LeetCode 2221. Find Triangular Sum of an Array Solution in Java, C++, Python & More | Explanation + Code

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 nums, updating the array nums according to the rules described in the problem for each round. Finally, return the only remaining element in the array nums.

The time complexity is O(n²), where n is the length of the array nums. The space complexity is O(1).

PythonJavaC++GoTypeScriptRustC#

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](code-box)

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]; } }(code-box)

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]; } };(code-box)

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] }(code-box)

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]; }(code-box)

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] } }(code-box)

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]; } }(code-box)