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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
nums
comprise ofn
elements. Ifn == 1
, end the process. Otherwise, create a new 0-indexed integer arraynewNums
of 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
nums
withnewNums
. - 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
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] }