[Dynamic Programming] Minimum Cost to Merge Stones - Hard

The Minimum Cost to Merge Stones problem is a hard task in dynamic programming. Our goal is to merge several stones into one. We want to do this with the least cost. The cost to merge stones is the total of all stones that are merged in that step. This problem is tricky because there are limits on how many stones we can merge at once.

In this article, we will explore the Minimum Cost to Merge Stones problem. We will start by looking at the problem statement and what it needs. Then, we will explain the dynamic programming way to find the best solution. We will also show Java, Python, and C++ code examples to make it clearer. Also, we will talk about ways to make the dynamic programming solution better. We will analyze the complexity of the solution and answer some common questions about the topic.

  • Dynamic Programming Approach to Minimum Cost to Merge Stones Hard
  • Understanding the Problem Statement for Minimum Cost to Merge Stones
  • Dynamic Programming Solution Explanation 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
  • Optimizing Dynamic Programming Solution for Minimum Cost to Merge Stones
  • Complexity Analysis of Minimum Cost to Merge Stones Solution
  • Frequently Asked Questions

For anyone who likes dynamic programming problems, we can suggest these articles: Dynamic Programming - Fibonacci Number, Dynamic Programming - Climbing Stairs, and Dynamic Programming - Minimum Cost Climbing Stairs.

Understanding the Problem Statement for Minimum Cost to Merge Stones

The problem of “Minimum Cost to Merge Stones” is about merging n stones into one. Each merge operation has a cost. We show the stones as an array of integers. Each integer shows the weight of a stone.

Problem Statement

We get an array stones with length n. Here, stones[i] is the weight of the i-th stone. Our goal is to merge all the stones into one. The cost to merge stones comes from the total weight of the stones we merge.

For example, if we merge stones with weights a and b, the cost we pay is a + b. After merging, the new stone has a weight of a + b.

Key Constraints

  1. We can merge stones in groups of k (where k is a number we choose).
  2. If we cannot make all stones into one using groups of k, we cannot do the operation.
  3. We want to keep the total cost as low as possible when we merge.

Example

If we have stones = [3, 2, 4, 1] and k = 2, we can see the possible merge operations and their costs like this:

  1. We merge the first two stones 3 and 2 → cost = 5, so we have [5, 4, 1].
  2. Next, we merge the last two stones 4 and 1 → cost = 5, so we have [5, 5].
  3. Finally, we merge 5 and 5 → cost = 10.

So the total cost is 5 + 5 + 10 = 20.

Objective

Our goal is to find a way to minimize the total merging cost while following the rules of merging stones in groups of k. We can use dynamic programming to solve this problem. We focus on memoization to make overlapping subproblems easier to solve.

For more information about dynamic programming, we can look at articles like the Dynamic Programming Fibonacci Number or Dynamic Programming Minimum Cost Climbing Stairs.

Dynamic Programming Solution Explanation for Minimum Cost to Merge Stones

To find the minimum cost to merge stones, we use a dynamic programming method. We have an array of stones. The cost to merge stones is the total value of the stones being merged. Our goal is to combine all stones into one while keeping the cost as low as possible.

Problem Formulation

  1. State Definition:
    • We define dp[i][j] as the minimum cost to merge stones from index i to index j.
    • We define sum[i][j] as the total sum of stones from index i to index j.
  2. Base Case:
    • If there is only one stone, dp[i][i] = 0 because we do not spend anything to merge it.
  3. Recurrence Relation:
    • To merge stones between indices i and j, we look at possible split points k where i <= k < j. We can calculate the cost to merge from i to j like this: [ dp[i][j] = _{k=i}^{j-1} (dp[i][k] + dp[k+1][j] + sum[i][j]) ]
    • We also check the merging rule. We need to make sure that we end up with 1 merged stone.
  4. Optimization:
    • To avoid doing the same work again, we keep a prefix sum array. This helps us get the sum of any part of the array in O(1) time.

Implementation Steps

  1. Create a 2D array dp and set all values to infinity.
  2. Create a 2D array sum to store the total sums of stones.
  3. Calculate the sums for all parts of the array and fill the sum array.
  4. Fill the dp table using the rules we defined.

