How do you solve Minimum Cost to Split an Array on LeetCode?

Minimum Cost to Split an Array (LeetCode #2547) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Understanding the importance calculation is crucial for optimizing the solution.

What is the brute force solution for Minimum Cost to Split an Array?

The brute force approach for Minimum Cost to Split an Array is Brute Force, with O(n^3) time complexity and O(1) space complexity. The brute force approach considers all possible ways to split the array into subarrays and calculates the importance value for each split. This method is straightforward but inefficient due to its exhaustive nature. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Minimum Cost to Split an Array?

Minimum Cost to Split an Array uses the following concepts: Array, Hash Table, Dynamic Programming, Counting. The recommended patterns to study are: Hash Map, Dynamic Programming, Array.

What are the interview tips for Minimum Cost to Split an Array?

Clearly explain your thought process and how you arrived at each step. Be prepared to discuss the trade-offs between different approaches. Practice writing clean and efficient code under time constraints.

What are common mistakes when solving Minimum Cost to Split an Array?

Not correctly handling the counting of elements when calculating importance. Overlooking the need to store intermediate results in dynamic programming.

#2547

Minimum Cost to Split an Array

Hard

ArrayHash TableDynamic ProgrammingCountingHash MapDynamic ProgrammingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n^3)	O(n²)
Space	O(1)	O(n)

💡

Intuition

Time O(n²)Space O(n)

The optimal solution uses dynamic programming to build up the minimum cost for each subarray. By storing previously computed results, we avoid redundant calculations, significantly improving efficiency.

⚙️

Algorithm

4 steps

1Step 1: Initialize a dp array where dp[i] represents the minimum cost to partition the first i elements.
2Step 2: Iterate through the array and for each end index, calculate the importance of all possible subarrays ending at that index.
3Step 3: Update the dp array based on the minimum cost found for each subarray.
4Step 4: Return dp[n] as the final answer.

solution.py16 lines

1def minCost(nums, k):
2    n = len(nums)
3    dp = [float('inf')] * (n + 1)
4    dp[0] = 0
5    for r in range(1, n + 1):
6        count = {}
7        trimmed_length = 0
8        for l in range(r - 1, -1, -1):
9            count[nums[l]] = count.get(nums[l], 0) + 1
10            if count[nums[l]] == 2:
11                trimmed_length += 2
12            elif count[nums[l]] > 2:
13                trimmed_length += 1
14            importance = k + trimmed_length
15            dp[r] = min(dp[r], dp[l] + importance)
16    return dp[n]

ℹ

Complexity note: The time complexity is O(n²) due to the nested loop for calculating importance values, while the space complexity is O(n) for storing the dp array.

1Understanding the importance calculation is crucial for optimizing the solution.
2Dynamic programming can significantly reduce the number of redundant calculations.

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