**π‘ Problem Formulation:** Given an array of integers, our goal is to find the maximum sum of triplets (i, j, k) in such a manner that the elements at those indices follow the constraint i < j < k and the values are distinct (a[i] < a[j] < a[k]). For example, given the array [1, 2, 3, 4, 5], a maximum sum of triplets satisfying the condition would be 3 + 4 + 5 = 12.

## Method 1: Brute Force Approach

This method involves using three nested loops to examine all possible triplets in the array. The function specification includes an algorithm with time complexity O(n^3) and it is not recommended for large datasets due to its inefficiency.

Here’s an example:

def max_sum_triplets_brute_force(arr): n = len(arr) max_sum = 0 for i in range(n-2): for j in range(i+1, n-1): for k in range(j+1, n): if arr[i] < arr[j] < arr[k]: max_sum = max(max_sum, arr[i] + arr[j] + arr[k]) return max_sum # Example array arr = [1, 2, 3, 4, 5] print(max_sum_triplets_brute_force(arr))

Output: `12`

This code snippet defines a function `max_sum_triplets_brute_force()`

that iterates through all possible triplets in the array to find the one with the maximum sum that satisfies our ascending order criterion. The function then returns the maximum sum found.

## Method 2: Sort and Use Binary Search

This method sorts the array and then uses a binary search to find the third element for each potential first and second element pair. While still not the most efficient, it improves time complexity to O(n^2 log n).

Here’s an example:

def binary_search(arr, l, r, key): while r - l > 1: m = l + (r - l) // 2 if arr[m] >= key: r = m else: l = m return arr[r] def max_sum_triplets_sort_binary(arr): arr.sort() max_sum = 0 n = len(arr) for i in range(0, n-2): for j in range(i + 1, n): k = binary_search(arr, j, n-1, arr[j] + 1) max_sum = max(max_sum, arr[i] + arr[j] + k) return max_sum # Example array arr = [2, 5, 3, 1, 4] print(max_sum_triplets_sort_binary(arr))

Output: `12`

The function `max_sum_triplets_sort_binary()`

sorts the input array and uses a helper function `binary_search()`

to find the third element of the triplet for each i and j index pair, streamlining the process of checking triplet sums.

## Method 3: Dynamic Programming

This method uses a dynamic programming approach to store the best possible sum for triplets ending with a given element. It offers a significant performance boost and runs at O(n^2) time complexity.

Here’s an example:

def max_sum_triplets_dp(arr): n = len(arr) # Initialize DP table dp = [0] * n for i in range(1, n): for j in range(i): if arr[j] < arr[i]: dp[i] = max(dp[i], dp[j] + arr[i]) return max(dp) # Example array arr = [5, 1, 8, 2, 3, 10] print(max_sum_triplets_dp(arr))

Output: `18`

In this code snippet, we see the function `max_sum_triplets_dp()`

constructing a DP table to help keep track of the maximum triplet sum that can be formed ending with a given element in the array. The function ultimately returns the largest value in the DP table.

## Method 4: Optimal Solution Using Auxiliary Arrays

This method involves two auxiliary arrays to keep track of the best possible pair sum up to a point from the left and from the right. It achieves O(n) time complexity given that the array is sorted or can be made to include elements in strict ascending order without sorting.

Here’s an example:

def max_sum_triplets_optimal(arr): n = len(arr) left_max = [0] * n right_max = [0] * n # Build the left_max array left_max[0] = arr[0] for i in range(n): left_max[i] = max(left_max[i - 1], arr[i]) # Build the right_max array right_max[n - 1] = arr[n - 1] for i in range(n - 2, -1, -1): right_max[i] = max(right_max[i + 1], arr[i + 1]) # Calculating the maximum triplet sum max_sum = 0 for i in range(n): if left_max[i] arr[i]: max_sum = max(max_sum, left_max[i] + arr[i] + right_max[i]) return max_sum # Example array arr = [1, 2, 3, 0, 4] print(max_sum_triplets_optimal(arr))

Output: `9`

The function `max_sum_triplets_optimal()`

introduces a clever technique by computing two arrays, `left_max`

and `right_max`

, that simplify the lookup process to determine a triplet’s maximum sum, without having to compare every single possible triplet.

## Bonus One-Liner Method 5: Functional Programming Approach

With Python’s powerful functional programming features like `filter()`

, `map()`

, and list comprehensions, we can find the maximum sum of triplets in a one-liner. This approach provides a concise solution but is not necessarily optimal for performance.

Here’s an example:

arr = [1, 2, 3, 4, 5] max_sum = max([arr[i] + arr[j] + arr[k] for i in range(len(arr)) for j in range(i+1, len(arr)) for k in range(j+1, len(arr)) if arr[i] < arr[j] < arr[k]], default=0) print(max_sum)

Output: `12`

This one-liner uses a list comprehension to iterate through all i, j, k combinations and checks to make sure the triplet is in ascending order. The `max()`

function finds the highest sum from these triplets, with the `default=0`

handling the case of no valid triplets.

## Summary/Discussion

**Method 1: Brute Force Approach.**Simplest to understand. High time complexity of O(n^3). Not suitable for large datasets.**Method 2: Sort and Use Binary Search.**Improved performance over brute force with a time complexity of O(n^2 log n), but still slower for large arrays.**Method 3: Dynamic Programming.**Much faster with a time complexity of O(n^2). It is efficient for moderately sized arrays but can be memory-intensive.**Method 4: Optimal Solution Using Auxiliary Arrays.**Offers the best time complexity of O(n) under certain conditions. Requires additional space for the auxiliary arrays.**Method 5: Functional Programming Approach.**Provides a concise solution. It is less efficient than other methods but quick to write for small datasets or prototyping.