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2221. Find Triangular Sum of an Array
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:
- Let
numscomprise ofnelements. Ifn == 1, end the process. Otherwise, create a new 0-indexed integer arraynewNumsof lengthn - 1. - For each index
i, where0 <= i < n - 1, assign the value ofnewNums[i]as(nums[i] + nums[i+1]) % 10, where%denotes modulo operator. - Replace the array
numswithnewNums. - 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 <= 10000 <= nums[i] <= 9
Solutions
-
class Solution { public int triangularSum(int[] nums) { int n = nums.length; for (int i = n; i >= 0; --i) { for (int j = 0; j < i - 1; ++j) { nums[j] = (nums[j] + nums[j + 1]) % 10; } } return nums[0]; } } -
class Solution { public: int triangularSum(vector<int>& nums) { int n = nums.size(); for (int i = n; i >= 0; --i) for (int j = 0; j < i - 1; ++j) nums[j] = (nums[j] + nums[j + 1]) % 10; return nums[0]; } }; -
class Solution: def triangularSum(self, nums: List[int]) -> int: n = len(nums) for i in range(n, 0, -1): for j in range(i - 1): nums[j] = (nums[j] + nums[j + 1]) % 10 return nums[0] -
func triangularSum(nums []int) int { n := len(nums) for i := n; i >= 0; i-- { for j := 0; j < i-1; j++ { nums[j] = (nums[j] + nums[j+1]) % 10 } } return nums[0] } -
function triangularSum(nums: number[]): number { for (let k = nums.length - 1; k; --k) { for (let i = 0; i < k; ++i) { nums[i] = (nums[i] + nums[i + 1]) % 10; } } return nums[0]; } -
public class Solution { public int TriangularSum(int[] nums) { for (int k = nums.Length - 1; k > 0; --k) { for (int i = 0; i < k; ++i) { nums[i] = (nums[i] + nums[i + 1]) % 10; } } return nums[0]; } } -
impl Solution { pub fn triangular_sum(mut nums: Vec<i32>) -> i32 { let mut k = nums.len() as i32 - 1; while k > 0 { for i in 0..k as usize { nums[i] = (nums[i] + nums[i + 1]) % 10; } k -= 1; } nums[0] } }