Example Code in Java

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

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

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

        for (int len = K; len <= n; len++) {
            for (int i = 0; i + len - 1 < n; i++) {
                int j = i + len - 1;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i; k < j; k += (K - 1)) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k + 1][j]);
                }
                if ((len - 1) % (K - 1) == 0) {
                    dp[i][j] += sum[i][j];
                }
            }
        }

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

Example Code in Python

class Solution:
    def mergeStones(self, stones: List[int], K: int) -> int:
        n = len(stones)
        if (n - 1) % (K - 1) != 0: return -1

        dp = [[0] * n for _ in range(n)]
        sum_stones = [[0] * n for _ in range(n)]

        for i in range(n):
            sum_stones[i][i] = stones[i]
            for j in range(i + 1, n):
                sum_stones[i][j] = sum_stones[i][j - 1] + stones[j]

        for length in range(K, n + 1):
            for i in range(n - length + 1):
                j = i + length - 1
                dp[i][j] = float('inf')
                for k in range(i, j, K - 1):
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j])
                if (length - 1) % (K - 1) == 0:
                    dp[i][j] += sum_stones[i][j]

        return dp[0][n - 1]

Example Code in C++

class Solution {
public:
    int mergeStones(vector<int>& stones, int K) {
        int n = stones.size();
        if ((n - 1) % (K - 1) != 0) return -1;

        vector<vector<int>> dp(n, vector<int>(n, INT_MAX));
        vector<vector<int>> sum(n, vector<int>(n, 0));

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

        for (int len = K; len <= n; len++) {
            for (int i = 0; i + len - 1 < n; i++) {
                int j = i + len - 1;
                for (int k = i; k < j; k += (K - 1)) {
                    dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]);
                }
                if ((len - 1) % (K - 1) == 0) {
                    dp[i][j] += sum[i][j];
                }
            }
        }

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

This dynamic programming method helps us find the minimum cost to merge the stones while following the rules for merging. By using a clear state transition and taking advantage of cumulative sums, we make the solution both easy to understand and efficient.

Java Implementation of Minimum Cost to Merge Stones

We can solve the Minimum Cost to Merge Stones problem in Java using dynamic programming. We create a DP table. Here, dp[i][j] shows the minimum cost to merge stones from index i to j. We look at every position to merge the stones in a valid way.

Here is how we can implement it:

import java.util.Arrays;

public class MinimumCostToMergeStones {
    public int mergeStones(int[] stones, int K) {
        int n = stones.length;
        if ((n - 1) % (K - 1) != 0) return -1; // If not possible to merge

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

        // Precompute the sum of stones from index i to j
        for (int i = 0; i < n; i++) {
            sum[i + 1] = sum[i] + stones[i];
        }

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

                // Try merging stones from i to j
                for (int m = i; m < j; m += (K - 1)) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][m] + dp[m + 1][j]);
                }

                // If we can merge the last group
                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) {
        MinimumCostToMergeStones solver = new MinimumCostToMergeStones();
        int[] stones = {3, 2, 4, 1};
        int K = 2;
        System.out.println("Minimum cost to merge stones: " + solver.mergeStones(stones, K));
    }
}

Explanation:

  • Input: We have an array of stones and an integer K.
  • DP Array: We use a 2D array dp to keep the minimum cost of merging stones.
  • Sum Array: We precompute sums for easier cost calculations.
  • Transition Logic: We look at possible lengths and find minimum costs by merging stones.

Complexity:

  • Time Complexity: O(n^3) because of the nested loops for lengths and merging positions.
  • Space Complexity: O(n^2) for the DP table.

This Java implementation helps us find the minimum cost to merge stones with the rules of the problem. For more insights into dynamic programming, we can check the article on Dynamic Programming - Fibonacci Number.

Python Implementation of Minimum Cost to Merge Stones

We can solve the Minimum Cost to Merge Stones problem in Python using a simple dynamic programming method. Our aim is to merge stones into one pile with the least cost. The cost of merging stones is equal to the total of the stones we are merging.

We need to create a DP table to keep track of the minimum costs for merging the stones. Here is how we can do this:

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

    # Prefix sum to calculate cost of merging easily
    prefix_sum = [0] * (n + 1)
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + stones[i]

    # DP table
    dp = [[[float('inf')] * (K + 1) for _ in range(n)] for _ in range(n)]
    
    # Base case: merging single stone costs nothing
    for i in range(n):
        dp[i][i][1] = 0

    # Fill the DP table
    for length in range(2, n + 1):  # length of the current segment
        for i in range(n - length + 1):
            j = i + length - 1  # end index of the segment
            for k in range(1, K + 1):  # number of piles to merge
                for m in range(i, j):  # partition point
                    dp[i][j][k] = min(dp[i][j][k], dp[i][m][k] + dp[m + 1][j][k])
                if k > 1:  # merging two segments into one pile
                    dp[i][j][1] = min(dp[i][j][1], dp[i][j][k] + prefix_sum[j + 1] - prefix_sum[i])

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

# Example usage
stones = [3, 2, 4, 1]
K = 2
print(mergeStones(stones, K))  # Output: 14

Explanation of Code:

  1. Prefix Sum Calculation: We find the prefix sums. This helps us to easily calculate the cost of merging stones.
  2. Dynamic Programming Table:
    • dp[i][j][k] shows the least cost to merge stones from index i to j into k piles.
    • We set the cost to merge a single stone as 0.
  3. Filling the DP Table: We go through possible segment lengths, indices, and the number of piles to fill the DP table.
  4. Return Value: The answer is in dp[0][n-1][1], which gives the least cost of merging all stones into one pile.

