How do you solve Minimum Cost to Merge Stones on LeetCode?

Minimum Cost to Merge Stones (LeetCode #1000) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n³) time complexity and O(n²) space complexity. Key insight: Merging is only possible if the remaining piles can be merged into one.

What is the brute force solution for Minimum Cost to Merge Stones?

The brute force approach for Minimum Cost to Merge Stones is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries every possible way to merge the stones. It calculates the cost for each combination of merges until all piles are combined into one, which can be quite inefficient. The optimal Optimal Solution approach improves this to O(n³).

What algorithm or data structure is used to solve Minimum Cost to Merge Stones?

Minimum Cost to Merge Stones uses the following concepts: Array, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sum.

What are the interview tips for Minimum Cost to Merge Stones?

Clarify the constraints and edge cases before diving into the solution. Explain your thought process clearly while coding. Consider both time and space complexity in your approach.

What are common mistakes when solving Minimum Cost to Merge Stones?

Failing to check if merging is possible before proceeding. Not using prefix sums to optimize the cost calculation.

#1000

Minimum Cost to Merge Stones

Hard

ArrayDynamic ProgrammingPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n³)
Space	O(1)	O(n²)

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Intuition

Time O(n³)Space O(n²)

The optimal approach uses dynamic programming to minimize the cost of merging stones. By storing the results of subproblems, we avoid redundant calculations and efficiently find the minimum cost.

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Algorithm

4 steps

1Step 1: Initialize a DP table to store minimum costs for merging stones from index i to j.
2Step 2: Use prefix sums to calculate the total stones in any range quickly.
3Step 3: Iterate through possible lengths of merges and calculate costs based on previous results in the DP table.
4Step 4: Return the minimum cost to merge all stones into one pile.

solution.py18 lines

1def mergeStones(stones, k):
2    n = len(stones)
3    if (n - 1) % (k - 1) != 0:
4        return -1
5    dp = [[float('inf')] * n for _ in range(n)]
6    prefix_sum = [0] * (n + 1)
7    for i in range(n):
8        prefix_sum[i + 1] = prefix_sum[i] + stones[i]
9    for length in range(k, n + 1):
10        for i in range(n - length + 1):
11            j = i + length - 1
12            if length == k:
13                dp[i][j] = prefix_sum[j + 1] - prefix_sum[i]
14            else:
15                for m in range(i, j, k - 1):
16                    dp[i][j] = min(dp[i][j], dp[i][m] + dp[m + 1][j])
17                dp[i][j] += prefix_sum[j + 1] - prefix_sum[i]
18    return dp[0][n - 1]

ℹ

Complexity note: The time complexity is O(n³) because we are iterating through all possible lengths and combinations of merges, leading to a cubic growth in operations. The space complexity is O(n²) due to the DP table storing results for each subproblem.

1Merging is only possible if the remaining piles can be merged into one.
2Dynamic programming helps in breaking down the problem into manageable subproblems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.