**π‘ Problem Formulation:** We aim to solve the “maximum erasure value” problem, which involves finding the maximal sum of a contiguous subarray from a given list of positive integers, where all elements in the subarray are unique. For example, given the input `[4, 2, 4, 5, 6]`

, the desired output would be `17`

, corresponding to the sum of the subarray `[2, 4, 5, 6]`

.

## Method 1: Sliding Window Technique

This method uses the sliding window technique to solve the problem efficiently with a time complexity of O(n). The idea is to keep a dynamic window of the array that slides through, expanding to include unique elements and contracting when a duplicate is found, while calculating the sum of elements within the window.

Here’s an example:

def maxErasureValue(nums): seen = set() max_val = 0 current_sum = 0 left = 0 for right in range(len(nums)): while nums[right] in seen: seen.remove(nums[left]) current_sum -= nums[left] left += 1 seen.add(nums[right]) current_sum += nums[right] max_val = max(max_val, current_sum) return max_val print(maxErasureValue([4, 2, 4, 5, 6]))

The output of this code snippet:

17

This implementation keeps track of the current sum of the window and the maximum value found as it slides through the list. When a duplicate is detected, it shrinks the window from the left until the duplicate is removed. The solution works efficiently for large arrays.

## Method 2: Using an Ordered Dictionary

The ordered dictionary remembers the order in which its contents are added, allowing us to know which element to remove when a duplicate is encountered. This method makes it easier to update the subarray on finding a duplicate.

Here’s an example:

from collections import OrderedDict def maxErasureValue(nums): seen = OrderedDict() current_sum = max_sum = 0 for num in nums: if num in seen: popped = False while not popped: popped_num = seen.popitem(last=False)[0] current_sum -= popped_num popped = popped_num == num seen[num] = num current_sum += num max_sum = max(max_sum, current_sum) return max_sum print(maxErasureValue([4, 2, 4, 5, 6]))

The output of this code snippet:

17

This method uses `OrderedDict`

from Python’s `collections`

module to maintain the order of the elements in the window, making it straightforward to remove the oldest entry (furthest left) when a duplicate is found. The code has a simple logic flow, but its performance could degrade slightly compared to the sliding window for very large arrays due to rehashing.

## Method 3: Brute Force

The naive brute force method involves generating all possible subarrays, ensuring uniqueness of elements, and finding their sums. This method is easy to understand but has a high time complexity of O(n^3), making it impractical for large datasets.

Here’s an example:

def maxErasureValue(nums): max_val = 0 for i in range(len(nums)): for j in range(i+1, len(nums)+1): if len(set(nums[i:j])) == j-i: max_val = max(max_val, sum(nums[i:j])) return max_val print(maxErasureValue([4, 2, 4, 5, 6]))

The output of this code snippet:

17

This brute force approach iterates through all possible subarrays, checks for uniqueness of the elements, and calculates the sum. The main downside is its inefficient computational complexity, making it unsuitable for large inputs.

## Method 4: Two Pointers with HashMap

Using two pointers alongside a hashmap to keep track of the elements can optimize the process. The hashmap stores the most recent index of each element, enabling efficient updates to the left pointer when a duplicate is found.

Here’s an example:

def maxErasureValue(nums): num_index = {} max_val = 0 left = 0 for right, num in enumerate(nums): if num in num_index: left = max(left, num_index[num] + 1) num_index[num] = right max_val = max(max_val, sum(nums[left:right+1])) return max_val print(maxErasureValue([4, 2, 4, 5, 6]))

The output of this code snippet:

17

The code moves the right pointer across the array, uses the hashmap to adjust the left pointer when necessary, and calculates the current window’s sum. It is more space-efficient compared to the OrderedDict approach, but the summing part makes it O(n^2) in the worst case.

## Bonus One-Liner Method 5: Pythonic One-Liner with itertools

For those who love Python’s powerful one-liners, this method leverages `itertools`

to create a compact, though not the most efficient, solution. It’s a neat trick but should be treated more as a clever use of Python’s syntax rather than a practical approach.

Here’s an example:

from itertools import accumulate def maxErasureValue(nums): return max(accumulate((seen := set()).add(x) or x for x in nums if x not in seen, initial=0)) print(maxErasureValue([4, 2, 4, 5, 6]))

The output of this code snippet:

17

This code uses a walrus operator to update a set and chain it with accumulate to keep a running sum. It’s an elegant one-liner solution, but understanding and debugging it can be challenging, and performance may be suboptimal for large inputs due to accumulating from the start for every element encountered.

## Summary/Discussion

**Method 1:**Sliding Window Technique. Highly efficient with O(n) complexity. May require additional effort to understand the sliding window logic.**Method 2:**Ordered Dictionary. Easier to implement and maintain order of entries. Performance hit due to rehashing in large arrays.**Method 3:**Brute Force. Simple to comprehend. Extremely inefficient with O(n^3) complexity, not advisable for anything but the smallest input sets.**Method 4:**Two Pointers with HashMap. Space-efficient with faster lookups. Requires summing window elements on each iteration, leading to O(n^2) worst-case complexity.**Method 5:**Pythonic One-Liner with itertools. Compact and utilizes Python’s features. Not the most efficient and can be tricky to get right or debug.