# 5 Best Ways to Check for Equal Sum Partitions in Python

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π‘ Problem Formulation: A common problem in computer science is determining whether an array can be divided into two subsets with equal sums. This is relevant, for instance, in balancing loads or partitioning datasets. Given an input array, `[1, 5, 11, 5]`, we want to find out if it’s possible to split it into two groups, where the sum of the elements in each group is the same. In this case, the output should be `True`, since the array can be partitioned into `[1, 5, 5]` and `[11]`, both summing to `11`.

## Method 1: Recursive Approach

A recursive approach to the partition problem explores all possible combinations by including or excluding each element. It checks if there exists a subset with the sum equal to half the total sum of the array, since two equal partitions would each sum up to half of the total.

Here’s an example:

```def can_partition_recursively(arr, n, sum):
if sum == 0:
return True
if n == 0 and sum != 0:
return False
if arr[n-1] > sum:
return can_partition_recursively(arr, n-1, sum)
return can_partition_recursively(arr, n-1, sum) or can_partition_recursively(arr, n-1, sum-arr[n-1])

arr = [1, 5, 11, 5]
n = len(arr)
sum = sum(arr)
print(can_partition_recursively(arr, n, sum//2) if sum % 2 == 0 else False)```

Output: `True`

The function `can_partition_recursively()` checks whether we can partition the list into two groups with equal sum using a recursive approach. The base cases handle the termination conditions, and the recursive calls consider both possibilities for each element: being included in the subset or not. This example assumes the total sum is even, as two equal groups would not be possible otherwise.

## Method 2: Dynamic Programming – Top-Down Approach

The top-down dynamic programming approach (“memoization”) reduces the redundant calculations of the recursive approach by storing the results of subproblems. It dramatically improves efficiency, especially for larger arrays.

Here’s an example:

```def can_partition_memoization(arr):
total_sum = sum(arr)
if total_sum % 2 != 0:
return False
n = len(arr)
memo = [[-1 for _ in range(total_sum//2 + 1)] for _ in range(n + 1)]
def recurse(i, current_sum):
if current_sum == 0:
return True
if i < 0 or current_sum < 0:
return False
if memo[i][current_sum] != -1:
return memo[i][current_sum]
memo[i][current_sum] = recurse(i - 1, current_sum) or recurse(i - 1, current_sum - arr[i])
return memo[i][current_sum]
return recurse(n - 1, total_sum // 2)

arr = [1, 5, 11, 5]
print(can_partition_memoization(arr))```

Output: `True`

The function `can_partition_memoization()` uses a 2D array, `memo`, to store the result of subproblems and avoid recalculating them. It follows the same logic as the recursive approach but is more efficient thanks to memoization.

## Method 3: Dynamic Programming – Bottom-Up Approach

The bottom-up dynamic programming approach constructs the solution iteratively and is memory efficient since it does not require a recursive call stack.

Here’s an example:

```def can_partition_bottom_up(arr):
total_sum = sum(arr)
if total_sum % 2 != 0:
return False
subset_sum = total_sum // 2
n = len(arr)
dp = [[False for _ in range(subset_sum + 1)] for _ in range(n + 1)]
for i in range(n + 1):
dp[i][0] = True
for i in range(1, n + 1):
for j in range(1, subset_sum + 1):
if j < arr[i - 1]:
dp[i][j] = dp[i - 1][j]
else:
dp[i][j] = dp[i - 1][j] or dp[i - 1][j - arr[i - 1]]
return dp[n][subset_sum]

arr = [1, 5, 11, 5]
print(can_partition_bottom_up(arr))```

Output: `True`

The bottom-up approach, implemented in `can_partition_bottom_up()`, iteratively builds up a solution using a 2D array, `dp`. Each cell `dp[i][j]` is filled with `True` if the subset `{arr[0], ..., arr[i]}` has a subset with sum `j`, else `False`.

## Method 4: Using Bit Manipulation

Bit manipulation leverages bitwise operations to solve the partition problem with a clever approach that treats set membership as binary flags.

Here’s an example:

```def can_partition_bitwise(arr):
total_sum = sum(arr)
if total_sum % 2 != 0:
return False
subset_sum = total_sum // 2
possible_sums = 1
for num in arr:
possible_sums |= possible_sums <> subset_sum) & 1

arr = [1, 5, 11, 5]
print(can_partition_bitwise(arr))```

Output: `True`

In this method, `can_partition_bitwise()`, we use a bit vector `possible_sums` where each bit represents if a particular sum is possible with the subset. Through left shifts and bitwise OR operations, we populate this vector. Finally, we check whether the bit at position `subset_sum` is set, indicating whether the target subset sum is possible.

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

A succinct solution leveraging a powerful Python library that provides functionality for managing mathematical combinatorics and set operations (example with made-up library function).

Here’s an example:

```from set_combinatorics import can_split_to_equal_sum_subsets

arr = [1, 5, 11, 5]
print(can_split_to_equal_sum_subsets(arr))```

Output: `True`

This one-liner example assumes there’s a hypothetical library called `set_combinatorics` with a function `can_split_to_equal_sum_subsets()` that directly checks if the array can be split into two subsets with equal sums. This is a simple, clean, and highly abstracted solution.

## Summary/Discussion

• Method 1: Recursive Approach. Can be conceptually simple but may lead to exponential time complexity for large inputs.
• Method 2: Dynamic Programming – Top-Down. More efficient than recursive for big data due to memoization. It can still consume a significant amount of memory.
• Method 3: Dynamic Programming – Bottom-Up. Efficient and memory-conserving. Less intuitive than the recursive method and requires understanding of dynamic programming.
• Method 4: Using Bit Manipulation. Fast and memory-efficient. However, it can be harder to understand, and the approach is somewhat limited to integers.
• Bonus Method 5: Python Library Usage. The simplest implementation but relies on the availability of an external library that might not exist or be incomplete.