Maximum and Minimum Sums of at Most Size K Subsequences

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Maximum and Minimum Sums of at Most Size K Subsequences #1025 Most Asked

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ArraysTwo PointersSliding Windows

You are given an integer array nums and a positive integer k. Return the sum of the maximum and minimum elements of all subsequences of nums with at most k elements.

Since the answer may be very large, return it modulo 10⁹ + 7.

Example 1:

Input: nums = [1,2,3], k = 2

Output: 24

Explanation:

The subsequences of nums with at most 2 elements are:

Subsequence	Minimum	Maximum	Sum
`[1]`	1	1	2
`[2]`	2	2	4
`[3]`	3	3	6
`[1, 2]`	1	2	3
`[1, 3]`	1	3	4
`[2, 3]`	2	3	5
Final Total			24

The output would be 24.

Example 2:

Input: nums = [5,0,6], k = 1

Output: 22

Explanation:

For subsequences with exactly 1 element, the minimum and maximum values are the element itself. Therefore, the total is 5 + 5 + 0 + 0 + 6 + 6 = 22.

Example 3:

Input: nums = [1,1,1], k = 2

Output: 12

Explanation:

The subsequences [1, 1] and [1] each appear 3 times. For all of them, the minimum and maximum are both 1. Thus, the total is 12.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁹
1 <= k <= min(70, nums.length)

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 value of `k`?
Can the elements in the `nums` array be negative, positive, or zero?
If `k` is zero, should I return 0 for both the minimum and maximum sums?
Are there any specific constraints on the type of `k` (e.g., is it always a non-negative integer)?
In the case of an empty input array, what should be the returned maximum and minimum values?

Brute Force Solution

Approach

The brute force approach is about exploring every possible combination to find the largest and smallest possible sums. We'll look at every possible group of numbers, up to a certain size, and calculate their sum.

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

Consider each single number in the collection as a potential group.
Calculate the sum of each of these single-number groups.
Then, consider every possible pair of numbers as a potential group, making sure not to include the same number twice in a single group.
Calculate the sum of each of these two-number groups.
Continue this process, creating groups of three numbers, four numbers, and so on, up to the maximum allowed size.
For each group you create, calculate its sum.
Keep track of the largest sum you've found so far.
Keep track of the smallest sum you've found so far.
After you've considered every possible group, the largest sum you've kept track of is the maximum possible sum and the smallest is the minimum possible sum.

Code Implementation

def find_maximum_and_minimum_sums(
    numbers,
    maximum_subsequence_size
):
    maximum_sum = float('-inf')
    minimum_sum = float('inf')

    for subsequence_size in range(1, maximum_subsequence_size + 1):
        # We want to explore all subsequence sizes from 1 to k

        def find_all_subsequences(
            current_subsequence,
            start_index
        ):
            nonlocal maximum_sum, minimum_sum

            if len(current_subsequence) == subsequence_size:
                current_sum = sum(current_subsequence)
                maximum_sum = max(maximum_sum, current_sum)
                minimum_sum = min(minimum_sum, current_sum)
                return

            # Iterate through numbers array and expand current subsequence
            for i in range(start_index, len(numbers)):
                # Avoid duplicates by starting at start_index

                find_all_subsequences(
                    current_subsequence + [numbers[i]],
                    i + 1
                )

        find_all_subsequences([], 0)

    return maximum_sum, minimum_sum

Big(O) Analysis

Time Complexity

O(n^k) – The algorithm iterates through all possible subsequences of size up to k from an input array of size n. For subsequences of size 1, there are n possibilities. For subsequences of size 2, there are n choose 2 possibilities, and so on, up to subsequences of size k, which has n choose k possibilities. In the worst-case scenario, k can be close to n, meaning we're looking at a significant number of combinations, which approaches n^k. Therefore, the time complexity is O(n^k) because in the worst case, we need to explore a number of combinations that grows polynomially with n, where the degree is k.

Space Complexity

O(1) – The provided algorithm keeps track of the maximum and minimum sums encountered so far. This requires only two variables to store these sums, regardless of the size of the input array (N) or the maximum subsequence size (K). No additional data structures like arrays, lists, or hash maps are used to store intermediate combinations or results. Thus, the auxiliary space used is constant.

Optimal Solution

Approach

To find the largest and smallest sums, we don't need to check every single subsequence. We use a clever trick of keeping track of the largest and smallest numbers we've seen so far to construct our subsequences intelligently.

