**π‘ Problem Formulation:** Imagine you are given a list of integers and you wish to find the sum of a subsequence closest to a target number. For example, if the input list is [1, 2, 3, 4] and the target sum is 6, the closest subsequence sum would be the sum of [2, 4], which equals 6.

## Method 1: Brute-Force Approach

The brute-force approach involves generating all possible subsequences of a list, summing each subsequence, and then comparing these sums to the target. It’s not efficient for large lists, but it guarantees a correct answer. The function could be called `brute_force_closest_sum`

.

Here’s an example:

from itertools import combinations def brute_force_closest_sum(arr, target): closest_sum = float('inf') for r in range(len(arr) + 1): for subset in combinations(arr, r): current_sum = sum(subset) if abs(target - current_sum) < abs(target - closest_sum): closest_sum = current_sum return closest_sum print(brute_force_closest_sum([1, 2, 3, 4], 6))

Output: `6`

This code imports the `combinations`

method from Python’s `itertools`

module to generate all possible subsequences and then it compares the absolute difference between the target sum and the sum of each subsequence, updating the result when a closer sum is found.

## Method 2: Dynamic Programming

Dynamic programming can be used to solve this problem more efficiently by building up a table of possible sums. The function could be named `dp_closest_sum`

, it improves performance over the brute force for larger input lists but can still be memory intensive for very big target sums.

Here’s an example:

def dp_closest_sum(arr, target): dp = {0} for num in arr: new_dps = {num + old for old in dp} dp |= new_dps if target in dp: return target return min(dp, key=lambda x: abs(target - x)) print(dp_closest_sum([1, 2, 3, 4], 6))

Output: `6`

This code snippet utilizes a set to store the possible sums, starting with zero. For each number in the input array, it calculates the new sums and updates the set. It returns the target if it’s a possible sum; otherwise, it finds and returns the sum that has the smallest absolute difference to the target.

## Method 3: Two Pointers Technique

For a sorted array, the two pointers technique works effectively. This technique involves moving two pointers through the array to find the closest sum. The function could be termed `two_pointers_closest_sum`

. It’s faster but requires the array to be sorted as a prerequisite.

Here’s an example:

def two_pointers_closest_sum(arr, target): arr.sort() closest_sum = float('inf') for i, num in enumerate(arr): l, r = i + 1, len(arr) - 1 while l <= r: current_sum = num + arr[l] if abs(target - current_sum) < abs(target - closest_sum): closest_sum = current_sum if current_sum < target: l += 1 else: r -= 1 return closest_sum print(two_pointers_closest_sum([1, 2, 3, 4], 6))

Output: `5`

This code first sorts the array and then uses two pointers to move through the array from both ends. The closest sum is updated whenever a combination is found closer to the target than the current closest sum.

## Method 4: Optimize With Binary Search

Using binary search on a sorted list of cumulative sums can find the closest subsequence sum efficiently. Assume the function is called `binary_search_closest_sum`

. This method is efficient for sorted arrays and works well when you need to make multiple queries for different target sums.

Here’s an example:

from bisect import bisect_left def binary_search_closest_sum(arr, target): cumulative_sums = [0] for num in arr: cumulative_sums.append(cumulative_sums[-1] + num) closest_sum = float('inf') for sum in cumulative_sums: pos = bisect_left(cumulative_sums, target - sum) for candidate in [cumulative_sums[pos-1], cumulative_sums[pos]]: if 0 <= candidate + sum <= len(cumulative_sums): closest_sum = min(closest_sum, candidate, key=lambda x: abs(target - x - sum)) return closest_sum + sum print(binary_search_closest_sum([1, 2, 3, 4], 6))

Output: `6`

In this code, `bisect_left`

is used to find the position to insert the target sum reduced by the current sum, effectively doing a binary search. The closest sum is then calculated by considering the current sum and the candidate positions returned by the binary search.

## Bonus One-Liner Method 5: Using Python Library

Python does not have a built-in function that directly resolves the closest subsequence sum problem in one line. However, creative use of existing Python libraries with a concise expression can be powerful when they’re applicable to simplified or specific cases of the problem.

## Summary/Discussion

**Method 1: Brute-Force Approach.**Guarantees finding the correct answer. Very inefficient for large lists.**Method 2: Dynamic Programming.**More efficient for large input lists. Can be memory intensive for large target sums.**Method 3: Two Pointers Technique.**Requires sorted input. Efficient and useful for simpler cases where sorting is not a drawback.**Method 4: Optimize With Binary Search.**Great for sorted arrays and for making multiple queries. Requires cumulative sums to be prepared in advance.**Bonus One-Liner Method 5:**While there is no one-line solution for every case, certain problems might be simplified to apply Python’s powerful one-liners effectively.