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Partition Array Into Two Arrays to Minimize Sum Difference

Hard

Asked by:

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Topics:

ArraysRecursionDynamic Programming

You are given an integer array nums of 2 * n integers. You need to partition nums into two arrays of length n to minimize the absolute difference of the sums of the arrays. To partition nums, put each element of nums into one of the two arrays.

Return the minimum possible absolute difference.

Example 1:

Input: nums = [3,9,7,3]
Output: 2
Explanation: One optimal partition is: [3,9] and [7,3].
The absolute difference between the sums of the arrays is abs((3 + 9) - (7 + 3)) = 2.

Example 2:

Input: nums = [-36,36]
Output: 72
Explanation: One optimal partition is: [-36] and [36].
The absolute difference between the sums of the arrays is abs((-36) - (36)) = 72.

Example 3:

Input: nums = [2,-1,0,4,-2,-9]
Output: 0
Explanation: One optimal partition is: [2,4,-9] and [-1,0,-2].
The absolute difference between the sums of the arrays is abs((2 + 4 + -9) - (-1 + 0 + -2)) = 0.

Constraints:

1 <= n <= 15
nums.length == 2 * n
-10⁷ <= nums[i] <= 10⁷

Try coding it in LeetCode

Partition Array Into Two Arrays to Minimize Sum Difference

Hard

Asked by:

120 views

Topics:

ArraysRecursionDynamic Programming

Return the minimum possible absolute difference.

Example 1:

Input: nums = [3,9,7,3]
Output: 2
Explanation: One optimal partition is: [3,9] and [7,3].
The absolute difference between the sums of the arrays is abs((3 + 9) - (7 + 3)) = 2.

Example 2:

Input: nums = [-36,36]
Output: 72
Explanation: One optimal partition is: [-36] and [36].
The absolute difference between the sums of the arrays is abs((-36) - (36)) = 72.

Example 3:

Input: nums = [2,-1,0,4,-2,-9]
Output: 0
Explanation: One optimal partition is: [2,4,-9] and [-1,0,-2].
The absolute difference between the sums of the arrays is abs((2 + 4 + -9) - (-1 + 0 + -2)) = 0.

Constraints:

1 <= n <= 15
nums.length == 2 * n
-10⁷ <= nums[i] <= 10⁷

Try coding it in LeetCode

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 are the constraints on the size of the input array `nums` and the range of integer values within it?
Can the input array `nums` contain negative numbers, zeros, or duplicate values?
Is it guaranteed that the input array `nums` will always have an even length and be non-empty?
If multiple partitions result in the same minimum sum difference, does it matter which one is chosen?
What should be returned if the input array is empty or has an odd length (although the problem statement says even length, it's good to confirm expected behavior for invalid inputs)?

Brute Force Solution

Approach

We want to divide a collection of numbers into two groups so that the sum of numbers in each group is as close as possible. The brute force way to do this is to consider every single possible way to split the original collection into two parts and then see which split results in the smallest difference between the two group sums.

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

Think about each number in the original collection.
For every number, decide if it will go into the first group or the second group.
Go through all the possible combinations of assigning each number to one of the two groups.
For each combination, calculate the total sum for the first group and the total sum for the second group.
Find the difference between these two sums.
Keep track of the smallest difference you've found so far.
After trying every single possible assignment for every number, the smallest difference you recorded is your answer.

Code Implementation

def partition_array_minimize_sum_difference_brute_force(original_numbers):
    number_of_elements = len(original_numbers)
    smallest_difference_found = float('inf')

    # Iterate through all possible combinations of assigning elements to two groups
    for i in range(1 << number_of_elements):
        first_group_sum = 0
        second_group_sum = 0

        # Determine which group each number belongs to based on the bitmask
        for bit_position in range(number_of_elements):
            if (i >> bit_position) & 1:
                first_group_sum += original_numbers[bit_position]
            else:
                second_group_sum += original_numbers[bit_position]

        # Calculate the absolute difference between the two group sums
        current_difference = abs(first_group_sum - second_group_sum)