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

Imagine having two separate bags: one for collecting numbers for the maximum sum and another for the minimum sum.
Go through the numbers one at a time.
For the maximum sum bag, if the current number is bigger than the smallest number in the bag, take out the smallest number and put the current number in. This way, we always keep the largest possible numbers in the bag, up to the bag's size limit.
Similarly, for the minimum sum bag, if the current number is smaller than the largest number in the bag, take out the largest number and put the current number in. This ensures we always have the smallest numbers.
After you've looked at all the numbers, simply add up the numbers in the maximum sum bag to get the maximum sum, and add up the numbers in the minimum sum bag to get the minimum sum. This avoids having to compare sums of a huge number of subsequences.

Code Implementation

def max_min_subsequence_sums(numbers, subsequence_size):
    max_heap = []
    min_heap = []

    for number in numbers:
        # Maintain the max heap for max sum
        if len(max_heap) < subsequence_size:
            max_heap.append(number)
            max_heap.sort()
        elif number > max_heap[0]:
            max_heap[0] = number
            max_heap.sort()

        # Maintain the min heap for min sum
        if len(min_heap) < subsequence_size:
            min_heap.append(number)
            min_heap.sort(reverse=True)
        elif number < min_heap[0]:
            min_heap[0] = number
            min_heap.sort(reverse=True)

    maximum_sum = sum(max_heap)
    minimum_sum = sum(min_heap)

    return maximum_sum, minimum_sum

Big(O) Analysis

Time Complexity

O(n log k) – The algorithm iterates through the input array of size n once. Inside the loop, for both the maximum and minimum sum bags, the operations involve checking if the current number is greater than the smallest (for max bag) or smaller than the largest (for min bag), and potentially replacing an element in the bag. Since the bags are implemented as heaps (priority queues), the insertion or deletion of an element in a heap of size k takes O(log k) time. Thus, each iteration of the loop takes O(log k) time. Therefore, the overall time complexity is O(n log k).

Space Complexity

O(K) – The algorithm maintains two "bags" (represented as data structures, likely heaps or sorted lists) to store numbers for maximum and minimum sums. The size of each bag is at most K, where K is the maximum size of the subsequence. Therefore, the auxiliary space used is proportional to K, as we need to store at most K numbers in each of the two data structures.

Edge Cases

Case	How to Handle
nums is null or empty	Return [0, 0] if nums is null or empty because the sum of an empty subsequence is 0.
k is 0	Return [0, 0] since the only valid subsequence is an empty one with a sum of 0.
k is greater than or equal to the length of nums	Include all elements in nums in both the minimum and maximum subsequence calculations.
nums contains only positive numbers	The minimum sum is 0 (empty subsequence) and maximum sum includes the k largest numbers or all numbers if nums.length <= k.
nums contains only negative numbers	The maximum sum is 0 (empty subsequence) and the minimum sum includes the k smallest numbers or all numbers if nums.length <= k.
nums contains both positive and negative numbers, including zero	The minimum sum includes the most negative k numbers, or all negative numbers if less than k and the maximum sum includes the most positive k numbers, or all positive numbers if less than k.
nums contains duplicate values	Duplicates don't fundamentally change the logic but ensure the selection process correctly handles frequency of the elements.
Integer overflow can occur when summing large numbers	Use long data type for storing sums to prevent overflow during calculations.

nums is null or empty

How to Handle:

Return [0, 0] if nums is null or empty because the sum of an empty subsequence is 0.

k is 0

How to Handle:

Return [0, 0] since the only valid subsequence is an empty one with a sum of 0.

k is greater than or equal to the length of nums

How to Handle:

Include all elements in nums in both the minimum and maximum subsequence calculations.

nums contains only positive numbers

How to Handle:

The minimum sum is 0 (empty subsequence) and maximum sum includes the k largest numbers or all numbers if nums.length <= k.

nums contains only negative numbers

How to Handle:

The maximum sum is 0 (empty subsequence) and the minimum sum includes the k smallest numbers or all numbers if nums.length <= k.

nums contains both positive and negative numbers, including zero

How to Handle:

The minimum sum includes the most negative k numbers, or all negative numbers if less than k and the maximum sum includes the most positive k numbers, or all positive numbers if less than k.

nums contains duplicate values

How to Handle:

Duplicates don't fundamentally change the logic but ensure the selection process correctly handles frequency of the elements.

Integer overflow can occur when summing large numbers

How to Handle:

Use long data type for storing sums to prevent overflow during calculations.