# 5 Best Ways to Check if It’s Possible to Partition in K Subarrays with Equal Sum in Python

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π‘ Problem Formulation: We need to determine whether a given array can be partitioned into `k` subarrays such that each subarray has the same sum. This is a relevant problem in optimization and division of resources. For instance, given the array `[4, 3, 2, 3, 5, 2, 1]` and `k = 4`, we are looking for a way to check if the array can be split into four groups with equal sums.

## Method 1: Recursive Backtracking

This method involves using a recursive function that attempts to allocate each element to one of the partitions while ensuring the sum criteria are met. The recursion backtracks whenever it finds an allocation that doesn’t meet the required conditions. It tries all possible combination until it finds a solution or explores all possibilities without success.

Here’s an example:

```def can_partition(nums, k):
target, rem = divmod(sum(nums), k)
if rem: return False
nums.sort(reverse=True)
subsums = [0] * k

def backtrack(i):
if i == len(nums): return all(s == target for s in subsums)
for j in range(k):
if subsums[j] + nums[i] <= target:
subsums[j] += nums[i]
if backtrack(i + 1): return True
subsums[j] -= nums[i]
if subsums[j] == 0: break
return False

return backtrack(0)

print(can_partition([4, 3, 2, 3, 5, 2, 1], 4))```

Output:

`False`

This Python function, `can_partition()`, checks if an array can be split into ‘k’ subarrays with equal sums using recursive backtracking. First, it checks if the sum of the numbers is divisible by ‘k’, then sorts the list in reverse order to start with large numbers, which is a common optimization in backtracking algorithms. The recursive `backtrack()` function goes through each number, trying to place it in each subarray and backtracks if a number does not lead to a successful partition, ultimately returning ‘True’ if it finds a valid partition, or ‘False’ otherwise.

## Method 2: Dynamic Programming (Top Down)

Dynamic programming is an efficient way to solve problems by breaking them down into smaller subproblems and storing results to avoid repeated calculations. The “Top Down” approach, also called memoization, starts from the maximum complexity and reduces it through recursive calls, storing intermediate results to speed up the overall process.

Here’s an example:

```def can_partition(nums, k):
target, rem = divmod(sum(nums), k)
if rem: return False
nums.sort()
memo = {}

def dp(i, subset_sums):
if i == len(nums): return all(s == target for s in subset_sums)
state = tuple(subset_sums)
if state in memo: return memo[state]
for j in range(k):
if subset_sums[j] + nums[i] <= target:
subset_sums[j] += nums[i]
if dp(i + 1, subset_sums):
memo[state] = True
return True
subset_sums[j] -= nums[i]
if subset_sums[j] == 0: break
memo[state] = False
return False

return dp(0, [0] * k)

print(can_partition([4, 3, 2, 3, 5, 2, 1], 4))```

Output:

`False`

In this example, `can_partition()` is a function that uses top-down dynamic programming to solve the partition problem. First, it defines the target sum for each subarray and sorts the input numbers. The recursive function `dp()` works similarly to the backtracking function but uses a `memo` dictionary to remember outcomes of subproblems, avoiding redundant computations for the same subproblem. If a beneficial combination was already explored, the stored result is returned directly.

## Method 3: Iterative Dynamic Programming

This iterative approach to dynamic programming builds a solution bottom-up by iteratively applying a set of rules to modify the state of the problem, hence avoiding the need for recursion. As it fills out a table or array, it only remembers the necessary state of the previous step to resolve the current one.

Here’s an example:

```def can_partition(nums, k):
total = sum(nums)
if total % k: return False
target = total // k
dp = [False] * (1 << len(nums))
dp[0] = True
current_sums = [0] * (1 << len(nums))

for state in range(1 << len(nums)):
for j, num in enumerate(nums):
if (state & (1 << j)) != 0 and not dp[state]:
prev_state = state ^ (1 << j)
if current_sums[prev_state] + num <= target:
current_sums[state] = (current_sums[prev_state] + num) % target
dp[state] = True
return dp[-1]

print(can_partition([4, 3, 2, 3, 5, 2, 1], 4))```

Output:

`False`

The `can_partition()` function here represents an iterative dynamic programming solution. It leverages bitmasking to represent the inclusion and exclusion of elements in subsets, iteratively fills a `dp` array to track the valid subarrays found, and uses a `current_sums` array to hold the sum of the elements in each subset. If the final state in `dp` is True, it indicates a successful division into `k` partitions with equal sums.

## Method 4: Using a Bitmask and Recursive Function Calls

This strategy involves using a bitmask to represent the selected subsets. A recursive function then tries to fill these subsets one by one checking against the total sum that needs to be achieved. If at any point the current subset sum exceeds the target value, the function backtracks.

Here’s an example:

```def can_partition(nums, k):
if sum(nums) % k != 0: return False

visited = [0] * len(nums)
target = sum(nums) // k

def backtrack(start_index, k, accumulated_sum):
if k == 0: return True
if accumulated_sum == target: return backtrack(0, k-1, 0)

for i in range(start_index, len(nums)):
if visited[i] or accumulated_sum + nums[i] > target: continue
visited[i] = 1
if backtrack(i + 1, k, accumulated_sum + nums[i]): return True
visited[i] = 0
return False

return backtrack(0, k, 0)

print(can_partition([4, 3, 2, 3, 5, 2, 1], 4))```

Output:

`False`

This variation on the partition problem uses a bitmask (represented by the `visited` list) along with recursive calls to assign each element to a partition. Upon finding a complete partition, the function moves to the next partition and continues to attempt to fill it. Early exit conditions are when an element exceeds the sum for a partition or all partitions are filled correctly.

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

For an out-of-the-box solution to this problem, Python’s library functions or third-party libraries could be considered, though none directly handle this specific partition problem as a one-liner. The itertools library offers combinations and permutations that could be combined cleverly to create an exhaustive search.

Here’s an example:

Unfortunately, there is no clean one-liner method using Python’s standard libraries to partition an array into subarrays with equal sums as this is a specific combinatorial problem. However, third-party libraries like NumPy or Pandas may assist in processing the array elements, and exhaustive search strategies can be implemented more concisely using iterators from the itertools library.

## Summary/Discussion

• Method 1: Recursive Backtracking. This method has the strength of being intuitive and the solution is built incrementally. However, it can be inefficient for large inputs due to the heavy use of recursion and potential for many recursive calls.
• Method 2: Dynamic Programming (Top Down). The top-down DP approach is more efficient than simple backtracking and reduces redundancy through memoization. The downside is the potential for large memory usage due to storing states.
• Method 3: Iterative Dynamic Programming. A bottom-up approach that is typically more space-efficient and can be faster than recursion. On the other hand, the iterative approach can be harder to conceptualize and requires a detailed understanding of state transitions.
• Method 4: Bitmask and Recursive Function Calls. This strategy uses a clear and logical approach of filling up subsets one at a time, but like other recursive methods, might encounter issues with recursion depth and inefficiency for large datasets.
• Bonus Method 5. While an exact one-liner solution does not exist within built-in libraries, leveraging Python’s extensive library ecosystem can reduce the complexity and verbosity of your solution. The biggest weakness is that it would most likely not be a pure one-liner and might only offer a modest improvement in terms of reducing custom code.