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:
- Let
numscomprise ofnelements. Ifn == 1, end the process. Otherwise, create a new 0-indexed integer arraynewNumsof lengthn - 1. - For each index
i, where0 <= i < n - 1, assign the value ofnewNums[i]as(nums[i] + nums[i+1]) % 10, where%denotes modulo operator. - Replace the array
numswithnewNums. - 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 <= 10000 <= 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 $O(1)$.
Python3
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]
Java
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];
}
}
C++
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];
}
};
Go
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]
}
TypeScript
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];
}