LeetCode 2221: Find Triangular Sum of an Array

Explanation:

To solve this problem, we iterate through the array nums repeatedly until there is only one element left. At each iteration, we compute the new array newNums by applying the given operation (nums[i] + nums[i+1]) % 10 to each pair of adjacent elements. We then replace the original array nums with newNums and continue the process. Finally, when there is only one element left in the array, we return that element as the triangular sum.

• Time complexity: O(n^2) where n is the number of elements in the input array.
• Space complexity: O(n) for the new array created at each iteration.

:

class Solution {
    public int findTriangularSum(int[] nums) {
        while(nums.length > 1) {
            int[] newNums = new int[nums.length - 1];
            for(int i = 0; i < nums.length - 1; i++) {
                newNums[i] = (nums[i] + nums[i+1]) % 10;
            }
            nums = newNums;
        }
        return nums[0];
    }
}