**π‘ Problem Formulation:** We aim to solve a common algorithmic problem: finding a pair of indices (i, j) such that the value of nums[i] + nums[j] + i – j is maximized, where `nums`

is a given list of integers. The goal is to return the maximum sum obtained. For instance, if nums = [1, 3, 5], the optimal pair is (2, 1) resulting in 5 + 3 + 2 – 1 = 9.

## Method 1: Brute Force Approach

The brute force method involves checking all possible pairs of i and j, and calculating their sums according to the given formula. It is the simplest way to understand and implement, but not efficient for large arrays.

Here’s an example:

def max_pair_sum(nums): max_sum = float('-inf') for i in range(len(nums)): for j in range(len(nums)): if i != j: max_sum = max(max_sum, nums[i] + nums[j] + i - j) return max_sum print(max_pair_sum([1, 3, 5]))

Output: 9

This snippet defines a function that iterates through every possible pair (i, j) in the array to find the maximum possible sum. It uses a nested loop to traverse all index combinations, calculates their sums, and updates the maximum sum found.

## Method 2: Optimize with One Pass

This approach takes advantage of the fact that nums[i] + i and nums[j] – j can be maximized separately, scanning the list once to find the maximum possible values.

Here’s an example:

def max_pair_sum(nums): max_i = float('-inf') max_sum = float('-inf') for j in range(len(nums)): max_sum = max(max_sum, max_i + nums[j] - j) max_i = max(max_i, nums[j] + j) return max_sum print(max_pair_sum([1, 3, 5]))

Output: 9

The code example optimizes the search by maintaining a running maximum of nums[i] + i while iterating over j. This means it only requires a single pass through the input array, thus reducing the time complexity.

## Method 3: Dynamic Programming

Dynamic Programming can be used to solve this problem by breaking it down into subproblems. We keep track of the maximum value of nums[i] + i up to the current index.

Here’s an example:

def max_pair_sum(nums): dp = [0] * len(nums) dp[0] = nums[0] for i in range(1, len(nums)): dp[i] = max(dp[i-1], nums[i] + i) max_sum = float('-inf') for j in range(len(nums)): max_sum = max(max_sum, dp[j] + nums[j] - j * 2) return max_sum print(max_pair_sum([1, 3, 5]))

Output: 9

The code utilizes dynamic programming to store the maximum value of nums[i] + i at each step, avoiding redundant calculations. Then it scans the array to find the maximum sum possible using the computed values.

## Method 4: Greedy Approach

The greedy approach focuses on finding the locally optimal solution at each step which will lead to a globally optimal solution. In this case, we greedily update our result while traversing the array.

Here’s an example:

def max_pair_sum(nums): max_sum = float('-inf') max_val = nums[0] - 0 for j in range(1, len(nums)): max_sum = max(max_sum, max_val + nums[j] + j) max_val = max(max_val, nums[j] - j) return max_sum print(max_pair_sum([1, 3, 5]))

Output: 9

With the greedy method, the code keeps track of the highest nums[i] – i found so far and constantly updates the maximum sum with the current value of nums[j] + j. This ensures finding the global maximum by continuously selecting the best local option.

## Bonus One-Liner Method 5: Using Pythonβs Max and Zip Functions

For a concise one-liner solution, Python’s built-in `max`

function can be used in a smart way along with the `zip`

function to combine the necessary values and find the maximum sum.

Here’s an example:

max_pair_sum = lambda nums: max(i + x + j - y for (i, x), (j, y) in zip(enumerate(nums), enumerate(nums[::-1]))) print(max_pair_sum([1, 3, 5]))

Output: 9

This lambda function uses list comprehension to generate all possible sums from nums[i] + i and nums[j] – j and then selects the maximum. The `zip`

and `enumerate`

functions are used to combine indices and values.

## Summary/Discussion

**Method 1:**Brute Force Approach. It’s simple and easy to implement but inefficient for large datasets due to its O(n^2) time complexity.**Method 2:**Optimized One Pass. It improves efficiency by using a single pass, resulting in O(n) time complexity which is much better for larger datasets.**Method 3:**Dynamic Programming. Effective at avoiding repetitive work, thus it’s also of O(n) time complexity, yet more memory intensive due to additional storage requirements.**Method 4:**Greedy Approach. It’s similar in efficiency to method 2 with O(n) time complexity, while sometimes being more intuitive in terms of local vs global optima reasoning.**Bonus Method 5:**Using Pythonβs Max and Zip Functions. Provides a highly Pythonic and concise solution, although it might be less readable and comprehensible to new programmers.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.