2324. Find Triangular Sum Of An Array¶

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2324. 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:

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(int[] nums) {
        int n = nums.length;
        ArrayList<Integer> temp = new ArrayList<>();
        for (int i = 0; i < n; i++) 
            temp.add(nums[i]);
        while (true) {
            if (temp.size() == 1) 
                break;
            else {
                ArrayList<Integer> current = new ArrayList<>();
                for (int i = 0; i < temp.size() - 1; i++) 
                    current.add((temp.get(i) + temp.get(i + 1)) % 10);
                temp.clear();
                for (int ele : current)
                    temp.add(ele);
                if (temp.size() == 1) 
                    break;
            }
        }
        return temp.get(0);
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here