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:

  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

Solutions

Solution 1

Python Code
1
2
3
4
5
6
7

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]

Java Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11

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];
    }
}

C++ Code
 1
 2
 3
 4
 5
 6
 7
 8
 9
10

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];
    }
};

Go Code
1
2
3
4
5
6
7
8
9

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]
}