Is Minimum Cost to Split into Ones II easy or hard?

The problem is typically rated Medium because the brute-force idea is straightforward but inefficient. Recognizing that only the positions of '1's influence valid splits allows a linear-time solution, which is the key insight interviewers expect.

Minimum Cost to Split into Ones II Python/Java solution

Most implementations iterate through the binary array, record positions of '1's, and update the cost as gaps are processed. The logic translates directly to Python, Java, C++, or JavaScript because it relies on simple loops, counters, and optional DP arrays.

How to solve Minimum Cost to Split into Ones II in O(n)?

Track the previous position of a '1' while scanning the sequence. Each time a new '1' appears, compute the cost contribution from the gap between the current and previous '1'. Maintain a running minimum cost using dynamic programming or greedy accumulation, ensuring each element is processed exactly once.

What is the best approach for Minimum Cost to Split into Ones II?

The most efficient solution processes the positions of '1's and evaluates the gaps between them. By tracking these indices and updating the cost incrementally, you avoid testing every split boundary. This reduces the complexity from O(n^2) brute force to an O(n) dynamic programming or greedy scan.

Is Minimum Cost to Split into Ones II asked at Google/Amazon/Meta?

Problems involving binary segmentation, cost minimization, and prefix-based dynamic programming frequently appear in interviews at companies like Google, Amazon, and Meta. Variations often test whether candidates recognize patterns around processing positions of key elements such as '1's.

What data structure is used in Minimum Cost to Split into Ones II?

The solution primarily uses arrays and prefix tracking. A dynamic programming array or simple index variables track the previous '1' and accumulated cost. Some implementations also use prefix sums to quickly compute segment statistics.

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

The optimal solution runs in O(n) time because the array or string is scanned once while tracking the positions of '1's. Each index contributes constant work. Space complexity ranges from O(1) to O(n) depending on whether a DP array or simple index tracking is used.

Description Solution Video 0:51

3958. Minimum Cost to Split into Ones II

Medium69.2% acceptancePremium elsewhere, free here

You are given an integer n.

In one operation, you may split an integer x into two positive integers a and b such that a + b = x.

The cost of this operation is a * b.

Return the minimum total cost required to split the integer n into n ones.

Example 1:

Input: n = 3

Output: 3

Explanation:

One optimal set of operations is:

`x`	`a`	`b`	`a + b`	`a * b`	Cost
3	1	2	3	2	2
2	1	1	2	1	1

Thus, the minimum total cost is 2 + 1 = 3.

Example 2:

Input: n = 4

Output: 6

Explanation:

One optimal set of operations is:

`x`	`a`	`b`	`a + b`	`a * b`	Cost
4	2	2	4	4	4
2	1	1	2	1	1

Thus, the minimum total cost is 4 + 1 + 1 = 6.

Constraints:

1 <= n <= 5 * 10⁷

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