Problem Description

You are given an array nums containing distinct positive integers and an integer target.

Your task is to determine if you can split the array into two non-empty groups where:

• Every element from nums must belong to exactly one group
• The two groups cannot share any elements (disjoint)
• The product of all elements in the first group equals target
• The product of all elements in the second group also equals target

Return true if such a partition is possible, otherwise return false.

Example walkthrough:

• If nums = [2, 3, 4, 6] and target = 12
• One possible partition: Group 1 = [2, 6] with product = 2 × 6 = 12, Group 2 = [3, 4] with product = 3 × 4 = 12
• Since both groups have product equal to 12 (the target), return true

Key constraints:

• All numbers in nums are distinct and positive
• Both groups must be non-empty (each must contain at least one element)
• Each element must be used exactly once

The solution uses binary enumeration to explore all possible ways to partition the array. For each partition configuration represented by a binary number i, elements are assigned to either the first group (if the corresponding bit is 1) or the second group (if the bit is 0). The products of both groups are calculated and checked against the target value.

Intuition

Since we need to partition the array into two groups where both groups have the same product equal to target, we need to explore all possible ways to divide the elements between the two groups.

Think of it this way: for each element in the array, we have a binary choice - it either goes into group 1 or group 2. This naturally leads us to think about binary representation, where each bit position corresponds to an element's group assignment.

For an array of size n, there are 2^n total ways to assign elements to two groups. We can represent each partition using a number from 0 to 2^n - 1. In the binary representation of this number:

• If bit j is 1, element nums[j] goes to group 1
• If bit j is 0, element nums[j] goes to group 2

For example, with 4 elements, the number 5 (binary: 0101) means elements at indices 0 and 2 go to group 1, while elements at indices 1 and 3 go to group 2.

Why does this brute force approach work here? The key insight is that:

1. The array size is typically small enough for 2^n iterations to be feasible
2. We need to check a specific condition (both products equal to target) rather than optimize something
3. There's no obvious greedy strategy or dynamic programming state that would help us avoid checking all possibilities

The beauty of this approach is its simplicity - we systematically check every possible partition using bit manipulation (i >> j & 1 checks if bit j is set in number i), calculate the products of both groups, and return true as soon as we find a valid partition where both products equal the target.

Learn more about Recursion patterns.

Solution Approach

The solution uses binary enumeration to systematically check all possible subset partitions. Here's how the implementation works:

1. Iteration through all partitions: We iterate through all numbers from 0 to 2^n - 1 using for i in range(1 << n), where 1 << n equals 2^n. Each value of i represents a unique way to partition the array.

2. Binary representation for partition assignment: For each partition state i, we check each bit position j from 0 to n-1:

• i >> j & 1 extracts the j-th bit of i
• If this bit is 1, we assign nums[j] to the first subset (multiply with x)
• If this bit is 0, we assign nums[j] to the second subset (multiply with y)

3. Product calculation: We initialize both products x and y to 1. As we iterate through each element:

if i >> j & 1:
    x *= nums[j]  # Element goes to first subset
else:
    y *= nums[j]  # Element goes to second subset

4. Validation check: After processing all elements for a given partition, we check if both products equal the target:

if x == target and y == target:
    return True

5. Edge cases handled automatically:

• Empty subsets are naturally avoided because when i = 0 (all elements in second subset) or i = 2^n - 1 (all elements in first subset), one product will be 1 while the other will be the product of all elements
• These cases will only return true if target = 1 and the other product also equals 1

Time Complexity: O(n × 2^n) - We check 2^n partitions and for each partition, we iterate through n elements.

Space Complexity: O(1) - We only use a constant amount of extra space for variables x and y.

Example Walkthrough

Let's walk through the solution with nums = [2, 3, 4] and target = 6.

We need to check all possible ways to partition these 3 elements into two groups. With 3 elements, we have 2^3 = 8 possible partitions to check.

Let's trace through each iteration:

i = 0 (binary: 000):

• All bits are 0, so all elements go to group 2
• Group 1: empty → product = 1
• Group 2: [2, 3, 4] → product = 24
• Check: 1 ≠ 6 and 24 ≠ 6 → Continue

i = 1 (binary: 001):

• Bit 0 is 1, so nums[0]=2 goes to group 1
• Group 1: [2] → product = 2
• Group 2: [3, 4] → product = 12
• Check: 2 ≠ 6 and 12 ≠ 6 → Continue

i = 2 (binary: 010):

• Bit 1 is 1, so nums[1]=3 goes to group 1
• Group 1: [3] → product = 3
• Group 2: [2, 4] → product = 8
• Check: 3 ≠ 6 and 8 ≠ 6 → Continue

i = 3 (binary: 011):