        # Update the smallest difference if the current one is smaller
        smallest_difference_found = min(smallest_difference_found, current_difference)

    # This smallest difference represents the minimum achievable sum difference
    return smallest_difference_found

Big(O) Analysis

Time Complexity

O(2^n) – For each of the n numbers in the original collection, there are two independent choices: assign it to the first group or assign it to the second group. Since these choices are independent for each of the n numbers, the total number of possible assignments (combinations) is 2 multiplied by itself n times, which is 2^n. For each of these 2^n combinations, we perform a constant amount of work to sum the elements and calculate the difference. Therefore, the overall time complexity is dominated by the number of combinations, resulting in O(2^n).

Space Complexity

O(2^n) – The brute-force approach described involves iterating through all possible assignments of each of the N numbers to one of two groups. This leads to 2^n possible combinations. Implicitly, to explore these combinations, a recursive approach or an iterative approach using bit manipulation might be employed, where the call stack depth or an auxiliary data structure to store intermediate sums for each subset could grow up to 2^n in size. Therefore, the auxiliary space complexity is dominated by the need to store or manage these 2^n possibilities.

Optimal Solution

Approach

This problem involves splitting a collection of numbers into two groups so their sums are as close as possible. The key insight is to realize that finding the perfect split is equivalent to finding a subset of numbers that sums up to exactly half of the total sum, or as close to it as possible. We can then efficiently find this sum by exploring combinations.

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

First, calculate the total sum of all the numbers.
Our goal is to create two groups whose sums are as close as possible, meaning each group should ideally have a sum around half of the total sum.
Think about this as trying to find a subset of the original numbers that adds up to exactly half of the total sum, or the largest sum less than or equal to half the total sum that we can form.
We can systematically explore all possible combinations of numbers that can form a subset, keeping track of all the different sums we can achieve using these subsets.
Once we have all the achievable subset sums, we look for the one that is closest to, but not exceeding, half of the total sum.
If we find a subset that sums up to exactly half of the total sum, then the remaining numbers will automatically form the other group, and their sum will also be half the total, giving us a perfect split.
If we can't find a subset that sums up to exactly half, we choose the largest achievable subset sum that is less than half the total. The difference between the total sum and twice this chosen subset sum will be the minimum possible difference between the two groups.
This approach avoids trying every single way to split the numbers into two groups by focusing on finding a specific target sum for one of the groups.

Code Implementation

def minimum_difference(array_of_numbers):
    total_sum_of_elements = sum(array_of_numbers)
    target_sum_for_one_half = total_sum_of_elements // 2

    # Keep track of all achievable subset sums.
    achievable_sums = {0}

    for current_number in array_of_numbers:
        new_sums_to_add = set()
        for existing_sum in achievable_sums:
            new_sum = existing_sum + current_number
            new_sums_to_add.add(new_sum)
        achievable_sums.update(new_sums_to_add)

    # Find the largest achievable sum that is less than or equal to the target half sum.
    closest_sum_to_half = 0
    for subset_sum in achievable_sums:
        if subset_sum <= target_sum_for_one_half:
            closest_sum_to_half = max(closest_sum_to_half, subset_sum)

    # The minimum difference is the total sum minus twice the closest sum found.
    return total_sum_of_elements - 2 * closest_sum_to_half

Big(O) Analysis

Time Complexity

O(n * 2^n) – The core of the solution involves generating all possible subset sums. For an input array of size n, there are 2^n possible subsets. For each subset, we calculate its sum, which takes O(n) time in the worst case if we iterate through the elements for each subset. Therefore, generating all subset sums takes O(n * 2^n) time. Finding the closest sum to half the total sum within the generated sums takes at most O(2^n) time. The dominant factor is the generation of subset sums, resulting in an overall time complexity of O(n * 2^n).

Space Complexity

O(n*S) – The solution likely uses a data structure, such as a set or a boolean array, to store all achievable subset sums. If the total sum of the array elements is S, and there are N elements, the maximum number of distinct subset sums can be up to 2^N in the worst case, but practically bounded by N*S where S is half the total sum. This table/set stores intermediate results, hence contributing to the space complexity proportional to the number of achievable sums up to half of the total sum, which can be up to O(N * (TotalSum/2)).