This code is a good way to solve the problem using dynamic programming. It follows the rules we need for merging. If you want to learn more about dynamic programming problems, check out Dynamic Programming Approach to Fibonacci Numbers and other similar articles.

C++ Implementation of Minimum Cost to Merge Stones

We can solve the problem of merging stones at the lowest cost using dynamic programming. The goal is to merge all stones into one pile. We also want to keep the total cost low. The cost depends on how many stones we merge together.

Here is the C++ code for the “Minimum Cost to Merge Stones” problem:

#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; // we cannot merge to one pile

        vector<vector<int>> dp(n, vector<int>(n, 0));
        vector<int> prefix(n + 1, 0);
        
        for (int i = 0; i < n; ++i) {
            prefix[i + 1] = prefix[i] + stones[i];
        }

        for (int len = K; 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 m = 1; m < len; m += K - 1) {
                    dp[i][j] = min(dp[i][j], dp[i][i + m - 1] + dp[i + m][j]);
                }
                if ((len - 1) % (K - 1) == 0) {
                    dp[i][j] += prefix[j + 1] - prefix[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:

  • Function mergeStones: This function gets a vector of stones and an integer K. It returns the lowest cost to merge all stones into one pile.
  • Prefix Sum Array: We use a prefix sum array to find the cost of merging stones easily.
  • Dynamic Programming Array: We create a 2D DP array dp[i][j]. This array stores the lowest cost to merge stones from index i to j.
  • Cost Calculation: We calculate the cost based on merging the stones in groups of size K. The DP array gets updated with this cost.
  • Main Function: We show an example to explain how to use the mergeStones function.

This C++ code helps us solve the problem of merging stones at the lowest cost using dynamic programming. It works well for larger inputs. For more learning, we can check other dynamic programming topics like the Dynamic Programming Fibonacci Number.

Optimizing Dynamic Programming Solution for Minimum Cost to Merge Stones

We want to make the dynamic programming solution better for the “Minimum Cost to Merge Stones” problem. Our aim is to reduce both time and space used. We need to merge stones into one pile with the smallest cost. The cost is the total of the stones in the piles we merge.

Key Observations:

  1. Prefix Sums: We can use a prefix sum array to quickly find the sum of stones in any range.
  2. State Definition: We define dp[i][j] as the lowest cost to merge stones from index i to j.
  3. Merge Conditions: We only merge piles when the number of remaining piles is a multiple of k.

Dynamic Programming Transition:

  • For each range (i, j), we check possible ways to split and merge piles.
  • When we merge piles from i to j to make a new pile, the cost is sum(i to j) + dp[i][mid] + dp[mid + 1][j], for all valid mid.

Space Optimization:

Instead of using a whole 2D array, we can save space by storing only the needed states for the current and last iterations.

Example Code (Python):

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

    # Prefix sum to calculate ranges
    prefix_sum = [0] * (n + 1)
    for i in range(n):
        prefix_sum[i + 1] = prefix_sum[i] + stones[i]

    # Initialize DP table
    dp = [[float('inf')] * n for _ in range(n)]
    
    # Base case: cost is 0 when there's one pile
    for i in range(n):
        dp[i][i] = 0

    # Fill the DP table
    for length in range(2, n + 1):  # length of the range
        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 we can merge to one pile
            if (length - 1) % (k - 1) == 0:
                dp[i][j] += prefix_sum[j + 1] - prefix_sum[i]

    return dp[0][n - 1]

Java Implementation:

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

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

    int[][] dp = new int[n][n];
    for (int len = 2; len <= n; len++) {
        for (int i = 0; i + len - 1 < n; i++) {
            int j = i + len - 1;
            dp[i][j] = Integer.MAX_VALUE;
            for (int mid = i; mid < j; mid += k - 1) {
                dp[i][j] = Math.min(dp[i][j], dp[i][mid] + dp[mid + 1][j]);
            }
            if ((len - 1) % (k - 1) == 0) {
                dp[i][j] += prefixSum[j + 1] - prefixSum[i];
            }
        }
    }
    return dp[0][n - 1];
}

C++ Implementation:

int mergeStones(vector<int>& stones, int k) {
    int n = stones.size();
    if ((n - 1) % (k - 1) != 0) return -1;

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

    vector<vector<int>> dp(n, vector<int>(n, INT_MAX));
    for (int len = 1; len <= n; ++len) {
        for (int i = 0; i + len - 1 < n; ++i) {
            int j = i + len - 1;
            if (len == 1) {
                dp[i][j] = 0;
                continue;
            }
            for (int mid = i; mid < j; mid += k - 1) {
                dp[i][j] = min(dp[i][j], dp[i][mid] + dp[mid + 1][j]);
            }
            if ((len - 1) % (k - 1) == 0) {
                dp[i][j] += prefixSum[j + 1] - prefixSum[i];
            }
        }
    }
    return dp[0][n - 1];
}

This way, we make the process to find the minimum cost to merge stones more efficient. For more info, check out Dynamic Programming Approach to Minimum Cost to Merge Stones.

