# Maximizing the Sum with Optimal Pair Selection in Python

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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 Pythonβs 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 Pythonβs Max and Zip Functions. Provides a highly Pythonic and concise solution, although it might be less readable and comprehensible to new programmers.