Minimum Difference in Sums After Removal of Elements

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ArraysGreedy AlgorithmsDynamic Programming

You are given a 0-indexed integer array nums consisting of 3 * n elements.

You are allowed to remove any subsequence of elements of size exactly n from nums. The remaining 2 * n elements will be divided into two equal parts:

The first n elements belonging to the first part and their sum is sum_first.
The next n elements belonging to the second part and their sum is sum_second.

The difference in sums of the two parts is denoted as sum_first - sum_second.

For example, if sum_first = 3 and sum_second = 2, their difference is 1.
Similarly, if sum_first = 2 and sum_second = 3, their difference is -1.

Return the minimum difference possible between the sums of the two parts after the removal of n elements.

Example 1:

Input: nums = [3,1,2]
Output: -1
Explanation: Here, nums has 3 elements, so n = 1. 
Thus we have to remove 1 element from nums and divide the array into two equal parts.
- If we remove nums[0] = 3, the array will be [1,2]. The difference in sums of the two parts will be 1 - 2 = -1.
- If we remove nums[1] = 1, the array will be [3,2]. The difference in sums of the two parts will be 3 - 2 = 1.
- If we remove nums[2] = 2, the array will be [3,1]. The difference in sums of the two parts will be 3 - 1 = 2.
The minimum difference between sums of the two parts is min(-1,1,2) = -1.

Example 2:

Input: nums = [7,9,5,8,1,3]
Output: 1
Explanation: Here n = 2. So we must remove 2 elements and divide the remaining array into two parts containing two elements each.
If we remove nums[2] = 5 and nums[3] = 8, the resultant array will be [7,9,1,3]. The difference in sums will be (7+9) - (1+3) = 12.
To obtain the minimum difference, we should remove nums[1] = 9 and nums[4] = 1. The resultant array becomes [7,5,8,3]. The difference in sums of the two parts is (7+5) - (8+3) = 1.
It can be shown that it is not possible to obtain a difference smaller than 1.

Constraints:

nums.length == 3 * n
1 <= n <= 10⁵
1 <= nums[i] <= 10⁵

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`? Specifically, what is the maximum number of elements it can contain?
Can the integers in the input array `nums` be negative, positive, or zero?
Are there any duplicate numbers allowed in the input array `nums`?
What should I return if, after removing n elements, it is impossible to divide the remaining elements into two groups of size n such that the absolute difference of their sums is minimized? Should I return a specific error value or throw an exception?
Is `n` guaranteed to be an integer, and is it always equal to `nums.length / 3`?

Brute Force Solution

Approach

The brute force approach explores every possible combination of removing elements from the original collection. It's like trying every single way to pick and choose elements until we find the best possible arrangement based on the difference in sums. We want to find the removal of elements which result in a minimum difference between the sums of the two remaining groups.

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

Consider all possible groups of elements that could be removed from the original set.
For each of these removal groups, calculate the sum of the elements that are left in the collection after the removal.
Divide the remaining elements into two equal-sized groups (or as close to equal as possible if there are an odd number of elements left). Calculate the sum of each of these two subgroups.
Find the difference between the sums of the two subgroups.
Remember this difference and compare it to the smallest difference you've seen so far. Update the smallest difference if the current difference is smaller.
After examining all possible removal groups, the smallest difference you've remembered is the answer.

Code Implementation

def minimum_difference_in_sums_after_removal(numbers):
    number_of_elements = len(numbers)
    minimum_absolute_difference = float('inf')

    for i in range(2 ** number_of_elements):
        removed_elements = []
        remaining_elements = []

        for j in range(number_of_elements):
            if (i >> j) & 1:
                removed_elements.append(numbers[j])
            else:
                remaining_elements.append(numbers[j])

        # We proceed only if the number of removed elements is number_of_elements // 3
        if len(removed_elements) == number_of_elements // 3:

            remaining_elements.sort()
            group1 = remaining_elements[:len(remaining_elements) // 2]
            group2 = remaining_elements[len(remaining_elements) // 2:]

            # Calculate the sum of each of the two subgroups
            sum_group1 = sum(group1)
            sum_group2 = sum(group2)

            current_absolute_difference = abs(sum_group1 - sum_group2)

            # Update the minimum difference if necessary
            minimum_absolute_difference = min(minimum_absolute_difference, current_absolute_difference)

    return minimum_absolute_difference

Big(O) Analysis

Time Complexity

O(3^n) – The brute force approach iterates through all possible subsets of the input array of size n to determine which elements to remove. There are 2^n possible subsets. For each removed subset, we need to divide the remaining elements into two groups and find the minimum difference. Finding the optimal division of the remaining elements can take exponential time as well but it is dominated by the initial subset generation. Therefore, we iterate through each subset to potentially remove, and in the worst case, we need to consider the cases of keeping either n/3 elements to the left or the right of the remaining array which are 2^n also dominating the remaining elements division. Due to the overlapping possibilities, the total time complexity approximates O(3^n).

