the total cost as low as possible.

For more help with dynamic programming techniques, we can check related articles like Dynamic Programming - Minimum Cost Climbing Stairs.

Dynamic Programming Approach for Minimum Cost to Merge Stones

We can solve the Minimum Cost to Merge Stones problem well using dynamic programming. The main idea is to use a 2D DP array. This array helps us keep track of the minimum cost for merging stones in different ranges. We also use a cumulative sum to help us calculate costs faster.

Problem Definition

We have an array of stones. We can merge two stones that are next to each other. The cost to merge them is the sum of their weights. Our goal is to merge all stones into one stone with the least cost. There are rules about how many stones we can merge at once.

Dynamic Programming Table Setup

DP Definition: We define dp[i][j] as the minimum cost to merge stones from index i to index j.
Cumulative Sum: We create an extra array called prefixSum. This array stores cumulative sums of stones. It helps us calculate costs quickly.

Transition Formula

For every segment of stones from i to j, if we want to merge them at a size k (where k is between 1 and j-i), we can write the cost like this:
```
dp[i][j] = min(dp[i][j], dp[i][m] + dp[m+1][j] + prefixSum[j+1] - prefixSum[i])
```
Here, we look at all possible splits m between i and j.

Base Case

If there is only one stone, then dp[i][i] = 0. This is because we do not have to spend any money.

Implementation

Here is a simple way to implement this in Python:

def mergeStones(stones, K):
    n = len(stones)
    if (n - 1) % (K - 1) != 0: return -1

    prefixSum = [0] * (n + 1)
    for i in range(n):
        prefixSum[i + 1] = prefixSum[i] + stones[i]

    dp = [[float('inf')] * n for _ in range(n)]
    
    for i in range(n):
        dp[i][i] = 0  # Base case

    for length in range(2, n + 1):  # Length of the segment
        for i in range(n - length + 1):
            j = i + length - 1
            for m in range(i, j, K - 1):  # m is the last stone in the group
                dp[i][j] = min(dp[i][j], dp[i][m] + dp[m + 1][j])
            if (length - 1) % (K - 1) == 0:  # If we can merge all
                dp[i][j] += prefixSum[j + 1] - prefixSum[i]

    return dp[0][n - 1]

Complexity Analysis

Time Complexity: O(n^3). This is due to three loops that calculate the DP values.
Space Complexity: O(n^2). This is for the DP table.

This dynamic programming way to solve the Minimum Cost to Merge Stones problem finds the best solution by checking and saving costs for different parts of the input array. For more information about dynamic programming strategies, you can check other articles like Dynamic Programming - Fibonacci Number.

Java Implementation of Minimum Cost to Merge Stones

We can solve the Minimum Cost to Merge Stones problem in Java using a simple dynamic programming method. We need a DP table to track the lowest cost of merging stones in different parts. Here is a clear implementation:

public class MergeStones {
    public int mergeStones(int[] stones, int K) {
        int n = stones.length;
        if (n == 0 || (n - 1) % (K - 1) != 0) return -1;

        int[][] dp = new int[n][n];
        int[] sum = new int[n + 1];

        // Precompute prefix sums
        for (int i = 0; i < n; i++) {
            sum[i + 1] = sum[i] + stones[i];
        }

        // Fill the DP table
        for (int len = K; len <= n; len++) {
            for (int i = 0; i + len <= n; i++) {
                int j = i + len - 1;
                dp[i][j] = Integer.MAX_VALUE;

                // Try merging at different points
                for (int m = 1; m < len; m++) {
                    int left = dp[i][i + m - 1];
                    int right = dp[i + m][j];
                    if (left != Integer.MAX_VALUE && right != Integer.MAX_VALUE) {
                        dp[i][j] = Math.min(dp[i][j], left + right);
                    }
                }

                // If we can merge all stones from i to j
                if ((len - 1) % (K - 1) == 0) {
                    dp[i][j] += sum[j + 1] - sum[i];
                }
            }
        }
        return dp[0][n - 1];
    }

    public static void main(String[] args) {
        MergeStones ms = new MergeStones();
        int[] stones = {3, 2, 4, 1};
        int K = 2;
        System.out.println("Minimum cost to merge stones: " + ms.mergeStones(stones, K));
    }
}

Explanation of the Code:

Input: The method mergeStones takes an array of stones and a number K. K shows how many stones we merge at once.
Prefix Sum Calculation: We calculate prefix sums to quickly find the cost of merging stones.
Dynamic Programming Table: We fill the dp table using values we computed before. We check all possible merges.
Final Cost Calculation: If we can merge all stones from i to j, we add the cost to the DP table.

