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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: Simulation

We can directly simulate the operations described in the problem. Perform \(n - 1\) rounds of operations on the array \(\textit{nums}\), updating the array \(\textit{nums}\) according to the rules described in the problem for each round. Finally, return the only remaining element in the array \(\textit{nums}\).

The time complexity is \(O(n^2)\), where \(n\) is the length of the array \(\textit{nums}\). The space complexity is \(O(1)\).

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class Solution:
    def triangularSum(self, nums: List[int]) -> int:
        for k in range(len(nums) - 1, 0, -1):
            for i in range(k):
                nums[i] = (nums[i] + nums[i + 1]) % 10
        return nums[0]

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

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

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func triangularSum(nums []int) int {
    for k := len(nums) - 1; k > 0; k-- {
        for i := 0; i < k; i++ {
            nums[i] = (nums[i] + nums[i+1]) % 10
        }
    }
    return nums[0]
}

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function triangularSum(nums: number[]): number {
    for (let k = nums.length - 1; k; --k) {
        for (let i = 0; i < k; ++i) {
            nums[i] = (nums[i] + nums[i + 1]) % 10;
        }
    }
    return nums[0];
}

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impl Solution {
    pub fn triangular_sum(mut nums: Vec<i32>) -> i32 {
        let mut k = nums.len() as i32 - 1;
        while k > 0 {
            for i in 0..k as usize {
                nums[i] = (nums[i] + nums[i + 1]) % 10;
            }
            k -= 1;
        }
        nums[0]
    }
}

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public class Solution {
    public int TriangularSum(int[] nums) {
        for (int k = nums.Length - 1; k > 0; --k) {
            for (int i = 0; i < k; ++i) {
                nums[i] = (nums[i] + nums[i + 1]) % 10;
            }
        }
        return nums[0];
    }
}

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