#3098

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Find the Sum of Subsequence Powers

Compute the sum of powers for all subsequences of length k using state transition dynamic programming efficiently.

Problem Statement

Given an integer array nums of length n and a positive integer k, calculate the power of all subsequences of length k. The power of a subsequence is defined as the minimum absolute difference between any two elements within it.

Return the total sum of powers of every subsequence of length k in nums. Constraints are 2 <= n <= 50, -10^8 <= nums[i] <= 10^8, and 2 <= k <= n. Sorting nums can help simplify dynamic programming state transitions.

Examples

Example 1

Input: nums = [1,2,3,4], k = 3

Output: 4

There are 4 subsequences in nums which have length 3: [1,2,3] , [1,3,4] , [1,2,4] , and [2,3,4] . The sum of powers is |2 - 3| + |3 - 4| + |2 - 1| + |3 - 4| = 4 .

Example 2

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

Output: 0

The only subsequence in nums which has length 2 is [2,2] . The sum of powers is |2 - 2| = 0 .

Example 3

Input: nums = [4,3,-1], k = 2

Output: 10

There are 3 subsequences in nums which have length 2: [4,3] , [4,-1] , and [3,-1] . The sum of powers is |4 - 3| + |4 - (-1)| + |3 - (-1)| = 10 .

Constraints

2 <= n == nums.length <= 50
-108 <= nums[i] <= 108
2 <= k <= n

Solution Approach

Sort and Prepare DP Table

Sort the array nums in ascending order. Initialize a DP table dp[i][j] representing the sum of powers for subsequences of length j ending at index i.

State Transition for Subsequences

For each index i and length j, update dp[i][j] by adding the minimum absolute difference between nums[i] and previous elements. Transition uses dp[i-1][j-1] to include nums[i] efficiently.

Aggregate the Results

Sum dp[i][k] for all valid indices i to get the total sum of powers of all subsequences of length k. Return this total as the final answer.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on iterating through n elements for each subsequence length up to k, with potential nested loops for difference calculations. Space complexity depends on maintaining a DP table of size n by k.

What Interviewers Usually Probe

Sorting the array hints at simplifying subsequence power calculation.
Using a DP table suggests a state transition approach for subsequences of length k.
Watch for overlapping subsequences to avoid redundant computations.

Common Pitfalls or Variants

Common pitfalls

Not sorting nums, which complicates minimum absolute difference calculation.
Incorrect DP state transition, leading to missed subsequences or wrong sums.
Failing to aggregate only subsequences of exact length k.

Follow-up variants

Compute maximum instead of minimum absolute difference for subsequences of length k.
Return sum of powers for all subsequences up to length k instead of exactly k.
Handle subsequences with additional constraints, such as even sum or bounded elements.

FAQ

What is the best strategy to handle minimum difference for subsequences of length k?

Sort nums first, then use a DP table to track the sum of powers for subsequences of length k efficiently.

Can this approach handle arrays with negative numbers?

Yes, sorting works for negative numbers and the DP state transitions calculate absolute differences correctly.

Why is state transition dynamic programming important here?

It avoids recomputation by reusing sums of powers for smaller subsequences and correctly builds sequences of length k.

Are all subsequences of length k considered in the sum?

Yes, the DP ensures every valid subsequence of exact length k is included in the total sum of powers.

Does the algorithm scale for maximum n = 50?

Yes, with DP and careful state transition, the computation avoids exponential growth and completes efficiently.

terminal

Solution

Solution 1: Memoization Search

Given the problem involves the minimum difference between elements of a subsequence, we might as well sort the array $\textit{nums}$, which facilitates the calculation of the minimum difference between subsequence elements.

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class Solution:
    def sumOfPowers(self, nums: List[int], k: int) -> int:
        @cache
        def dfs(i: int, j: int, k: int, mi: int) -> int:
            if i >= n:
                return mi if k == 0 else 0
            if n - i < k:
                return 0
            ans = dfs(i + 1, j, k, mi)
            if j == n:
                ans += dfs(i + 1, i, k - 1, mi)
            else:
                ans += dfs(i + 1, i, k - 1, min(mi, nums[i] - nums[j]))
            ans %= mod
            return ans

        mod = 10**9 + 7
        n = len(nums)
        nums.sort()
        return dfs(0, n, k, inf)

Continue Practicing

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