LeetCode in Kotlin

2221. 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 {
    fun triangularSum(nums: IntArray): Int {
        var len = nums.size
        while (len-- > 1) {
            for (i in 0 until len) {
                nums[i] += nums[i + 1]
                nums[i] %= 10
            }
        }
        return nums[0]
    }
}

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