Space Complexity

O(1) – The brute force approach described explores all possible combinations by iteratively calculating sums and differences. It maintains a variable to store the minimum difference found so far. No auxiliary data structures that scale with the input size N (the number of elements in the original collection) are created, thus the extra space used is constant. The algorithm only uses a fixed number of variables to store intermediate sums and the minimum difference.

Optimal Solution

Approach

The key idea is to find the best way to split the numbers into two groups after removing some elements. We'll use a clever trick of thinking about what must always be true for any possible split to avoid checking every possibility.

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

First, find the total sum of all the numbers.
Next, understand that after removing some numbers, we are left with two groups, and we want the difference between the sums of these groups to be as small as possible.
Think about what the possible sums of one of the groups could be. The smallest this sum could be is if we chose the 'smallest' values, and the largest is if we chose the 'largest' values.
Instead of trying all combinations to form the groups, find what 'target sum' makes the two groups' sums closest together (think halfway point).
The best difference between the sums will happen when one group's sum is closest to the halfway point.
Use a strategy like keeping track of all the possible sums that we can make, and then choose the one that's closest to this target sum.
From that 'chosen' sum, you can calculate the second group's sum and their difference.

Code Implementation

def minimum_difference_in_sums(
    numbers_list
):
    total_sum = sum(numbers_list)
    list_length = len(numbers_list)
    subset_size = list_length // 2
    target_sum = total_sum // 2

    possible_sums = {0}

    for number in numbers_list:
        new_possible_sums = set()

        # Iterate through all the possible sums we've seen so far
        for possible_sum in possible_sums:
            new_sum = possible_sum + number
            new_possible_sums.add(new_sum)

        possible_sums.update(new_possible_sums)

    min_difference = float('inf')

    # We need to check possible sums to minimize the diff
    for possible_sum in possible_sums:
        if possible_sum <= target_sum:

            # Calculates the difference between the two subsets
            other_subset_sum = total_sum - possible_sum
            difference = abs(possible_sum - other_subset_sum)
            min_difference = min(min_difference, difference)

    # Find the minimum difference in sums.
    return min_difference

Big(O) Analysis

Time Complexity

O(n * k * k) – The algorithm involves iterating through the input array of size n and considering all possible combinations to form a group of size k (where k is n/3). The number of such combinations can grow significantly. A dynamic programming approach to store all possible sums using k numbers contributes another factor of k. Finding the closest sum to the target also involves iterating through possible sums of size up to k. Therefore, the time complexity is approximately O(n * k * k).

Space Complexity

O(N * S) – The algorithm keeps track of all possible sums we can make to find a sum closest to the target sum. The number of possible sums can grow linearly with the input size N and the total sum of the numbers, which we can represent as S. Therefore, the algorithm uses space proportional to the product of N and S to store these possible sums. Thus, the space complexity is O(N * S).

Edge Cases

Case	How to Handle
Empty array	Return 0 if the input array is empty, as there are no elements to partition and the difference is zero.
Array size not divisible by 3	If the array size is not divisible by 3, return an appropriate error value (e.g., Infinity or -1) since the problem requires partitioning the array into three equal-sized parts.
Array with only one distinct value	If all elements are the same, the minimum difference will always be 0; compute and return immediately.
Array with a mix of positive, negative, and zero values	The partitioning logic must correctly handle negative numbers when calculating the sums and comparing differences.
Large input array causing potential integer overflow during summation	Use long data type or appropriate overflow handling techniques during sum calculation.
Array with extremely large numbers (close to the maximum integer value)	Guard against potential integer overflow in intermediate calculations by using a data type with a larger range or appropriate modulo arithmetic.
Maximum array size leading to memory constraints in dynamic programming approaches	Consider optimizing the memory usage by using iterative dynamic programming instead of recursive or memoization techniques to avoid stack overflow or excessive memory consumption.
The sums of the three subarrays are equal	The minimum difference will be zero; return the value immediately without any further operations.

Empty array

How to Handle:

Return 0 if the input array is empty, as there are no elements to partition and the difference is zero.

Array size not divisible by 3

How to Handle:

If the array size is not divisible by 3, return an appropriate error value (e.g., Infinity or -1) since the problem requires partitioning the array into three equal-sized parts.

Array with only one distinct value

How to Handle:

If all elements are the same, the minimum difference will always be 0; compute and return immediately.

Array with a mix of positive, negative, and zero values

How to Handle:

The partitioning logic must correctly handle negative numbers when calculating the sums and comparing differences.

Large input array causing potential integer overflow during summation

How to Handle:

Use long data type or appropriate overflow handling techniques during sum calculation.

Array with extremely large numbers (close to the maximum integer value)

How to Handle:

Guard against potential integer overflow in intermediate calculations by using a data type with a larger range or appropriate modulo arithmetic.

Maximum array size leading to memory constraints in dynamic programming approaches

How to Handle:

Consider optimizing the memory usage by using iterative dynamic programming instead of recursive or memoization techniques to avoid stack overflow or excessive memory consumption.

The sums of the three subarrays are equal

How to Handle:

The minimum difference will be zero; return the value immediately without any further operations.