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알고리즘 문제/Leetcode

2221. Find Triangular Sum of an Array

BEstyle 2023. 5. 2. 14:50

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

class Solution:
    def triangularSum(self, nums: List[int]) -> int:
        def dfs(t):
            if len(t)==1:
                return t[0]
            temp=[]
            for i in range(len(t)-1):
                temp.append(t[i]+t[i+1])
            return dfs(temp)
        return dfs(nums) % 10