5 Best Ways to Check if is Possible to Get a Given Sum from a Set of Elements in Python

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π‘ Problem Formulation: This article explores various methods to determine whether a particular sum can be obtained from a set of given numbers. For example, given the set [3, 34, 4, 12, 5, 2] and the sum 9, we want to confirm whether there are any subsets whose sum equals 9.

Method 1: Recursive Approach

The recursive approach involves breaking down the problem into smaller subproblems. This method checks whether the target sum can be reached by either including or excluding a particular element. Function signature can typically be `can_get_sum_recursive(set_elements, total_sum)`.

Here’s an example:

```def can_get_sum_recursive(nums, sum):
if sum == 0:
return True
if len(nums) == 0 or sum < 0:
return False
return can_get_sum_recursive(nums[1:], sum) or can_get_sum_recursive(nums[1:], sum - nums[0])

# Example usage
print(can_get_sum_recursive([3, 34, 4, 12, 5, 2], 9))```

Output: `True`

The function checks for the sum either by including or excluding the first element of the list. If including or excluding the element results in the sum, it returns `True`, else it continues the search recursively. This is a simple approach, but it isn’t efficient for larger input lists due to its exponential time complexity.

Method 2: Dynamic Programming (Subset Sum Problem)

Dynamic programming can be used to solve the subset sum problem more effectively. It uses memorization to avoid recalculating subproblems. The function can be defined as `can_get_sum_dp(set_elements, total_sum)`.

Here’s an example:

```def can_get_sum_dp(nums, sum):
dp = [[False] * (sum + 1) for _ in range(len(nums) + 1)]
for i in range(len(nums) + 1):
dp[i][0] = True
for i in range(1, len(nums) + 1):
for j in range(1, sum + 1):
if j < nums[i - 1]:
dp[i][j] = dp[i - 1][j]
else:
dp[i][j] = dp[i - 1][j] or dp[i - 1][j - nums[i - 1]]
return dp[len(nums)][sum]

# Example usage
print(can_get_sum_dp([3, 34, 4, 12, 5, 2], 9))```

Output: `True`

This code initializes a 2D array `dp` to store the results of subproblems. It iteratively constructs the array by using the results of previously solved subproblems, eventually determining whether the `sum` can be obtained from the set. This method is more efficient than the pure recursive approach, especially for larger sets.

Method 3: Greedy Algorithm

The greedy algorithm attempts to solve the problem by always choosing the best immediate or local solution, though it doesn’t guarantee an optimal solution for all cases. It can be particularly useful when the elements of the set are in some sorted order and the problem has some specific constraints.

Here’s an example:

```def can_get_sum_greedy(nums, sum):
nums.sort(reverse=True)
for num in nums:
if sum >= num:
sum -= num
return sum == 0

# Example usage
print(can_get_sum_greedy([3, 34, 4, 12, 5, 2], 9))```

Output: `False`

This code sorts the elements in descending order and iteratively subtracts the largest element from the sum until it is no longer possible. If the remaining sum is zero, it found a subset; otherwise, it means such a subset does not exist. Be aware that this method may fail for certain sets where a greedy approach is not applicable.

Method 4: Using Backtracking

Backtracking is a more systematic way to go through all the subsets than the pure recursive approach. It “backtracks” when it determines that a set of choices does not lead to a solution. A typical function might look like this: `can_get_sum_backtracking(set_elements, total_sum)`.

Here’s an example:

```def can_get_sum_backtracking(nums, target, partial=[]):
current_sum = sum(partial)

if current_sum == target:
return True
if current_sum > target:
return False

for i in range(len(nums)):
n = nums[i]
remaining = nums[i+1:]
if can_get_sum_backtracking(remaining, target, partial + [n]):
return True
return False

# Example usage
print(can_get_sum_backtracking([3, 34, 4, 12, 5, 2], 9))```

Output: `True`

This code uses the current partial subset and tries to add the next element in the set. If the sum of the partial set exceeds the target, it returns false. If it equals the target, it returns true. Otherwise, it recursively explores each possibility with backtracking. This approach can be more efficient than a naive recursive approach since it eliminates many unnecessary paths early.

Bonus One-Liner Method 5: Using itertools.combinations

A concise one-liner technique using Python’s `itertools.combinations` function can examine all possible combinations with varying lengths to check for the target sum. This is however not the most efficient method for large datasets.

Here’s an example:

```from itertools import combinations

def can_get_sum_itertools(nums, target_sum):
return any(sum(comb) == target_sum for r in range(len(nums) + 1) for comb in combinations(nums, r))

# Example usage
print(can_get_sum_itertools([3, 34, 4, 12, 5, 2], 9))```

Output: `True`

This one-liner uses list comprehensions and `itertools.combinations` to generate all possible subsets of the list of numbers, checking whether any of these subsets sum to the target value. While this method is incredibly concise, it is also brute force and therefore inefficient as it checks every possible combination.

Summary/Discussion

• Method 1: Recursive Approach. Easy to understand and implement. Exponential time complexity makes it inefficient for large datasets.
• Method 2: Dynamic Programming. Efficient for larger datasets with polynomial time complexity. More complex to implement and understand.
• Method 3: Greedy Algorithm. Quick and simple. Does not guarantee an optimal solution for all cases and can fail when the problem doesn’t have the right constraints.
• Method 4: Backtracking. Systematically explores subsets. More efficient than recursive approach for some cases. Can still be relatively slow for large sets.
• Bonus Method 5: itertools.combinations. Extremely concise implementation. Inefficient for large datasets due to its brute-force nature.