30_Find_Triangular_Sum_of_an_Array.cpp

// 2221. Find Triangular Sum of an Array

// 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: 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
{
public:
    int triangularSum(vector<int> &nums)
    {
        int n = nums.size();
        for (int size = n; size > 1; --size)
        {
            for (int i = 0; i < size - 1; ++i)
            {
                nums[i] = (nums[i] + nums[i + 1]) % 10;
            }
        }
        return nums[0];
    }
};