Is Maximize Cyclic Partition Score easy or hard?

Maximize Cyclic Partition Score is classified as a Hard problem because it combines two tricky ideas: cyclic array handling and optimal partitioning with dynamic programming. Efficient solutions require recognizing the array duplication trick and structuring correct DP transitions.

Maximize Cyclic Partition Score Python/Java solution

The standard implementation uses dynamic programming over a doubled array to simulate cyclic rotations. The same algorithm works across Python, Java, C++, and Go by maintaining a DP array and iterating through candidate partition boundaries.

How to solve Maximize Cyclic Partition Score in O(n)?

Most implementations do not achieve strict O(n) time because each partition endpoint may require scanning previous cut positions. The practical optimized approach runs in O(n^2) using dynamic programming and efficient segment score computation such as prefix sums or precomputed statistics.

What is the best approach for Maximize Cyclic Partition Score?

The most effective solution uses dynamic programming combined with a circular-array trick. Duplicate the array to simulate rotations and compute optimal partition scores within a window of length n. The DP state tracks the best score ending at each index while evaluating possible previous cut positions.

Is Maximize Cyclic Partition Score asked at Google/Amazon/Meta?

Problems involving cyclic arrays and partition-based dynamic programming appear frequently in interviews at companies like Google, Amazon, and Meta. While the exact problem may vary, the core techniques—circular array handling and partition DP—are common interview topics.

What data structure is used in Maximize Cyclic Partition Score?

The primary technique relies on dynamic programming arrays along with standard array preprocessing like prefix sums or range statistics. These help evaluate segment scores efficiently while transitioning between partition states.

What is the time complexity of Maximize Cyclic Partition Score?

The optimized dynamic programming solution runs in O(n^2) time and O(n) space. Each index acts as a partition endpoint, and the algorithm checks previous cut positions to compute the best score while respecting the circular window constraint.

Description Hints 4 Video 13:36

3743. Maximize Cyclic Partition Score

Hard14.5% acceptance2 companies

You are given a cyclic array nums and an integer k.

Partition nums into at most k subarrays. As nums is cyclic, these subarrays may wrap around from the end of the array back to the beginning.

The range of a subarray is the difference between its maximum and minimum values. The score of a partition is the sum of subarray ranges.

Return the maximum possible score among all cyclic partitions.

Example 1:

Input: nums = [1,2,3,3], k = 2

Output: 3

Explanation:

Partition nums into [2, 3] and [3, 1] (wrapped around).
The range of [2, 3] is max(2, 3) - min(2, 3) = 3 - 2 = 1.
The range of [3, 1] is max(3, 1) - min(3, 1) = 3 - 1 = 2.
The score is 1 + 2 = 3.

Example 2:

Input: nums = [1,2,3,3], k = 1

Output: 2

Explanation:

Partition nums into [1, 2, 3, 3].
The range of [1, 2, 3, 3] is max(1, 2, 3, 3) - min(1, 2, 3, 3) = 3 - 1 = 2.
The score is 2.

Example 3:

Input: nums = [1,2,3,3], k = 4

Output: 3

Explanation:

Identical to Example 1, we partition nums into [2, 3] and [3, 1]. Note that nums may be partitioned into fewer than k subarrays.

Constraints:

1 <= nums.length <= 1000
1 <= nums[i] <= 10⁹
1 <= k <= nums.length

Asked at 2 companies

Google Amazon

Topics

Array Dynamic Programming

Hints open one at a time

Hint 1

Use dynamic programming on the number of extreme picks: allow up to 2 * k picks (each partition can supply a + and a -), so the problem becomes choosing which elements are + or -.

Hint 2

Model a partition by its max (+nums[i]) and min (-nums[i]) contributions; selecting an element as a max adds +nums[i], and selecting it as a min adds -nums[i].

Hint 3

Use a DP state (picks, balance) where picks is the total number of + and - chosen so far (<= 2 * k), and balance represents the difference between the counts of + and - currently; at each i, you conceptually take +, take -, or skip.

Hint 4

Handle cyclicity by limiting balance to the range [0, 2]. You can show that such a balance is always achievable.

LanguageLoading editor…

1class Solution:
2    def maximumScore(self, nums: List[int], k: int) -> int:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[1,3,4]
2

Input

[1,1000000000,1000000000]
3

Input

[1]
1

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