Edge Cases

Case	How to Handle
Input array is null or empty	The problem statement implies an array of even length, so null or empty inputs should ideally be validated upfront to prevent runtime errors.
Input array has odd length	The problem statement guarantees an even length array; if this constraint is violated, the partitioning into equal halves is impossible.
Input array contains only two elements	This is the smallest valid input, and the solution should correctly partition these two elements into two arrays of length one.
All elements in the input array are identical	Regardless of the value, if all elements are the same, any partition will result in a sum difference of zero, which the algorithm should find.
Input array contains negative numbers	The algorithm should handle negative numbers correctly in sum calculations, as the goal is the absolute difference.
Input array contains zeros	Zeros do not affect the sum difference, and the algorithm should integrate them into partitions without special handling.
Large input array size (e.g., 2N elements where N is large)	The chosen algorithm's time complexity must be efficient enough to handle large inputs without exceeding typical time limits.
Integer overflow during sum calculation	Using a data type that can accommodate the maximum possible sum of elements is crucial to prevent overflow errors.

Input array is null or empty

How to Handle:

The problem statement implies an array of even length, so null or empty inputs should ideally be validated upfront to prevent runtime errors.

Input array has odd length

How to Handle:

The problem statement guarantees an even length array; if this constraint is violated, the partitioning into equal halves is impossible.

Input array contains only two elements

How to Handle:

This is the smallest valid input, and the solution should correctly partition these two elements into two arrays of length one.

All elements in the input array are identical

How to Handle:

Regardless of the value, if all elements are the same, any partition will result in a sum difference of zero, which the algorithm should find.

Input array contains negative numbers

How to Handle:

The algorithm should handle negative numbers correctly in sum calculations, as the goal is the absolute difference.

Input array contains zeros

How to Handle:

Zeros do not affect the sum difference, and the algorithm should integrate them into partitions without special handling.

Large input array size (e.g., 2N elements where N is large)

How to Handle:

The chosen algorithm's time complexity must be efficient enough to handle large inputs without exceeding typical time limits.

Integer overflow during sum calculation

How to Handle:

Using a data type that can accommodate the maximum possible sum of elements is crucial to prevent overflow errors.

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 are the constraints on the size of the input array `nums` and the range of integer values within it?
Can the input array `nums` contain negative numbers, zeros, or duplicate values?
Is it guaranteed that the input array `nums` will always have an even length and be non-empty?
If multiple partitions result in the same minimum sum difference, does it matter which one is chosen?
What should be returned if the input array is empty or has an odd length (although the problem statement says even length, it's good to confirm expected behavior for invalid inputs)?

Brute Force Solution

Approach

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

Think about each number in the original collection.
For every number, decide if it will go into the first group or the second group.
Go through all the possible combinations of assigning each number to one of the two groups.
For each combination, calculate the total sum for the first group and the total sum for the second group.
Find the difference between these two sums.
Keep track of the smallest difference you've found so far.
After trying every single possible assignment for every number, the smallest difference you recorded is your answer.

Code Implementation

def partition_array_minimize_sum_difference_brute_force(original_numbers):
    number_of_elements = len(original_numbers)
    smallest_difference_found = float('inf')

    # Iterate through all possible combinations of assigning elements to two groups
    for i in range(1 << number_of_elements):
        first_group_sum = 0
        second_group_sum = 0

        # Determine which group each number belongs to based on the bitmask
        for bit_position in range(number_of_elements):
            if (i >> bit_position) & 1:
                first_group_sum += original_numbers[bit_position]
            else:
                second_group_sum += original_numbers[bit_position]

        # Calculate the absolute difference between the two group sums
        current_difference = abs(first_group_sum - second_group_sum)

        # Update the smallest difference if the current one is smaller
        smallest_difference_found = min(smallest_difference_found, current_difference)

