Maximum and Minimum Sums of at Most Size K Subsequences Interview Question for Meta

## Problem: Sum of Maximum and Minimum Elements of Subsequences

Given an integer array `nums` and a positive integer `k`, the task is to 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 `10^9 + 7`.

**Example:**

`nums = [1, 2, 3], k = 2`

Subsequences with at most 2 elements:

*   `[1]`: min = 1, max = 1, sum = 2
*   `[2]`: min = 2, max = 2, sum = 4
*   `[3]`: min = 3, max = 3, sum = 6
*   `[1, 2]`: min = 1, max = 2, sum = 3
*   `[1, 3]`: min = 1, max = 3, sum = 4
*   `[2, 3]`: min = 2, max = 3, sum = 5

Total sum = 2 + 4 + 6 + 3 + 4 + 5 = 24

## Solution

Here's a Python solution to calculate the sum of the maximum and minimum elements of all subsequences of `nums` with at most `k` elements.

```python
def solve():
    nums = [1, 2, 3]
    k = 2
    print(sum_of_max_min_subsequences(nums, k))

def sum_of_max_min_subsequences(nums, k):
    n = len(nums)
    total_sum = 0
    MOD = 10**9 + 7

    for i in range(1, min(n + 1, k + 1)):
        for mask in range(2**n):
            subsequence = []
            for j in range(n):
                if (mask >> j) & 1:
                    subsequence.append(nums[j])

            if len(subsequence) == i:
                if subsequence:
                    total_sum = (total_sum + min(subsequence) + max(subsequence)) % MOD

    return total_sum

Brute Force Approach:

Generate all possible subsequences of the input array nums.
Filter subsequences based on the condition that their length is at most k.
For each valid subsequence, find the minimum and maximum elements.
Add the minimum and maximum elements to the total sum.
Return the total sum modulo 10^9 + 7.

def sum_of_max_min_subsequences_brute_force(nums, k):
    n = len(nums)
    total_sum = 0
    MOD = 10**9 + 7

    for i in range(1, min(n + 1, k + 1)):
        for mask in range(2**n):
            subsequence = []
            for j in range(n):
                if (mask >> j) & 1:
                    subsequence.append(nums[j])

            if len(subsequence) == i:
                if subsequence:
                    total_sum = (total_sum + min(subsequence) + max(subsequence)) % MOD

    return total_sum

Optimal Approach (Using Combinations):

Sort the input array nums in ascending order.
Initialize an array power2 where power2[i] stores the value of 2^i % MOD.
Iterate through the sorted array nums.
- For each element nums[i], calculate the number of subsequences where nums[i] is the minimum element.
- This is done by considering the i elements to the left of nums[i] and forming subsequences of at most k - 1 elements from them. The number of such subsequences can be calculated using combinations.
- Similarly, calculate the number of subsequences where nums[i] is the maximum element by considering the n - i - 1 elements to the right of nums[i] and forming subsequences of at most k - 1 elements.
Calculate the total sum by adding the contributions of each element as both the minimum and maximum.
Return the total sum modulo 10^9 + 7.

def sum_of_max_min_subsequences_optimal(nums, k):
    n = len(nums)
    nums.sort()
    MOD = 10**9 + 7
    power2 = [1] * (n + 1)
    for i in range(1, n + 1):
        power2[i] = (power2[i - 1] * 2) % MOD

    total_sum = 0

    for i in range(n):
        # nums[i] as minimum
        if i > 0:
            num_subsequences_min = 0
            for j in range(min(i, k - 1) + 1):
                num_subsequences_min = (num_subsequences_min + combination(i, j, MOD)) % MOD
            total_sum = (total_sum + nums[i] * num_subsequences_min) % MOD
        else:
            if k >= 1:
                total_sum = (total_sum + nums[i]) % MOD


        # nums[i] as maximum
        if n - i - 1 > 0:
            num_subsequences_max = 0
            for j in range(min(n - i - 1, k - 1) + 1):
                num_subsequences_max = (num_subsequences_max + combination(n - i - 1, j, MOD)) % MOD
            total_sum = (total_sum + nums[i] * num_subsequences_max) % MOD
        else:
            if k >= 1:
                total_sum = (total_sum + nums[i]) % MOD


    return total_sum % MOD

def combination(n, k, MOD):
    if k < 0 or k > n:
        return 0
    if k == 0 or k == n:
        return 1
    if k > n // 2:
        k = n - k
    
    result = 1
    for i in range(k):
        result = (result * (n - i)) % MOD
        result = (result * pow(i + 1, MOD - 2, MOD)) % MOD
    return result

Time Complexity Analysis

Brute Force Approach:
- Generating all subsequences takes O(2^n) time.
- Finding the minimum and maximum elements of each subsequence takes O(n) time in the worst case.
- Therefore, the overall time complexity is O(n * 2^n).
Optimal Approach:
- Sorting the array takes O(n log n) time.
- Calculating the combinations takes O(k) time for each element.
- Iterating through the array takes O(n) time.
- Therefore, the overall time complexity is O(n log n + n * k), where k is the maximum subsequence length.

Space Complexity Analysis

Brute Force Approach:
- The space complexity is O(n) in the worst case, to store the subsequence.
Optimal Approach:
- The space complexity is O(n) to store the sorted array and the power2 array.

Edge Cases

Empty Input Array:
- If the input array nums is empty, the sum of the maximum and minimum elements is 0.
k = 0:
- If k is 0, then there are no valid subsequences. In this case the sum is 0.
k >= n:
- If k is greater than or equal to the length of the input array, then we need to consider all subsequences of length 1 up to n.
Duplicate elements:
- The array may contain duplicate elements, which can be handled without any specific logic.
Negative Elements:
- The array may contain negative elements. The algorithm should handle both positive and negative elements correctly.