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2221. Find Triangular Sum of an Array

Difficulty Topics

Description

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

find-triangular-sum-of-an-array.py
class Solution:
    def triangularSum(self, nums: List[int]) -> int:
        n = len(nums)

        for _ in range(n - 1):
            curr = []

            for i in range(1, len(nums)):
                curr.append((nums[i - 1] + nums[i]) % 10)

            nums = curr

        return nums[0]