Solution Explanation for LeetCode 1330: Reverse Subarray To Maximize Array Value

This problem asks to find the maximum possible value of an array after reversing a single subarray. The value of an array is defined as the sum of absolute differences between consecutive elements. The solution employs a clever approach involving careful case analysis and optimization to achieve linear time complexity.

Core Idea:

The solution explores several scenarios to maximize the array's value after a single sub-array reversal. Instead of brute-forcing all possible subarray reversals (which would be O(n^3)), it cleverly categorizes the reversals and optimizes the calculations.

Case Analysis and Optimization:

1. No Reversal: The initial array value is calculated and stored as a baseline.

2. Reversal Including the First Element: The algorithm iterates through possible subarrays starting from the first element. For each subarray ending at index i, it calculates the change in the array's value after reversing that subarray. This change is efficiently computed by considering only the affected differences: the difference between the first element and the new element following the reversed subarray, and the difference between the last element of the reversed subarray and its original neighbor.

3. Reversal Including the Last Element: Similar to the previous case, this iterates through subarrays ending at the last element, calculating the value change efficiently.

4. Reversal Excluding First and Last Elements: This is the most intricate part. The algorithm doesn't explicitly reverse every possible inner subarray. Instead, it observes that the effect of reversing an inner subarray involves changes to only two pairs of adjacent elements. Let's say we reverse a subarray from i to j. The changes in the absolute differences will be:

   • |nums[i-1] - nums[j]| replaces |nums[i-1] - nums[i]|
   • |nums[i] - nums[j+1]| replaces |nums[j] - nums[j+1]|

   The core insight is to identify the maximum potential increase in the overall value. This is achieved by considering different combinations of signs (+ or -) for the elements involved in these changes. The code uses a clever loop with dirs array to cover all these sign combinations systematically. The mx and mi variables keep track of the maximum potential increase and minimum potential decrease, leading to an efficient calculation of the maximum possible value increase from this type of reversal.

Time Complexity: O(n)

The algorithm iterates through the array a constant number of times (a few loops). The operations within the loops are all constant time. Therefore, the overall time complexity is linear.

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space, regardless of the input array's size.

Code Explanation (Python):

from itertools import pairwise  # Efficient pairwise iteration
 
class Solution:
    def maxValueAfterReverse(self, nums: List[int]) -> int:
        ans = s = sum(abs(x - y) for x, y in pairwise(nums)) #Initial value
        for x, y in pairwise(nums): # Iterating through pairs
            ans = max(ans, s + abs(nums[0] - y) - abs(x - y)) #Case 2
            ans = max(ans, s + abs(nums[-1] - x) - abs(x - y)) #Case 3
        for k1, k2 in pairwise((1, -1, -1, 1, 1)): # clever sign combinations
            mx, mi = -inf, inf
            for x, y in pairwise(nums): # Iterate through pairs again
                a = k1 * x + k2 * y # Calculating potential increase/decrease
                b = abs(x - y)
                mx = max(mx, a - b)
                mi = min(mi, a + b)
            ans = max(ans, s + max(mx - mi, 0)) #Update ans with max change from Case 4
        return ans

The code in other languages (Java, C++, Go, TypeScript) follows the same logic with minor syntax variations. The pairwise iterator (Python) is a convenient tool; other languages might use a standard for loop with index manipulation to achieve the same pairwise iteration.