    # This smallest difference represents the minimum achievable sum difference
    return smallest_difference_found

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, calculate the total sum of all the numbers.
Our goal is to create two groups whose sums are as close as possible, meaning each group should ideally have a sum around half of the total sum.
Think about this as trying to find a subset of the original numbers that adds up to exactly half of the total sum, or the largest sum less than or equal to half the total sum that we can form.
We can systematically explore all possible combinations of numbers that can form a subset, keeping track of all the different sums we can achieve using these subsets.
Once we have all the achievable subset sums, we look for the one that is closest to, but not exceeding, half of the total sum.
If we find a subset that sums up to exactly half of the total sum, then the remaining numbers will automatically form the other group, and their sum will also be half the total, giving us a perfect split.
If we can't find a subset that sums up to exactly half, we choose the largest achievable subset sum that is less than half the total. The difference between the total sum and twice this chosen subset sum will be the minimum possible difference between the two groups.
This approach avoids trying every single way to split the numbers into two groups by focusing on finding a specific target sum for one of the groups.

Code Implementation

def minimum_difference(array_of_numbers):
    total_sum_of_elements = sum(array_of_numbers)
    target_sum_for_one_half = total_sum_of_elements // 2

    # Keep track of all achievable subset sums.
    achievable_sums = {0}

    for current_number in array_of_numbers:
        new_sums_to_add = set()
        for existing_sum in achievable_sums:
            new_sum = existing_sum + current_number
            new_sums_to_add.add(new_sum)
        achievable_sums.update(new_sums_to_add)

    # Find the largest achievable sum that is less than or equal to the target half sum.
    closest_sum_to_half = 0
    for subset_sum in achievable_sums:
        if subset_sum <= target_sum_for_one_half:
            closest_sum_to_half = max(closest_sum_to_half, subset_sum)

    # The minimum difference is the total sum minus twice the closest sum found.
    return total_sum_of_elements - 2 * closest_sum_to_half

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input array is null or empty	The problem statement implies an array of even length, so null or empty inputs should ideally be validated upfront to prevent runtime errors.
Input array has odd length	The problem statement guarantees an even length array; if this constraint is violated, the partitioning into equal halves is impossible.
Input array contains only two elements	This is the smallest valid input, and the solution should correctly partition these two elements into two arrays of length one.
All elements in the input array are identical	Regardless of the value, if all elements are the same, any partition will result in a sum difference of zero, which the algorithm should find.
Input array contains negative numbers	The algorithm should handle negative numbers correctly in sum calculations, as the goal is the absolute difference.
Input array contains zeros	Zeros do not affect the sum difference, and the algorithm should integrate them into partitions without special handling.
Large input array size (e.g., 2N elements where N is large)	The chosen algorithm's time complexity must be efficient enough to handle large inputs without exceeding typical time limits.
Integer overflow during sum calculation	Using a data type that can accommodate the maximum possible sum of elements is crucial to prevent overflow errors.

Input array is null or empty

How to Handle:

The problem statement implies an array of even length, so null or empty inputs should ideally be validated upfront to prevent runtime errors.

Input array has odd length

How to Handle:

The problem statement guarantees an even length array; if this constraint is violated, the partitioning into equal halves is impossible.

Input array contains only two elements

How to Handle:

This is the smallest valid input, and the solution should correctly partition these two elements into two arrays of length one.

All elements in the input array are identical

How to Handle:

Regardless of the value, if all elements are the same, any partition will result in a sum difference of zero, which the algorithm should find.

Input array contains negative numbers

How to Handle:

The algorithm should handle negative numbers correctly in sum calculations, as the goal is the absolute difference.

Input array contains zeros

How to Handle:

Zeros do not affect the sum difference, and the algorithm should integrate them into partitions without special handling.

Large input array size (e.g., 2N elements where N is large)

How to Handle:

The chosen algorithm's time complexity must be efficient enough to handle large inputs without exceeding typical time limits.

Integer overflow during sum calculation

How to Handle:

Using a data type that can accommodate the maximum possible sum of elements is crucial to prevent overflow errors.