Formatted question description: https://leetcode.ca/all/2221.html

# 2221. Find Triangular Sum of an Array (Medium)

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 of`n`

elements. If`n == 1`

,**end**the process. Otherwise,**create**a new**0-indexed**integer array`newNums`

of length`n - 1`

. - 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. **Replace**the array`nums`

with`newNums`

.**Repeat**the entire process starting from step 1.

Return *the triangular sum of* `nums`

.

**Example 1:**

Input:nums = [1,2,3,4,5]Output:8Explanation:The above diagram depicts the process from which we obtain the triangular sum of the array.

**Example 2:**

Input:nums = [5]Output:5Explanation: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`

**Similar Questions**:

## Solution 1. Pascal’s Triangle

```
// OJ: https://leetcode.com/problems/find-triangular-sum-of-an-array/
// Time: O(N^2)
// Space: O(1) extra space
class Solution {
public:
int triangularSum(vector<int>& A) {
for (int i = A.size(); i >= 1; --i) {
for (int j = 0; j < i - 1; ++j) {
A[j] = (A[j] + A[j + 1]) % 10;
}
}
return A[0];
}
};
```

## Discuss

https://leetcode.com/problems/find-triangular-sum-of-an-array/discuss/1907038/