Is Maximum Pulse Value After One Subarray Rotation easy or hard?

It is a medium-difficulty problem with a 38.9% acceptance rate. The brute force is straightforward, but getting to the O(n) DP requires defining states carefully and handling the rotation boundaries correctly.

How to solve Maximum Pulse Value After One Subarray Rotation in O(n)?

Precompute prefix sums, then iterate once while maintaining DP states for before, during, and after the rotated subarray. At each index, update each state using the previous index's best values and keep the global maximum. Return the maximum value seen across valid final states.

Maximum Pulse Value After One Subarray Rotation Python and Java solution?

Both languages use the same single-pass DP. In Python, build the prefix array with itertools.accumulate and update three state variables in a loop. In Java, use a long prefix array to avoid overflow and update the same three states with a for loop.

What is the best approach for Maximum Pulse Value After One Subarray Rotation?

The best approach is a single-pass dynamic programming solution using prefix sums. Track the best value for each state (no rotation yet, inside the rotated subarray, rotation finished) as you scan once through the array.

What data structure is used in Maximum Pulse Value After One Subarray Rotation?

The solution uses an array for input and prefix sums to get segment contributions in constant time. Dynamic programming stores a few scalar state values per index, so extra space is O(n) for the prefix array or O(1) if you roll the state variables.

What is the time complexity of Maximum Pulse Value After One Subarray Rotation?

The optimal solution runs in O(n) time. The brute force that checks every subarray and rotation runs in O(n^3), and a prefix-sum enumeration runs in O(n^2).

Is Maximum Pulse Value After One Subarray Rotation asked at Google, Amazon, or Meta?

Company tags for LeetCode problems change over time and vary by source, so check the company filter on FleetCode for current data. The problem's DP and prefix sum tags make it a common medium-level interview question at large tech companies.

Description Hints 3 Video 31:03

4058. Maximum Pulse Value After One Subarray Rotation

Medium38.9% acceptance

You are given an integer array nums of length n.

Define the pulse value of an integer array arr as the alternating sum starting at index 0: pulse(arr) = arr[0] - arr[1] + arr[2] - arr[3] + ....

You may perform at most one operation on nums:

Choose two indices l and r such that 0 <= l < r < n.
Left-rotate the subarray nums[l..r] by exactly one position. For example, [a, b, c, d] becomes [b, c, d, a].

Return the maximum pulse value that can be obtained after performing at most one such operation.

Example 1:

Input: nums = [1,5,2]

Output: 6

Explanation:

The original pulse value is 1 - 5 + 2 = -2.
Rotate the subarray nums[0..1] from [1, 5] to [5, 1].
The resulting array is [5, 1, 2] and its pulse value is 5 - 1 + 2 = 6, which is the maximum possible.

Example 2:

Input: nums = [6,4,3]

Output: 7

Explanation:

The original pulse value is 6 - 4 + 3 = 5.
Rotate the subarray nums[1..2] from [4, 3] to [3, 4].
The resulting array is [6, 3, 4] and its pulse value is 6 - 3 + 4 = 7, which is the maximum possible.

Example 3:

Input: nums = [9,7]

Output: 2

Explanation:

The original pulse value is 9 - 7 = 2, which is already maximum. Thus, no rotation is required.

Constraints:

1 <= n == nums.length <= 10⁵
-10⁹ <= nums[i] <= 10⁹

Topics

Hints open one at a time

Hint 1

After rotating nums[l..r], every element originally at an index from l + 1 through r changes its contribution's sign. The element nums[l] moves to index r.

Hint 2

Let P[t] be the alternating sum of the first t elements, with P[0] = 0. The change in pulse value is 2 * (P[l + 1] - P[r + 1]) when l and r have the same parity, and 2 * (P[l] - P[r + 1]) otherwise.

Hint 3

Scan r from left to right. Among indices l < r, maintain the maximum values of P[l] and P[l + 1] separately for each parity of l. Use these maxima to find the best gain in constant time per index, allowing a gain of zero for skipping the operation.

4058. Maximum Pulse Value After One Subarray Rotation

Topics

Hints open one at a time

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