This Java code gives a good solution for the Minimum Cost to Merge Stones problem. It uses dynamic programming ideas. To learn more, we can check other dynamic programming topics like Climbing Stairs and Minimum Cost Climbing Stairs.

Python Implementation of Minimum Cost to Merge Stones

We can solve the Minimum Cost to Merge Stones problem using Python with a simple approach called dynamic programming. Our goal is to merge stones into one pile while keeping the cost as low as possible. The cost of merging depends on the sum of the stones we are merging.

Code Implementation

def minCostToMergeStones(stones, K):
    if len(stones) == 0:
        return 0
    if (len(stones) - 1) % (K - 1) != 0:
        return -1

    n = len(stones)
    dp = [[[float('inf')] * (K + 1) for _ in range(n)] for _ in range(n)]
    prefix_sum = [0] * (n + 1)

    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + stones[i]

    for i in range(n):
        dp[i][i][1] = 0

    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            for k in range(1, K + 1):
                for m in range(i, j):
                    dp[i][j][k] = min(dp[i][j][k], dp[i][m][k] + dp[m + 1][j][k - 1])
                if k == 1:
                    dp[i][j][1] += prefix_sum[j + 1] - prefix_sum[i]

    return dp[0][n - 1][1]

# Example Usage
stones = [3, 2, 4, 1]
K = 2
result = minCostToMergeStones(stones, K)
print(f"The minimum cost to merge the stones is: {result}")

Explanation of the Code

Input Parameters:
- stones: A list of numbers showing the sizes of stones.
- K: A number showing how many stones we can merge at once.
Dynamic Programming Table:
- dp[i][j][k]: This shows the minimum cost to merge stones from index i to j when k stones are left.
Prefix Sum Array:
- prefix_sum: This helps us to quickly calculate the sum of stones.
Base Case:
- When we merge one stone, the cost is 0.
Filling the DP Table:
- We go through all lengths of the stone groups. We calculate the minimum cost using the results from before.

Complexity Analysis

Time Complexity: O(N^3), where N is the number of stones. This happens because of the loops that go through the lengths of stones, starting points, and split points.
Space Complexity: O(N^3) for the DP table and O(N) for the prefix sum array.

This code calculates the minimum cost to merge stones into one pile using dynamic programming in Python. For more information about dynamic programming, we can read the Dynamic Programming Fibonacci Number article.

C++ Implementation of Minimum Cost to Merge Stones

We can solve the “Minimum Cost to Merge Stones” problem using C++. We will use a method called dynamic programming. The goal is to reduce the cost of merging stones into one pile. The cost of merging k stones is the total of those stones.

Here is the C++ code for this solution:

#include <iostream>
#include <vector>
#include <limits>

using namespace std;

class Solution {
public:
    int mergeStones(vector<int>& stones, int K) {
        int n = stones.size();
        if ((n - 1) % (K - 1) != 0) return -1; // Check if we can merge into one pile

        vector<vector<int>> dp(n, vector<int>(n, 0));
        vector<int> prefixSum(n + 1, 0);

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

        for (int len = 2; len <= n; ++len) {
            for (int i = 0; i + len - 1 < n; ++i) {
                int j = i + len - 1;
                dp[i][j] = numeric_limits<int>::max();
                for (int k = 1; k < len; ++k) {
                    if ((len - k) % (K - 1) == 0) {
                        dp[i][j] = min(dp[i][j], dp[i][i + k - 1] + dp[i + k][j]);
                    }
                }
                if ((len - 1) % (K - 1) == 0) {
                    dp[i][j] += prefixSum[j + 1] - prefixSum[i];
                }
            }
        }
        return dp[0][n - 1];
    }
};

int main() {
    Solution sol;
    vector<int> stones = {3, 2, 4, 1};
    int K = 2;
    cout << "Minimum cost to merge stones: " << sol.mergeStones(stones, K) << endl;
    return 0;
}

Explanation of the Code:

Input Handling: We take a vector of integers. This vector shows the stone piles. The integer K tells us how many piles we can merge at once.
Prefix Sum: We find the prefix sum array. This helps us to calculate the cost of merging stones quickly.
DP Table: We use a 2D DP table called dp[i][j]. This table stores the minimum cost to merge stones from index i to j.
Merging Logic: We go through all possible lengths of the subarray. We calculate the minimum cost for merging using the results we got before.
Final Result: The final answer is in dp[0][n-1]. This shows the minimum cost to merge all stones into one pile.

