Introduction

In this blog post, we will explore how to solve the LeetCode problem "2221" 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 a 0-indexed integer array nums, where nums[i] is a digit between 0 and 9 (inclusive). The triangular sum of nums is the value of the only element present in nums after the following process terminates: Let nums comprise of n elements. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n - 1. For each index i, where 0 <= i < n - 1, assign the value of newNums[i] as (nums[i] + nums[i+1]) % 10, where % denotes modulo operator. Replace the array nums with newNums. Repeat the entire process starting from step 1. Return the triangular sum of nums. Example 1: Input: nums = [1,2,3,4,5] Output: 8 Explanation: The above diagram depicts the process from which we obtain the triangular sum of the array. Example 2: Input: nums = [5] Output: 5 Explanation: Since there is only one element in nums, the triangular sum is the value of that element itself. Constraints: 1 <= nums.length <= 1000 0 <= nums[i] <= 9

Explanation

Here's the breakdown:

• Pascal's Triangle Connection: The core idea is to recognize that the final triangular sum is a weighted sum of the original numbers, where the weights correspond to binomial coefficients (specifically, the elements of Pascal's Triangle). We only need the middle row of Pascal's triangle coefficients.
• Binomial Coefficient Calculation: Efficiently compute the relevant binomial coefficients modulo 10. Using direct calculation of factorials can lead to overflow and is unnecessary. We can compute the coefficients incrementally.

• Weighted Sum: Calculate the weighted sum of the input array elements, taking the modulo 10 at each step to prevent overflow.

• Runtime Complexity: O(n), where n is the length of the input array.

• Storage Complexity: O(1) - constant extra space.

Code

    def triangularSum(nums):
    n = len(nums)
    if n == 1:
        return nums[0]

    coeff = 1
    result = 0
    for i in range(n):
        result = (result + coeff * nums[i]) % 10
        if i < n - 1:
            coeff = coeff * (n - 1 - i) * pow(i + 1, 8, 10) % 10

    return result