• Bits 0 and 1 are 1, so nums[0]=2 and nums[1]=3 go to group 1
• Group 1: [2, 3] → product = 6
• Group 2: [4] → product = 4
• Check: 6 = 6 ✓ but 4 ≠ 6 → Continue

i = 4 (binary: 100):

• Bit 2 is 1, so nums[2]=4 goes to group 1
• Group 1: [4] → product = 4
• Group 2: [2, 3] → product = 6
• Check: 4 ≠ 6 but 6 = 6 → Continue

i = 5 (binary: 101):

• Bits 0 and 2 are 1, so nums[0]=2 and nums[2]=4 go to group 1
• Group 1: [2, 4] → product = 8
• Group 2: [3] → product = 3
• Check: 8 ≠ 6 and 3 ≠ 6 → Continue

i = 6 (binary: 110):

• Bits 1 and 2 are 1, so nums[1]=3 and nums[2]=4 go to group 1
• Group 1: [3, 4] → product = 12
• Group 2: [2] → product = 2
• Check: 12 ≠ 6 and 2 ≠ 6 → Continue

i = 7 (binary: 111):

• All bits are 1, so all elements go to group 1
• Group 1: [2, 3, 4] → product = 24
• Group 2: empty → product = 1
• Check: 24 ≠ 6 and 1 ≠ 6 → Continue

After checking all 8 partitions, we found no valid partition where both groups have product equal to 6. Therefore, the function returns false.

Note how the bit manipulation i >> j & 1 efficiently determines group assignment: for i=5 (binary: 101), checking bit positions gives us 1, 0, 1 from right to left, placing elements [2, 4] in group 1 and [3] in group 2.

Solution Implementation

Time and Space Complexity

Time Complexity: O(2^n × n)

The algorithm uses bit manipulation to iterate through all possible partitions of the array into two subsets. The outer loop runs 2^n times (from 0 to 2^n - 1), representing all possible subset combinations. For each iteration, the inner loop runs n times to check each element and decide which subset it belongs to based on the corresponding bit value. Therefore, the total time complexity is O(2^n × n).

Space Complexity: O(1)

The algorithm only uses a constant amount of extra space. The variables i, j, x, and y are the only additional variables created, and they don't depend on the input size n. No additional data structures like arrays or recursion stacks are used, resulting in constant space complexity.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow with Large Products

The most critical pitfall in this solution is that products can grow exponentially large, leading to integer overflow issues. When multiplying many numbers together, the result can exceed standard integer limits, causing incorrect comparisons.

Example scenario:

• nums = [100, 200, 300, 400, 500]
• Products can reach values like 100 × 200 × 300 = 6,000,000
• With more elements, products can easily exceed typical integer bounds

Solution: Instead of computing products directly, use logarithms to convert multiplication to addition:

import math

def checkEqualPartitions(self, nums: List[int], target: int) -> bool:
    n = len(nums)
    log_target = math.log(target)
  
    for partition_mask in range(1 << n):
        log_subset1 = 0
        log_subset2 = 0
      
        for element_index in range(n):
            if (partition_mask >> element_index) & 1:
                log_subset1 += math.log(nums[element_index])
            else:
                log_subset2 += math.log(nums[element_index])
      
        # Use epsilon for floating-point comparison
        epsilon = 1e-9
        if abs(log_subset1 - log_target) < epsilon and abs(log_subset2 - log_target) < epsilon:
            return True
  
    return False

2. Early Termination Optimization Missing

The current solution always checks all 2^n partitions even when an early partition might already exceed the target. This wastes computational resources.

Solution: Add early termination when a product exceeds the target:

for partition_mask in range(1 << n):
    subset1_product = 1
    subset2_product = 1
  
    for element_index in range(n):
        if (partition_mask >> element_index) & 1:
            subset1_product *= nums[element_index]
            if subset1_product > target:  # Early termination
                break
        else:
            subset2_product *= nums[element_index]
            if subset2_product > target:  # Early termination
                break
  
    if subset1_product == target and subset2_product == target:
        return True

3. Inefficient for Special Cases

The solution doesn't handle special cases efficiently, such as when the target cannot be formed by any subset due to prime factorization constraints.

Solution: Pre-check if the target can be factorized using the given numbers:

def checkEqualPartitions(self, nums: List[int], target: int) -> bool:
    # First check if target^2 can be formed by the product of all numbers
    total_product = 1
    for num in nums:
        total_product *= num
  
    if total_product != target * target:
        return False  # Impossible to form two groups with product = target
  
    # Continue with the original algorithm...