Find Triangular Sum of an Array

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:

1. 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.
2. 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.
3. Replace the array nums with newNums.
4. 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

SOLUTION:

class Solution:
    def triangularSum(self, nums: List[int]) -> int:
        k = len(nums)
        curr = [1]
        n = 1
        for i in range(k - 1):
            curr = [0] + curr
            n += 1
            for i in range(n - 1):
                curr[i] = curr[i] + curr[i + 1]
        total = 0
        for i in range(k):
            total += nums[i] * curr[i]
        return total % 10