Introduction

In this blog post, we will explore how to solve the LeetCode problem "3098" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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 length equal to k. Since the answer may be large, return it modulo 109 + 7. Example 1: Input: nums = [1,2,3,4], k = 3 Output: 4 Explanation: 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 Explanation: 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 Explanation: 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

Explanation

Here's the breakdown of the solution and the code:

• High-Level Approach:

  • Sort the input array nums to efficiently find minimum absolute differences.
  • Iterate through all possible subsequences of length k using combinations.
  • For each subsequence, calculate the minimum absolute difference between any two elements and add it to the running sum.

• Complexity:

  • Runtime: O(n log n + nCk * k^2), where nCk is the binomial coefficient "n choose k". The n log n comes from sorting. The nCk comes from calculating all subsequences. The k^2 comes from finding the min absolute difference within each subsequence. Note: this is a worst-case runtime complexity, as if k is small relative to n, n log n dominates. If k is close to n (n/2), nCk is much larger.
  • Storage: O(k) – primarily used for storing each subsequence.

Code

    from itertools import combinations

def sum_of_powers(nums, k):
    """
    Calculates the sum of powers of all subsequences of nums with length k.

    Args:
        nums: An integer array of length n.
        k: A positive integer.

    Returns:
        The sum of powers of all subsequences of nums with length k, modulo 10^9 + 7.
    """
    nums.sort()
    n = len(nums)
    total_power = 0
    MOD = 10**9 + 7

    for subsequence in combinations(nums, k):
        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 = (total_power + min_diff) % MOD

    return total_power