Maximum Pulse Value After One Subarray Rotation - Solution & Explanation
Problem Statement
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
landrsuch that0 <= 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 is5 - 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 is6 - 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
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Leetcode 4058 | Maximum Pulse Value After One Subarray Rotation | Leetcode weekly contest 520 • CodeWithMeGuys • 992 views views
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