Minimum Difference in Sums After Removal of Elements Interview Question for Amazon

Problem Description

Given a 0-indexed integer array nums consisting of 3 * n elements, the task is to remove a 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 forming the first part (sum denoted as sum_first) and the next n elements forming the second part (sum denoted as sum_second). The goal is to find the minimum possible difference between sum_first and sum_second after removing n elements.

For example, if nums = [3, 1, 2] (where n = 1), we can remove one element at a time and calculate the difference:

Remove 3: [1, 2], difference = 1 - 2 = -1
Remove 1: [3, 2], difference = 3 - 2 = 1
Remove 2: [3, 1], difference = 3 - 1 = 2

The minimum difference is min(-1, 1, 2) = -1.

Naive Approach

A brute-force solution would involve generating all possible subsequences of size n from the input array nums. For each subsequence, we remove it from nums, divide the remaining 2 * n elements into two equal parts, compute the difference between the sums of the two parts, and keep track of the minimum difference encountered so far.

import itertools
import math

def min_difference_naive(nums):
    n = len(nums) // 3
    min_diff = float('inf')

    for indices in itertools.combinations(range(3 * n), n):
        temp_nums = []
        removed_indices = set(indices)
        for i in range(3 * n):
            if i not in removed_indices:
                temp_nums.append(nums[i])

        sum_first = sum(temp_nums[:n])
        sum_second = sum(temp_nums[n:])
        min_diff = min(min_diff, sum_first - sum_second)

    return min_diff

# Example usage
nums = [3, 1, 2]
print(min_difference_naive(nums))
# Expected output: -1

nums = [7, 9, 5, 8, 1, 3]
print(min_difference_naive(nums))
# Expected output: 1

Optimal Approach

We can utilize prefix and suffix sums combined with min-heaps and max-heaps to optimize the process. The main idea is to precompute the sums of the first n elements after removing some of the smallest elements (prefix sums) and the sums of the last n elements after removing some of the largest elements (suffix sums).

Here's the algorithm:

Prefix Sums: Calculate prefix sums for all possible first parts after removing n elements from the first 2n elements of the array. We use a max-heap of size n to keep track of the n largest elements encountered so far.
Suffix Sums: Calculate suffix sums for all possible second parts after removing n elements from the last 2n elements of the array. We use a min-heap of size n to keep track of the n smallest elements encountered so far.
Minimum Difference: Iterate from n to 2n. At each index i, the minimum difference will be prefix[i] - suffix[i].

import heapq

def min_difference(nums):
    n = len(nums) // 3

    # Prefix sums
    prefix = [0] * (2 * n + 1)
    max_heap = []
    prefix[0] = 0
    for i in range(n):
        prefix[n] += nums[i]
        heapq.heappush(max_heap, -nums[i])

    for i in range(n + 1, 2 * n + 1):
        heapq.heappush(max_heap, -nums[i - 1])
        prefix[i] = prefix[i - 1] + nums[i - 1] + heapq.heappop(max_heap)

    # Suffix sums
    suffix = [0] * (2 * n + 1)
    min_heap = []
    suffix[2 * n] = 0
    for i in range(3 * n - 1, 3 * n - n - 1, -1):
        suffix[2 * n] += nums[i]
        heapq.heappush(min_heap, nums[i])

    for i in range(2 * n - 1, n - 1, -1):
        heapq.heappush(min_heap, nums[i])
        suffix[i] = suffix[i + 1] + nums[i] - heapq.heappop(min_heap)

    # Minimum difference
    min_diff = float('inf')
    for i in range(n, 2 * n + 1):
        min_diff = min(min_diff, prefix[i] - suffix[i])

    return min_diff


# Example usage
nums = [3, 1, 2]
print(min_difference(nums))
# Expected output: -1

nums = [7, 9, 5, 8, 1, 3]
print(min_difference(nums))
# Expected output: 1

Big(O) Time Complexity Analysis

Optimal Solution:
- Creating the prefix sums involves iterating n times to initially populate the max-heap and computing the sum for the first segment. Then, we iterate n times, performing heap operations (push and pop) inside the loop. Heap operations take O(log n) time each. Hence, the prefix sum calculation takes O(n log n) time.
- Similarly, creating the suffix sums also involves O(n log n) time due to heap operations.
- Calculating the minimum difference iterates n times, taking O(n) time.
- Therefore, the overall time complexity is O(n log n + n log n + n) = O(n log n).
Naive Solution:
- The brute-force solution generates all combinations of size n, which takes O(C(3n, n)) time, where C(3n, n) represents the binomial coefficient (combinations of 3n choose n). For each combination, summing the first and second parts takes O(n) time. Thus, the overall complexity is O(C(3n, n) * n). This is much slower.

Big(O) Space Complexity Analysis

Optimal Solution:
- The prefix and suffix arrays each use O(n) space.
- The max-heap and min-heap each use O(n) space.
- Therefore, the overall space complexity is O(n + n + n + n) = O(n).
Naive Solution:
- The brute-force approach uses O(n) space to store the temporary array temp_nums, contributing to a space complexity of O(n).

Edge Cases and Handling

Empty Input: If the input array nums is empty, we can return 0, as there's no difference to calculate. However, the problem statement specifies nums.length == 3 * n and 1 <= n <= 10^5, so an empty input is not a valid edge case based on the constraints.
Single Element: When n = 1, the algorithm still works correctly because we can remove any single element, and the minimum difference is simply the minimum difference achievable by removing each element individually.
Large Numbers: The numbers in nums can be as large as 10^5. Since we are summing up to n such numbers, the prefix and suffix sums can be up to n * 10^5. We should ensure that the data type used to store these sums can handle such large values (e.g., using int in Python, which has arbitrary precision, or long in languages where int has a fixed size).
Negative Numbers: The problem statement does not specify whether the numbers are non-negative. If negative numbers are allowed, the algorithm still works correctly, because the heaps will properly handle negative numbers during the prefix and suffix sum calculations.