Find the Sum of Subsequence Powers Interview Question for Google

def solve():
    nums = [4, 3, -1]
    k = 2
    print(f"Input: nums = {nums}, k = {k}")

    n = len(nums)
    subsequences = []

    def find_subsequences(index, current_subsequence):
        if len(current_subsequence) == k:
            subsequences.append(current_subsequence[:])
            return
        
        if index == n:
            return

        # Include the current element
        current_subsequence.append(nums[index])
        find_subsequences(index + 1, current_subsequence)
        current_subsequence.pop()

        # Exclude the current element
        find_subsequences(index + 1, current_subsequence)

    find_subsequences(0, [])
    
    total_power = 0
    for subsequence in subsequences:
        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]))
        total_power += min_diff

    print(f"Output: {total_power}")
    print(f"Explanation:")
    print(f"There are {len(subsequences)} subsequences in nums which have length 2: {[str(sub) for sub in subsequences]}. The sum of powers is calculated as the sum of the minimum absolute differences between elements in each subsequence.")
    print("\n")


    nums = [1, 2, 3, 4]
    k = 3
    print(f"Input: nums = {nums}, k = {k}")

    n = len(nums)
    subsequences = []

    def find_subsequences(index, current_subsequence):
        if len(current_subsequence) == k:
            subsequences.append(current_subsequence[:])
            return
        
        if index == n:
            return

        # Include the current element
        current_subsequence.append(nums[index])
        find_subsequences(index + 1, current_subsequence)
        current_subsequence.pop()

        # Exclude the current element
        find_subsequences(index + 1, current_subsequence)

    find_subsequences(0, [])
    
    total_power = 0
    for subsequence in subsequences:
        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]))
        total_power += min_diff

    print(f"Output: {total_power}")
    print(f"Explanation:")
    print(f"There are {len(subsequences)} subsequences in nums which have length 3: {[str(sub) for sub in subsequences]}. The sum of powers is calculated as the sum of the minimum absolute differences between elements in each subsequence.")

    print("\n")

    nums = [2, 2]
    k = 2
    print(f"Input: nums = {nums}, k = {k}")

    n = len(nums)
    subsequences = []

    def find_subsequences(index, current_subsequence):
        if len(current_subsequence) == k:
            subsequences.append(current_subsequence[:])
            return
        
        if index == n:
            return

        # Include the current element
        current_subsequence.append(nums[index])
        find_subsequences(index + 1, current_subsequence)
        current_subsequence.pop()

        # Exclude the current element
        find_subsequences(index + 1, current_subsequence)

    find_subsequences(0, [])
    
    total_power = 0
    for subsequence in subsequences:
        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]))
        total_power += min_diff

    print(f"Output: {total_power}")
    print(f"Explanation:")
    print(f"There is {len(subsequences)} subsequence in nums which has length 2: {[str(sub) for sub in subsequences]}. The sum of powers is calculated as the sum of the minimum absolute differences between elements in each subsequence.")


solve()

Brute Force Solution

The provided Python code implements a brute-force solution to the problem. It first generates all possible subsequences of length k from the input array nums. Then, for each subsequence, it computes the minimum absolute difference between any two elements. Finally, it sums up these minimum differences across all subsequences and returns the result.

Here's a breakdown:

Generate Subsequences: A recursive function find_subsequences generates all subsequences of length k. It explores all possible combinations of elements by either including or excluding each element in the input array.
Compute Minimum Difference: For each generated subsequence, the code iterates through all pairs of elements and computes the absolute difference between them. It keeps track of the minimum difference found.
Sum the Powers: The minimum difference calculated for each subsequence is added to a running total, which represents the sum of powers.

Optimal Solution

To avoid recomputing the subsequences, we could compute the number of times a particular pair of elements would be present in a subsequence of size k. Let's say we sort the array. Then, for each pair (nums[i], nums[j]) where i < j, we compute how many subsequences of size k contain both nums[i] and nums[j]. The remaining positions are k - 2 positions, and these can be filled by any of the remaining n - 2 elements. Thus, there are C(n-2, k-2) such subsequences. So the contribution of this pair to the final answer is abs(nums[i] - nums[j]) * C(n-2, k-2). So the overall complexity would be O(n^2) for going through each pair, and O(k) to compute the combination using dynamic programming.

def solve_optimal():
    nums = [4, 3, -1]
    k = 2
    print(f"Input: nums = {nums}, k = {k}")
    n = len(nums)
    nums.sort()
    
    MOD = 10**9 + 7
    
    def combinations(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
            
        dp = [0] * (k + 1)
        dp[0] = 1
        
        for i in range(1, n + 1):
            for j in range(min(i, k), 0, -1):
                dp[j] = (dp[j] + dp[j-1]) % mod
                
        return dp[k]

    total_power = 0
    num_combinations = combinations(n - 2, k - 2, MOD)
    for i in range(n):
        for j in range(i + 1, n):
            total_power = (total_power + abs(nums[i] - nums[j]) * num_combinations) % MOD
    
    print(f"Output: {total_power}")

    nums = [1, 2, 3, 4]
    k = 3
    print(f"Input: nums = {nums}, k = {k}")
    n = len(nums)
    nums.sort()
    
    total_power = 0
    num_combinations = combinations(n - 2, k - 2, MOD)
    for i in range(n):
        for j in range(i + 1, n):
            total_power = (total_power + abs(nums[i] - nums[j]) * num_combinations) % MOD
    
    print(f"Output: {total_power}")

    nums = [2, 2]
    k = 2
    print(f"Input: nums = {nums}, k = {k}")
    n = len(nums)
    nums.sort()

    total_power = 0
    num_combinations = combinations(n - 2, k - 2, MOD)
    for i in range(n):
        for j in range(i + 1, n):
            total_power = (total_power + abs(nums[i] - nums[j]) * num_combinations) % MOD
    
    print(f"Output: {total_power}")


solve_optimal()

Big(O) Time Complexity Analysis

Brute Force Solution

Generating Subsequences: The find_subsequences function explores all possible subsequences, which is 2ⁿ in the worst case. Thus, the time complexity is O(2ⁿ).
Compute Minimum Difference: For each subsequence, the code iterates through all pairs of elements (size k), which is O(k²).
Overall: O(2ⁿ * k²) since we need to compute the minimum differences for each of the 2ⁿ possible subsequences.

Optimal Solution

Sorting the array: O(n log n)
Compute combinations: O(k)
Nested loop to compute total power: O(n^2)
Overall: O(n log n + n^2), which simplifies to O(n^2) for n > log n

Big(O) Space Complexity Analysis

Brute Force Solution

Generating Subsequences: The subsequences list can store up to O(2ⁿ) subsequences in the worst case.
Recursion Depth: The recursive calls of find_subsequences can go as deep as O(n) in the call stack.
Overall: O(2ⁿ), dominated by the storage of all subsequences.

Optimal Solution

Sorting in place: O(1)
DP table for combinations: O(k)
Overall: O(k). Since k <= n, we can also say that space complexity is O(n) in the worst case.