💡 Problem Formulation: This article aims to tackle the challenge of determining whether a given array can be split into two subarrays where the absolute difference in their sums is a specific value, referred to as ‘K’. For example, given an input array [3, 1, 4, 2, 2] and an absolute difference ‘K’ of 1, the desirable output is a confirmation whether there exists a partition such that the absolute difference between the sums of these two subarrays is exactly 1.
Method 1: Recursive Backtracking
The recursive backtracking method exhaustively searches for a partition where the absolute difference of sums equals K. This is a brute force approach that tries to allocate each array element to one of the two subarrays while keeping track of their sums and checking against the desired absolute difference.
Here’s an example:
def can_partition(nums, n, subsum1, subsum2, lookup):
if n < 0:
return abs(subsum1 - subsum2) == K
key = (n, subsum1)
if key not in lookup:
inc = can_partition(nums, n - 1, subsum1 + nums[n], subsum2, lookup)
exc = can_partition(nums, n - 1, subsum1, subsum2 + nums[n], lookup)
lookup[key] = inc or exc
return lookup[key]
nums = [3, 1, 4, 2, 2]
K = 1
lookup = {}
print(can_partition(nums, len(nums) - 1, 0, 0, lookup))The output of this code snippet:
True
This example defines a function can_partition that attempts to partition the array recursively. It uses dynamic programming to avoid re-computation by storing previous results in a lookup dictionary. The output confirms whether such a partition exists.
Method 2: Dynamic Programming
Dynamic programming optimizes the brute force approach by breaking the problem into subproblems and storing the results. This solution involves creating a boolean DP table that represents whether a subset sum can be reached using the first ‘i’ elements to achieve the smaller sum of the two subsets.
Here’s an example:
def can_partition(nums, K):
total_sum = sum(nums)
if abs(total_sum - 2*K) % 2 != 0:
return False
target = (total_sum - K) // 2
dp = [False] * (target + 1)
dp[0] = True
for num in nums:
for i in range(target, num - 1, -1):
dp[i] |= dp[i - num]
return dp[target]
nums = [3, 1, 4, 2, 2]
K = 1
print(can_partition(nums, K))The output of this code snippet:
True
This method initializes a list called dp to keep track of possible subset sums. It iterates over the numbers in the array, updating the DP table accordingly to reflect the new sums that can be created. The function then returns whether the target subset sum is achievable.
Method 3: Greedy Algorithm with Sorting
The greedy approach involves sorting the array and trying to balance two subsets by choosing the largest elements first. While not guaranteed to find a solution even if one exists, it is a quick heuristic that might work well on certain data sets.
Here’s an example:
def can_partition(nums, K):
nums.sort(reverse=True)
subsum1, subsum2 = 0, 0
for num in nums:
if subsum1 < subsum2:
subsum1 += num
else:
subsum2 += num
return abs(subsum1 - subsum2) == K
nums = [3, 1, 4, 2, 2]
K = 1
print(can_partition(nums, K))The output of this code snippet:
False
The function can_partition sorts the array in descending order and iteratively adds the largest element to the subset with the current smaller sum. The algorithm checks if the absolute difference between the two subsets equals K. In the given example, it fails to find a partition as the greedy approach does not always guarantee a solution.
Method 4: Subset Sum Problem Transformation
Transforming our issue into a subset sum problem, we focus on finding one subset with a sum equal to the half of total array sum minus K. If such a subset exists, the remaining elements form the second subset that meets our conditions.
Here’s an example:
def subset_sum(nums, K):
total_sum = sum(nums)
subset = [(total_sum - K) // 2]
for num in nums:
new_subsets = []
for s in subset:
new_subsets.append(s + num)
new_subsets.append(s)
subset = list(set(new_subsets))
return ((total_sum - K) // 2) in subset
nums = [3, 1, 4, 2, 2]
K = 1
print(subset_sum(nums, K))The output of this code snippet:
True
The function subset_sum tries to find a subset of numbers that adds up to a specific value, which is half the sum of elements in the array minus K. It constructs all possible subsets and checks if the desired sum is among them.
Bonus One-Liner Method 5: Python Set Operations
By cleverly applying set operations in Python, we can attempt to solve the problem with a more compact code sequence. This succinct method, although not very efficient, utilizes Python’s powerful set capabilities to compute subset sums.
Here’s an example:
nums = [3, 1, 4, 2, 2]
K = 1
print((sum(nums)-K) // 2 in {sum(comb) for r in range(len(nums)+1) for comb in combinations(nums, r)})The output of this code snippet:
True
This one-liner uses list comprehension along with Python’s built-in combinations function from the itertools module to generate all possible subsets and computes their sums. It then checks if the target sum is in the set of sums.
Summary/Discussion
- Method 1: Recursive Backtracking. Strengths: Conceptually simple; explores all possibilities. Weaknesses: Highly inefficient for large arrays; exponential time complexity.
- Method 2: Dynamic Programming. Strengths: Optimized with polynomial time complexity; practical for larger datasets. Weaknesses: Requires additional memory for the DP table; might be complex to implement for beginners.
- Method 3: Greedy Algorithm with Sorting. Strengths: Fast and easy to understand. Weaknesses: Does not guarantee a solution; not a complete method.
- Method 4: Subset Sum Problem Transformation. Strengths: Turns the problem into a well-known computational problem; utilizes existing subset sum solutions. Weaknesses: Can become inefficient as the number of possible subsets grows exponentially with array size.
- Method 5: Python Set Operations. Strengths: Extremely concise and elegant; utilizes Python’s built-in capabilities. Weaknesses: Not efficient; set operations on all combinations is computationally expensive.