Introduction

In this blog post, we will explore how to solve the LeetCode problem "3428" 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 and a positive integer k. Return the sum of the maximum and minimum elements of all subsequences of nums with at most k elements. Since the answer may be very large, return it modulo 109 + 7. Example 1: Input: nums = [1,2,3], k = 2 Output: 24 Explanation: The subsequences of nums with at most 2 elements are: Subsequence Minimum Maximum Sum [1] 1 1 2 [2] 2 2 4 [3] 3 3 6 [1, 2] 1 2 3 [1, 3] 1 3 4 [2, 3] 2 3 5 Final Total 24 The output would be 24. Example 2: Input: nums = [5,0,6], k = 1 Output: 22 Explanation: For subsequences with exactly 1 element, the minimum and maximum values are the element itself. Therefore, the total is 5 + 5 + 0 + 0 + 6 + 6 = 22. Example 3: Input: nums = [1,1,1], k = 2 Output: 12 Explanation: The subsequences [1, 1] and [1] each appear 3 times. For all of them, the minimum and maximum are both 1. Thus, the total is 12. Constraints: 1 <= nums.length <= 105 0 <= nums[i] <= 109 1 <= k <= min(70, nums.length)

Explanation

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

• High-Level Approach:

  • Sort the input array nums. This enables us to efficiently find minimums and maximums.
  • Iterate through the sorted array. For each element, determine how many times it will be the minimum and maximum element in a subsequence of at most k elements. Use combinations (nCr) to calculate the number of such subsequences.
  • Calculate the contribution of each element to the total sum as both the minimum and maximum, considering the modulo.

• Complexity:

  • Runtime: O(n log n) due to sorting, and O(n*k) due to the generation of the combinations table. Thus, we can say O(n log n + nk).
  • Storage: O(k) primarily for storing combinations and intermediate results.

def solve():
    def subsequenceSum(nums, k):
        n = len(nums)
        nums.sort()
        MOD = 10**9 + 7

        # Precompute combinations (nCr) using dynamic programming
        C = [[0] * (k + 1) for _ in range(n + 1)]
        for i in range(n + 1):
            C[i][0] = 1
            for j in range(1, min(i, k) + 1):
                C[i][j] = (C[i-1][j-1] + C[i-1][j]) % MOD

        total_sum = 0
        for i in range(n):
            # Count subsequences where nums[i] is the minimum
            for length in range(1, k + 1):
                if length > 1:
                    if i > 0 and length -1 <= i:
                        total_sum = (total_sum + nums[i] * C[i][length - 1]) % MOD
                else:
                    total_sum = (total_sum + nums[i]) % MOD

            # Count subsequences where nums[i] is the maximum
            for length in range(1, k+1):
                if length > 1:
                    if n - 1 - i > 0 and length - 1 <= n - 1 - i:
                        total_sum = (total_sum + nums[i] * C[n - 1 - i][length - 1]) % MOD
                else:
                    total_sum = (total_sum + nums[i]) % MOD

        return total_sum % MOD

    nums = [int(x) for x in input().split()]
    k = int(input())
    print(subsequenceSum(nums, k))


    # Code
    ```python
    def solve():    def subsequenceSum(nums, k):
        n = len(nums)
        nums.sort()
        MOD = 10**9 + 7

        # Precompute combinations (nCr) using dynamic programming
        C = [[0] * (k + 1) for _ in range(n + 1)]
        for i in range(n + 1):
            C[i][0] = 1
            for j in range(1, min(i, k) + 1):
                C[i][j] = (C[i-1][j-1] + C[i-1][j]) % MOD

        total_sum = 0
        for i in range(n):
            # Count subsequences where nums[i] is the minimum
            for length in range(1, k + 1):
                if length > 1:
                    if i > 0 and length -1 <= i:
                        total_sum = (total_sum + nums[i] * C[i][length - 1]) % MOD
                else:
                    total_sum = (total_sum + nums[i]) % MOD

            # Count subsequences where nums[i] is the maximum
            for length in range(1, k+1):
                if length > 1:
                    if n - 1 - i > 0 and length - 1 <= n - 1 - i:
                        total_sum = (total_sum + nums[i] * C[n - 1 - i][length - 1]) % MOD
                else:
                    total_sum = (total_sum + nums[i]) % MOD

        return total_sum % MOD

    nums = [int(x) for x in input().split()]
    k = int(input())
    print(subsequenceSum(nums, k))