Partition Array Into Two Arrays to Minimize Sum Difference

Hard

Partition Array Into Two Arrays to Minimize Sum Difference

Hard

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum size of the input array `nums`?
Can the integers in `nums` be negative, positive, or zero?
Are duplicate numbers allowed in the input array `nums`?
Do the two resulting arrays need to have the same number of elements, or is any partition valid?
Can you provide an example input and the expected output to illustrate the desired behavior?

Brute Force Solution

Approach

The main idea is to explore all possible ways to divide the numbers into two groups. We find the best split by exhaustively evaluating the difference between the sums of the two groups for every possible arrangement. This ensures we don't miss any potential solutions.

Here's how the algorithm would work step-by-step:

Imagine we create two empty buckets to hold our numbers.
Start by picking a subset of the numbers and placing them in the first bucket. All the remaining numbers go into the second bucket.
Calculate the sum of the numbers in the first bucket and the sum of the numbers in the second bucket.
Find the difference between these two sums.
Remember this difference, as it represents how balanced this particular split is.
Now, try a different combination of numbers in the first bucket. For example, add one number, remove a number, or completely change the numbers in the first bucket.
Again, calculate the sums of both buckets and the difference between them.
Keep doing this for absolutely every possible combination of numbers you can put in the first bucket. This means trying every single possibility, without skipping any.
After you've checked every single combination, look at all the differences you've remembered.
Choose the combination that gave you the smallest difference. This is the closest you can get to having the two buckets have the same total value.

Code Implementation

def partition_array_brute_force(numbers):
    minimum_difference = float('inf')
    best_partition = None
    array_length = len(numbers)

    # Iterate through all possible subsets of the array
    for i in range(2**array_length):
        first_subset = []
        second_subset = []

        # Construct the subsets based on the bits of 'i'
        for j in range(array_length):
            if (i >> j) & 1:
                first_subset.append(numbers[j])
            else:
                second_subset.append(numbers[j])

        # Calculate the sums of the two subsets
        first_subset_sum = sum(first_subset)
        second_subset_sum = sum(second_subset)

        # Calculate the absolute difference between the sums
        current_difference = abs(first_subset_sum - second_subset_sum)

        # Update minimum difference if a better partition is found
        # This conditional statment holds the key logic.
        if current_difference < minimum_difference:

            minimum_difference = current_difference
            best_partition = (first_subset, second_subset)

    return minimum_difference

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores all possible subsets of the input array of size n. This is equivalent to generating the power set of the array. A power set contains 2^n subsets. For each subset, the algorithm calculates the sum of the elements in the subset and the sum of the remaining elements. Therefore, the time complexity is directly proportional to the number of subsets, which is 2^n.

Space Complexity

O(2^N) – The described solution explores all possible subsets of the input array of size N. Implicitly, this requires generating and storing information about each subset. Considering the plain English explanation uses words like 'remember this difference' and 'checked every single combination', this indicates a need to store the differences for all 2^N possible subsets. Therefore, the space complexity is O(2^N), as the number of possible combinations grows exponentially with the input size.

Optimal Solution

Approach

The goal is to divide a group of numbers into two smaller groups so that the sums of each group are as close as possible. The optimal approach involves considering all possible combinations of half the numbers and cleverly using these combinations to find the closest sums without checking every possible split.

Here's how the algorithm would work step-by-step:

First, let's calculate the total sum of all the numbers.
Then, think about all possible combinations of picking half of the numbers. For example, if you have 6 numbers, consider all combinations of picking 3.
For each of these combinations, calculate the sum of the chosen numbers. Remember this sum.
Since we want the two groups' sums to be as close as possible, we are searching for a sum that is closest to half of the total sum.
Go through all the sums you calculated from the combinations. Find the sum that is closest to half of the total sum. This represents the sum of one of the groups.
The sum of the other group is simply the total sum minus the sum of the first group. The difference between these two group sums is the smallest possible difference you can achieve.
Choosing the combination that resulted in that closest sum gives you one of the two groups. The remaining numbers form the other group.

Code Implementation

def min_abs_difference(array):
    total_sum = sum(array)
    number_of_elements = len(array)
    half_length = number_of_elements // 2

    possible_subset_sums = set()

    def calculate_subset_sums(index, count, current_sum):
        if count == half_length:
            possible_subset_sums.add(current_sum)
            return
        if index == number_of_elements:
            return

        # Explore including the current element.
        calculate_subset_sums(index + 1, count + 1, current_sum + array[index])

        # Explore excluding the current element.
        calculate_subset_sums(index + 1, count, current_sum)

    calculate_subset_sums(0, 0, 0)

    minimum_difference = float('inf')

    # We iterate through all possible sums of subsets.
    for subset_sum in possible_subset_sums:
        # Calculating the sum of the other subset.
        other_subset_sum = total_sum - subset_sum
        difference = abs(subset_sum - other_subset_sum)

        # Determine the minimum difference.
        minimum_difference = min(minimum_difference, difference)

    return minimum_difference

Big(O) Analysis

Time Complexity

O(C(n, n/2)) which simplifies to O(2^n) in the worst case – The dominant operation is generating combinations of n/2 elements from the n input elements. The number of such combinations is given by the binomial coefficient C(n, n/2), also denoted as "n choose n/2". In the worst-case, calculating and iterating through these combinations takes exponential time. While calculating the sum for each combination is O(n/2) or O(n), it's less significant than generating the combinations. Thus the overall time complexity is dominated by the combination generation, approximating O(C(n, n/2)), and in the worst-case scenario, this behaves like O(2^n).

Space Complexity

O(2^(N/2)) – The algorithm considers all possible combinations of picking half of the numbers (N/2). It implicitly stores the sums of these combinations. In the worst case, it may need to store all possible sums of combinations, leading to a space complexity proportional to the number of such combinations. The number of combinations of picking N/2 items from N items is approximately 2^(N/2), which dominates the auxiliary space requirement. Therefore, the auxiliary space used is O(2^(N/2)).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 as the minimum difference when the array is null or empty, since there are no elements to partition.
Array with a single element	Return the absolute value of the single element, as one partition contains that element and the other is empty.
Array with two elements	Return the absolute difference between the two elements, as these will form the two partitions.
Array containing all zeros	Return 0, as both partitions will have a sum of 0.
Array with extremely large positive and negative numbers that could cause integer overflow	Use long data type to store intermediate sums to prevent potential integer overflows.
Array containing duplicate numbers	The algorithm should handle duplicate numbers correctly, as each number contributes to the overall sum and is considered in the dynamic programming or other partitioning approach.
Array with a large number of elements (scaling issue)	Choose an efficient algorithm like dynamic programming with space optimization to handle large input sizes within reasonable time and memory constraints.
Array containing all identical values	The minimum difference will always be 0, regardless of how the array is partitioned, so the algorithm should handle this efficiently.