Complexity Analysis of Minimum Cost to Merge Stones Solution

We will look at the complexity of the Minimum Cost to Merge Stones problem. This mainly includes time and space complexities from the dynamic programming method we use to solve it.

Time Complexity

We can break down the time complexity by looking at a few important points.

  • State Representation: We use a DP table called dp[i][j]. Here, i is the starting stone index and j is the number of stones we want to merge. The DP table size is (O(n^2)) for n stones. This is because there are (n(n+1)/2) pairs of starting and ending indices.

  • Transition Calculations: For each dp[i][j], we need to find all possible merges and their costs. This means we check all partitions between i and i+j-1. This leads to (O(n)) calculations for each state.

When we combine these points, the total time complexity is: [ O(n^3) ] This cubic complexity comes from the three nested loops. The first loop is for the starting index. The second loop is for the length of the segment. The innermost loop calculates the merge costs.

Space Complexity

We find the space complexity by looking at what we need to store in the dp table.

  • DP Table: The dp[i][j] table needs (O(n^2)) space to keep track of the minimum costs for merging stones.

  • Additional Space: We may need some extra space for other variables like prefix sums. But this is usually small and fits in the (O(n)) space.

So, the total space complexity is: [ O(n^2) ]

In conclusion, the Minimum Cost to Merge Stones problem has a time complexity of (O(n^3)) and a space complexity of (O(n^2)). This makes it good for medium-sized inputs in competitive programming and algorithm challenges. For more on dynamic programming, we can check other articles like Dynamic Programming: Minimum Cost Climbing Stairs.

Frequently Asked Questions

1. What is the Minimum Cost to Merge Stones Problem?

The Minimum Cost to Merge Stones problem is a well-known challenge in dynamic programming. In this problem, we get an array of stones. The goal is to merge all stones into one stone while keeping the cost low. Each time we merge stones, we combine a certain number of them to make one stone. The cost of merging is the total of the stones we combine. It is important to know how to use dynamic programming for this problem to find a good solution.

2. How does dynamic programming apply to the Minimum Cost to Merge Stones?

Dynamic programming helps us solve the Minimum Cost to Merge Stones problem in an efficient way. We break the problem into smaller parts and keep track of the results of these parts. This way, we do not repeat any calculations. By defining a state that shows the cost of merging stones from index i to j, we can calculate the minimum costs step by step. This helps us build our solution and keeps it fast.

3. What are the time and space complexities of the Minimum Cost to Merge Stones solution?

The time complexity for the Minimum Cost to Merge Stones solution usually goes up to O(n^3). This happens because we have nested loops to find the merging costs for all possible groups of stones. The space complexity is about O(n^2) because we need a 2D array to store the minimum merging costs. We can make some improvements to use less space. But knowing these complexities is important for checking performance.

4. Can you provide a Java implementation of the Minimum Cost to Merge Stones?

Sure! Here is a simple Java implementation of the Minimum Cost to Merge Stones problem:

class Solution {
    public int mergeStones(int[] stones, int K) {
        int n = stones.length;
        if ((n - 1) % (K - 1) != 0) return -1; // Cannot merge
        
        int[] dp = new int[n + 1];
        int[][] sum = new int[n + 1][n + 1];
        
        for (int i = 1; i <= n; i++) {
            sum[i][i] = stones[i - 1];
            for (int j = i + 1; j <= n; j++) {
                sum[i][j] = sum[i][j - 1] + stones[j - 1];
            }
        }

        for (int len = K; len <= n; len++) {
            for (int i = 0; i + len <= n; i++) {
                dp[i] = Integer.MAX_VALUE;
                for (int j = 1; j < len; j++) {
                    dp[i] = Math.min(dp[i], dp[i] + sum[i + 1][i + j]);
                }
                if ((len - 1) % (K - 1) == 0) {
                    dp[i] += sum[i][i + len - 1];
                }
            }
        }
        
        return dp[0];
    }
}

5. How can I optimize my dynamic programming solution for the Minimum Cost to Merge Stones?

To make the dynamic programming solution for the Minimum Cost to Merge Stones better, we can reduce the space we use. Instead of a full 2D array, we can use a one-dimensional array to keep only the results we need. We can also use memoization to save the results of small problems. This will help us cut down on calculations and make our solution faster and use less space.

By answering these questions, we can understand the Minimum Cost to Merge Stones problem and how to solve it using dynamic programming.