Brute Force Solution

The most straightforward approach is to generate all possible subsequences of length k from the input array nums, calculate the power (minimum absolute difference between any two elements) for each subsequence, and then sum up these powers.

This involves:

1. Generating all combinations of k elements from nums.
2. For each subsequence, calculate the minimum absolute difference between any two elements.
3. Sum these minimum differences, applying modulo at each step.

from itertools import combinations

def calculate_power(subsequence):
    min_diff = float('inf')
    for i in range(len(subsequence)):
        for j in range(i + 1, len(subsequence)):
            min_diff = min(min_diff, abs(subsequence[i] - subsequence[j]))
    return min_diff

def sum_of_powers_brute_force(nums, k):
    subsequences = combinations(nums, k)
    total_power = 0
    for sub in subsequences:
        total_power = (total_power + calculate_power(sub)) % (10**9 + 7)
    return total_power

Time Complexity

• Generating combinations: O(C(n, k)), where C(n, k) is "n choose k," which equals n! / (k! * (n-k)!).
• Calculating the minimum difference for each subsequence of size k: O(k^2) in the worst case.
• Overall: O(C(n, k) * k^2)

Space Complexity

• O(C(n, k)) to store all subsequences, which can be significant. The combinations function in itertools creates an iterator, so it doesn't store all combinations in memory at once, but the algorithm still needs to iterate through them. The space used is dominated by the size of the subsequences being processed, which contributes to an effective space complexity related to the number of combinations in flight.

Optimal Solution

The key optimization is to realize that we only need to consider adjacent elements after sorting the subsequence. Here's how it works:

1. Sort the Input Array: This is crucial. Sorting allows us to identify minimum absolute differences more efficiently.
2. Generate Subsequences: Similar to the brute force method, generate all subsequences of length k.
3. Calculate Power Efficiently: For each subsequence, since the original array is sorted, the minimum absolute difference will always be between two adjacent elements in the sorted subsequence. This saves computation time.

from itertools import combinations

def sum_of_powers_optimal(nums, k):
    nums.sort()
    total_power = 0
    for sub in combinations(nums, k):
        min_diff = float('inf')
        for i in range(len(sub) - 1):
            min_diff = min(min_diff, abs(sub[i] - sub[i+1]))
        total_power = (total_power + min_diff) % (10**9 + 7)
    return total_power

Edge Cases

• k = 2: In this case, the power is simply the absolute difference between the two elements in each subsequence.
• Duplicate Numbers: The algorithm should correctly handle duplicate numbers in the input array.
• Negative Numbers: The algorithm should work correctly with negative numbers.

Time Complexity

• Sorting the input array: O(n log n)
• Generating combinations: O(C(n, k))
• Calculating the minimum difference for each subsequence: O(k)
• Overall: O(n log n + C(n, k) * k)

Space Complexity

• O(1) excluding space for storing the combinations themselves, as we process them one by one. Sorting is done in place.