How do you solve Minimum Cost to Split into Ones on LeetCode?

Minimum Cost to Split into Ones (LeetCode #3857) 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: Always split into 1 and n-1 for minimal cost.

What is the time complexity of Minimum Cost to Split into Ones?

The optimal solution for Minimum Cost to Split into Ones runs in O(n²) time and uses O(n) space. We compute results for each integer up to n, storing them, leading to O(n²) time and O(n) space.

What is the brute force solution for Minimum Cost to Split into Ones?

The brute force approach for Minimum Cost to Split into Ones is Brute Force, with O(n²) time complexity and O(1) space complexity. We can explore all possible ways to split the integer n into smaller integers recursively. Each split incurs a cost, and we track the minimum cost across all splits. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Minimum Cost to Split into Ones?

Minimum Cost to Split into Ones uses the following concepts: Math, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Minimum Cost to Split into Ones?

Explain your thought process clearly. Discuss edge cases like n=2. Practice similar dynamic programming problems.

What are common mistakes when solving Minimum Cost to Split into Ones?

Forgetting to initialize the dp array properly. Not considering all possible splits.

#3857

Minimum Cost to Split into Ones

Medium

MathDynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n²)Space O(n)

Using dynamic programming, we can build solutions for smaller integers up to n, storing the minimum costs to avoid redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Initialize an array dp where dp[i] represents the minimum cost to split integer i.
2Step 2: For each integer i from 2 to n, calculate the minimum cost by trying all splits j from 1 to i-1.
3Step 3: Update dp[i] with the minimum value found for all splits.

solution.py7 lines

1def minCost(n):
2    dp = [0] * (n + 1)
3    for i in range(2, n + 1):
4        dp[i] = float('inf')
5        for j in range(1, i):
6            dp[i] = min(dp[i], j * (i - j) + dp[j] + dp[i - j])
7    return dp[n]

ℹ

Complexity note: We compute results for each integer up to n, storing them, leading to O(n²) time and O(n) space.

1Always split into 1 and n-1 for minimal cost.
2Dynamic programming avoids redundant calculations.

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