This C++ code solves the problem well. It uses dynamic programming ideas and works good for the limits on stones and K.

For more about dynamic programming, we can check this article: Dynamic Programming - Minimum Cost Climbing Stairs.

Optimization Techniques for Minimum Cost to Merge Stones

To make the solution for the Minimum Cost to Merge Stones problem better, we can use some simple methods. These methods help us save time and work faster. Here are the main techniques we can use:

Dynamic Programming with Prefix Sums:
- We can use prefix sums. They help us find the sum of stones in any part quickly. This means we do not have to find sums again and again. It makes our work faster.
- We define prefix[i] to be the sum of stones from the start to index i.
```
prefix = [0] * (n + 1)
for i in range(n):
    prefix[i + 1] = prefix[i] + stones[i]
```
Space Optimization:
- Rather than keeping a big DP table, we can use a rolling array. We can also keep only the states we need. This way, we save space.
Iterative DP Approach:
- We can change the recursive DP solution to an iterative one. This helps us avoid stack overflow problems. It also makes the process faster.
Dividing the Problem:
- We can break the problem into smaller parts. By solving these smaller parts alone, we can use the results again. This helps us avoid doing extra work.
Memoization:
- We can use memoization. This means we save the results of our smaller problems. This way, we calculate each unique state only once.
Using Binary Indexed Trees or Segment Trees:
- When we need to find sums quickly, we can use these data structures. They help us keep track of sums and find them easily.

Example Implementation

Here is a simple Python code that shows how to use these techniques:

def minCostToMergeStones(stones, K):
    n = len(stones)
    if (n - 1) % (K - 1) != 0:
        return -1

    prefix = [0] * (n + 1)
    for i in range(n):
        prefix[i + 1] = prefix[i] + stones[i]

    dp = [[float('inf')] * n for _ in range(n)]
    for i in range(n):
        dp[i][i] = 0

    for length in range(2, n + 1):  # length of the subarray
        for i in range(n - length + 1):
            j = i + length - 1
            for mid in range(i, j, K - 1):
                dp[i][j] = min(dp[i][j], dp[i][mid] + dp[mid + 1][j])
            if (length - 1) % (K - 1) == 0:
                dp[i][j] += prefix[j + 1] - prefix[i]

    return dp[0][n - 1]

This code finds the minimum cost to merge stones. It uses dynamic programming, prefix sums, and keeps the states managed well. For more information on similar dynamic programming problems, we can check out the Dynamic Programming techniques for Fibonacci numbers or Dynamic Programming for Minimum Cost Climbing Stairs.

Analyzing Time and Space Complexity for Minimum Cost to Merge Stones

In the Minimum Cost to Merge Stones problem, we want to find the lowest cost to merge all stones into one pile. The cost of merging piles is based on the total number of stones being merged. We can solve this problem well using dynamic programming.

Time Complexity

Let n be the number of stones. The dynamic programming method uses a three-dimensional DP table.
We look at all possible groups of stones. This makes the complexity (O(n^3)).
For each group ((i, j)), we check all places to merge stones. This gives us (O(n^2)) for each group.

So, the total time complexity for the Minimum Cost to Merge Stones is:

[ O(n^3) ]

Space Complexity

The space complexity mostly comes from the DP table. This table keeps the minimum costs and cumulative sums.
If we use a 2D array for the DP results, the space complexity will be (O(n^2)).
We might also need another array for cumulative sums. This adds another (O(n)) space.

