Problem Description

You are given an integer array nums of length n, and a positive integer k. The power of a subsequence is defined as the minimum absolute difference between any two elements in the subsequence. Return the sum of powers of all subsequences of nums which have a length equal to k. Since the answer may be large, return it modulo 10^9 + 7.

Intuition

The problem involves finding the sum of minimum differences between elements for all subsequences of a specified length k from a given array nums. To efficiently calculate the power of subsequences, we first sort the array. Sorting aids us in quickly finding the minimum difference since consecutive elements in a sorted array have the smallest differences.

We use a recursive depth-first search (DFS) approach with memoization to explore potential subsequences. The function dfs(i, j, k, mi) is designed to determine the power when processing the i-th element of the array. The parameters include the last selected element j, the number of elements k still needed to complete the subsequence, and the current minimum difference mi.

The function's logic is:

• If all elements have been processed (i >= n) and k elements have been successfully selected, return mi. Otherwise, return 0.
• If the remaining elements (n - i) are fewer than needed (k), return 0.
• Without selecting the i-th element, the resultant power is calculated as dfs(i + 1, j, k, mi).
• If the i-th element is chosen:
  • If no elements were chosen before (j = n), update power with dfs(i + 1, i, k - 1, mi).
  • Otherwise, update power with dfs(i + 1, i, k - 1, min(mi, nums[i] - nums[j])).
• Aggregate these results, ensuring results fit within the constraints by using modulo 10^9 + 7.

Memoization optimizes this approach by storing previously computed results, thus avoiding redundant computations.

Learn more about Dynamic Programming and Sorting patterns.

Solution Approach

Solution 1: Memoization Search

To solve this problem efficiently, we employ the technique of memoization along with a depth-first search (DFS) strategy. Here's how the solution is structured:

1. Sorting the Array:
   First, we sort the array nums. Sorting is crucial because it allows us to calculate the minimum difference between elements in a subsequence easily. For any subsequence, the smallest difference in a sorted array will occur between consecutive elements.

2. Recursive DFS Function:
   We define a recursive function dfs(i, j, k, mi) where:

   • i is the current index in the array we are considering.
   • j is the index of the last element that was included in our subsequence.
   • k is the number of elements still needed to form a complete subsequence.
   • mi is the current minimum difference observed within the selected elements of the subsequence.

   The goal is to compute the sum of subsequences' powers systematically using the following logic:

   • Base Case Handling:
     1. If i >= n, meaning all elements have been considered:
        • If k is zero (successfully formed a subsequence of length k), return mi.
        • Otherwise, return 0 since the subsequence is incomplete.
     2. If the remaining elements (n - i) are fewer than needed (k), return 0 since completing the subsequence is impossible.
   • Recursive Exploration:
     1. Not Select i-th Element:
        • Compute result as dfs(i + 1, j, k, mi).
     2. Select i-th Element:
        • If j = n (no prior selections), result is dfs(i + 1, i, k - 1, mi).
        • Otherwise, calculate the power with the minimum difference update as dfs(i + 1, i, k - 1, min(mi, nums[i] - nums[j])).
   • Aggregation:
     Sum the results of both paths and return the result modulo 10^9 + 7 to handle large numbers effectively.

3. Memoization:
   By decorating the dfs function with @cache, we store already computed results during recursion. This optimization significantly reduces unnecessary recomputation of the same subproblems, enhancing efficiency.

The implementation begins by sorting the nums array, initializing parameters, and invoking the dfs function starting from the array's start index. The final return value is the solution to the problem, containing the sum of powers for all suitable subsequences.

Example Walkthrough

Let's walkthrough the solution with a small example.

Suppose we have the array nums = [4, 1, 3] and we want to find the sum of the powers of all subsequences of length k = 2.

1. Step 1: Sort the Array

   First, we sort the array: nums becomes [1, 3, 4].

2. Step 2: Understand the DFS Approach

   We aim to calculate the sum of the minimum differences between any two elements in all possible subsequences of length k = 2.

   • Recursive Function Parameters:
     • i: Current element index under consideration.
     • j: Index of the last selected element.
     • k: Number of elements still needed for the subsequence.
     • mi: Current minimum difference within the subsequence.

3. Step 3: Base Cases and Switching

   • Base Case 1: If we've reached the end of the array (i >= n):
     • If k == 0, it means we've formed a subsequence of required length k, so return mi.
     • Otherwise, return 0.
   • Base Case 2: If remaining elements are fewer than k, return 0.

4. Step 4: Recursive Calculation

   For each element, we decide whether to include it in the subsequence:

   • Skip Current Element: Call dfs(i + 1, j, k, mi).
   • Include Current Element:
     • If no previous elements were selected (j == n), call dfs(i + 1, i, k - 1, mi).
     • Otherwise, update the minimum difference and call dfs(i + 1, i, k - 1, min(mi, nums[i] - nums[j])).

5. Step 5: Use Memoization

   Leverage memoization to store results of already computed states and reuse them to avoid redundancy.

6. Example Execution:

   Following this approach, consider (i, j, k, mi) states:

   • Start: dfs(0, 3, 2, inf) for full exploration.

   • State-wise Analysis:

     • dfs(0, 3, 2, inf) considers choosing 1.
       • dfs(1, 0, 1, inf) chooses 3, resulting in mi = 2 (|3 - 1|).
       • dfs(2, 0, 1, inf) chooses 4, resulting in mi = 3 (|4 - 1|).
     • Skipping 1:
       • dfs(1, 3, 2, inf) continues with 3, 4, leading to:
         • dfs(2, 1, 1, inf) chooses 4, resulting in mi = 1 (|4 - 3|).

   • Sum Powers:

     • Powers: 2 from (3, 1) and 1 from (4, 3).
     • Total power = 2 + 1 = 3.

7. Conclusion

   Thus, the final output is 3 after calculating the sum and ensuring results are within modulo constraints.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n^4 * k), as the function dfs is called with three parameters i, j, and k that can independently take values across a possible range of [0, n], alongside computations involving min. Each recursive state explores multiple combinations related to these parameters in a nested manner, contributing to the high order of complexity.

The space complexity is O(n^4 * k) as well because the @cache decorator memoizes results for each distinct combination of i, j, k, and mi, resulting in a storage requirement proportional to the number of such states, where each state saves intermediate results to avoid recomputation.

Learn more about how to find time and space complexity quickly.