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
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]
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];
}
};
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]
}