Today I have done leetcode's September LeetCoding Challenge with cpp.

September LeetCoding Challenge 30

Description

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:

1. 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.
2. 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.
3. Replace the array nums with newNums.
4. 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

Solution

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

// Accepted
// 300/300 cases passed (53 ms)
// Your runtime beats 79.33 % of cpp submissions
// Your memory usage beats 60.31 % of cpp submissions (22.5 MB)