What is the triangular sum of an array?

The triangular sum of an array is the final element obtained by repeatedly summing adjacent pairs in the array until only one element remains.

How does the iterative process work in the "Find Triangular Sum of an Array" problem?

You repeatedly sum adjacent pairs of elements, updating the array with these sums, until you have only one element remaining.

What are the time and space complexities of solving this problem?

Time complexity is O(n^2) for the simulation, and space complexity is O(1) if done in-place, or O(n) if using an additional array.

How can I optimize my approach to solving the triangular sum problem?

Focus on minimizing the number of steps by efficiently handling array updates, and consider in-place array modifications to save space.

What is the best approach for large arrays in the "Find Triangular Sum of an Array" problem?

For large arrays, make sure to minimize time complexity by avoiding redundant operations, and reduce space complexity by reusing the array for updates.

#2221

Medium

auto_awesome数组·数学

LeetCode 题解工作台

数组的三角和

给你一个下标从 0 开始的整数数组 nums ，其中 nums[i] 是 0 到 9 之间（两者都包含）的一个数字。 nums 的三角和是执行以下操作以后最后剩下元素的值： nums 初始包含 n 个元素。如果 n == 1 ，终止操作。否则，创建一个新的下标从 0 开始的长度为 n -…

数组数学模拟组合数学

题目描述

给你一个下标从 0 开始的整数数组 nums ，其中 nums[i] 是 0 到 9 之间（两者都包含）的一个数字。

nums 的 三角和 是执行以下操作以后最后剩下元素的值：

nums 初始包含 n 个元素。如果 n == 1 ，终止操作。否则，创建一个新的下标从 0 开始的长度为 n - 1 的整数数组 newNums 。
对于满足 0 <= i < n - 1 的下标 i ，newNums[i] 赋值为 (nums[i] + nums[i+1]) % 10 ，% 表示取余运算。
将 newNums 替换数组 nums 。
从步骤 1 开始重复整个过程。

请你返回 nums 的三角和。

示例 1：

输入：nums = [1,2,3,4,5]
输出：8
解释：
上图展示了得到数组三角和的过程。

示例 2：

输入：nums = [5]
输出：5
解释：
由于 nums 中只有一个元素，数组的三角和为这个元素自己。

提示：

1 <= nums.length <= 1000
0 <= nums[i] <= 9

lightbulb

解题思路

方法一：模拟

我们可以直接模拟题目描述的操作，对数组 $\textit{nums}$ 进行 $n - 1$ 轮操作，每轮操作都按照题目描述的规则更新数组 $\textit{nums}$ 。最后返回数组 $\textit{nums}$ 中剩下的唯一元素即可。

时间复杂度 $O(n^2)$ ，其中 $n$ 是数组 $\textit{nums}$ 的长度。空间复杂度 $O(1)$ 。

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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]

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Look for understanding of iterative processes in arrays.
question_mark
Check for awareness of space optimization techniques in array manipulation.
question_mark
Evaluate familiarity with time complexity reductions and efficient problem-solving.

warning

常见陷阱

外企场景

error
Failing to correctly simulate the process by not updating the array properly in each step.
error
Not optimizing for space by using an additional array when the process can be done in-place.
error
Overcomplicating the solution by attempting to handle the problem in a more complex way when a simple iterative approach works.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Different array sizes could be tested, from the smallest case with one element to the largest with 1000 elements.
arrow_right_alt
Modifying the problem to use elements larger than 9 to test how the solution handles different ranges of numbers.
arrow_right_alt
Introducing more complex operations between elements (such as subtraction or multiplication) could test the adaptability of the algorithm.

help

常见问题

外企场景

继续练习

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