Introduction

In this blog post, we will explore how to solve the LeetCode problem "891" 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

The width of a sequence is the difference between the maximum and minimum elements in the sequence. Given an array of integers nums, return the sum of the widths of all the non-empty subsequences of nums. Since the answer may be very large, return it modulo 109 + 7. A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7]. Example 1: Input: nums = [2,1,3] Output: 6 Explanation: The subsequences are [1], [2], [3], [2,1], [2,3], [1,3], [2,1,3]. The corresponding widths are 0, 0, 0, 1, 1, 2, 2. The sum of these widths is 6. Example 2: Input: nums = [2] Output: 0 Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 105

Explanation

Here's the solution to calculate the sum of widths of all non-empty subsequences, along with explanations and code:

• Core Idea: Sort the input array. For each element nums[i], it will be the minimum element in 2^i subsequences and the maximum element in 2^(n-1-i) subsequences. Calculate the contribution of each element as both a minimum and a maximum, and sum these contributions.
• Modulo Arithmetic: Perform all calculations modulo 10^9 + 7 to prevent overflow.

• Power Calculation: Efficiently compute powers of 2 using the binary exponentiation technique.

• Complexity: Time: O(n log n) due to sorting. Space: O(1).

Code

    def sumSubseqWidths(nums: list[int]) -> int:
    """
    Calculates the sum of the widths of all non-empty subsequences of nums modulo 10^9 + 7.
    """
    nums.sort()
    n = len(nums)
    MOD = 10**9 + 7
    ans = 0
    for i in range(n):
        ans = (ans + nums[i] * pow(2, i, MOD) - nums[i] * pow(2, n - 1 - i, MOD)) % MOD
    return ans