Thus, the overall space complexity is:

[ O(n^2) ]

Summary

Time Complexity: (O(n^3))
Space Complexity: (O(n^2))

It is important to understand time and space complexity. This helps us to make the solution better for the Minimum Cost to Merge Stones problem. This is especially true for bigger inputs. If we want to learn more about dynamic programming, we can look at related problems like the Dynamic Programming Minimum Cost Climbing Stairs or Dynamic Programming Minimum Falling Path Sum.

Common Pitfalls in Minimum Cost to Merge Stones

When we work on the Minimum Cost to Merge Stones problem with dynamic programming, we can face some common mistakes. These mistakes can stop us from getting the right answer and make our solution slow. Here are some important things to keep in mind:

Incorrect Base Case Initialization:
We need to make sure that we set the base cases in the dynamic programming table correctly. If we do not set the cost for merging single stones right, we can get wrong results.
Ignoring Edge Cases:
Problems can happen when the number of stones is less than or equal to the merge size k. We should always check if we can merge the stones based on the rules in the problem.
Improper Range Calculation:
When we calculate the cost of merging stones, we must define the ranges of stones we are merging accurately. If we make off-by-one mistakes, we can access indices that are out of bounds.
Inefficient DP Table Updates:
We should update the dynamic programming table in an efficient way. We must avoid recalculating values by using results from states we computed before.
Not Considering All Merge Sizes:
Depending on the problem, we may need to look at different merge sizes. If we do not consider all possible splits, we can miss the best solutions.
Space Complexity Issues:
Dynamic programming can use a lot of space. We should think about using a rolling array or other space-saving methods to lower the space we need while still tracking important states.
Mismanagement of State Representation:
We need to clearly define how we show states in the DP table. The indices must correctly show the start and end of the stones we are thinking about merging.
Failing to Handle Large Input Sizes:
For large input sizes, we must make sure our solution is fast enough, usually around O(n^3) for simple DP methods. We might want to use pruning or memoization to make it faster.
Testing with Various Scenarios:
We should always test our work with many different cases. This includes edge cases like minimum and maximum input values and random scenarios to make sure our solution is strong.

By being careful of these common mistakes, we can better our chances of making an efficient solution for the Minimum Cost to Merge Stones problem. For more tips on dynamic programming, we can check articles like Dynamic Programming - Coin Change and Dynamic Programming - Minimum Path Sum in a Grid.

Frequently Asked Questions

1. What is the Minimum Cost to Merge Stones problem in dynamic programming?

The Minimum Cost to Merge Stones problem is about merging a list of stones into one stone at the lowest cost. The cost comes from adding up the stones in each merge. We can solve this problem well using dynamic programming methods to make the merging easier. For more details, you can look at the Dynamic Programming Solution for Minimum Cost to Merge Stones.

2. How does dynamic programming help in solving the Minimum Cost to Merge Stones problem?

Dynamic programming helps us solve the Minimum Cost to Merge Stones problem by breaking it into smaller problems and saving the results. This way, we do not do the same calculations again and again. By using memoization or tabulation, we can find the minimum cost for merging stones quickly. You can learn more about dynamic programming in the Dynamic Programming Fibonacci Number article.

3. What are the key steps in implementing the Minimum Cost to Merge Stones algorithm in Java?

To implement the Minimum Cost to Merge Stones algorithm in Java, we need to do these steps. First, we define the problem structure. Second, we use a 2D array to keep track of costs. Third, we use nested loops to go through possible merges. Lastly, we apply dynamic programming ideas to fill the cost array using values we already calculated. For a clear Java implementation, check the Java Implementation of Minimum Cost to Merge Stones.

4. Can the Minimum Cost to Merge Stones problem be solved using Python?

Yes, we can solve the Minimum Cost to Merge Stones problem well using Python. The method is to create a recursive function with memoization or to use a dynamic programming table to find the minimum cost for merging stones. This method gives a good solution and stops us from recalculating problems that overlap. For a full Python implementation, see the Python Implementation of Minimum Cost to Merge Stones.

5. What are common pitfalls when solving the Minimum Cost to Merge Stones problem?

Some common mistakes when solving the Minimum Cost to Merge Stones problem include forgetting the base cases in dynamic programming, getting the merging costs wrong, and not considering all possible merge options. It is very important to design the dynamic programming table carefully and make sure it starts correctly to avoid these problems. For more information on dynamic programming issues, visit the Dynamic Programming Maximum Product Subarray article.