find-triangular-sum-of-an-array

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Algorithms

You are given a 0-indexed integer array <code>nums</code>, where <code>nums[i]</code> is a digit between <code>0</code> and <code>9</code> (inclusive).

The triangular sum of <code>nums</code> is the value of the only element present in <code>nums</code> after the following process terminates:

<ol>
	<li>Let <code>nums</code> comprise of <code>n</code> elements. If <code>n == 1</code>, end the process. Otherwise, create a new 0-indexed integer array <code>newNums</code> of length <code>n - 1</code>.</li>
	<li>For each index <code>i</code>, where <code>0 &lt;= i &lt;&nbsp;n - 1</code>, assign the value of <code>newNums[i]</code> as <code>(nums[i] + nums[i+1]) % 10</code>, where <code>%</code> denotes modulo operator.</li>
	<li>Replace the array <code>nums</code> with <code>newNums</code>.</li>
	<li>Repeat the entire process starting from step 1.</li>
</ol>

Return the triangular sum of <code>nums</code>.

&nbsp;
Example 1:
<img alt="" src="https://assets.leetcode.com/uploads/2022/02/22/ex1drawio.png" style="width: 250px; height: 250px;" />
<pre>
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.</pre>

Example 2:

<pre>
Input: nums = [5]
Output: 5
Explanation:
Since there is only one element in nums, the triangular sum is the value of that element itself.</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 9</code></li>
</ul>

triangularSum

Find Triangular Sum of an Array

Pascal's Triangle II

pascals-triangle-ii

杨辉三角 II

Calculate Digit Sum of a String

calculate-digit-sum-of-a-string

计算字符串的数字和

Min Max Game

min-max-game

极大极小游戏

Array

Math

Combinatorics

Simulation

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

<code>nums</code>&nbsp;的 三角和&nbsp;是执行以下操作以后最后剩下元素的值：

<ol>
	<li><code>nums</code>&nbsp;初始包含&nbsp;<code>n</code>&nbsp;个元素。如果&nbsp;<code>n == 1</code>&nbsp;，终止&nbsp;操作。否则，创建&nbsp;一个新的下标从&nbsp;0&nbsp;开始的长度为 <code>n - 1</code>&nbsp;的整数数组&nbsp;<code>newNums</code>&nbsp;。</li>
	<li>对于满足&nbsp;<code>0 &lt;= i &lt;&nbsp;n - 1</code>&nbsp;的下标&nbsp;<code>i</code>&nbsp;，<code>newNums[i]</code> 赋值&nbsp;为&nbsp;<code>(nums[i] + nums[i+1]) % 10</code>&nbsp;，<code>%</code>&nbsp;表示取余运算。</li>
	<li>将&nbsp;<code>newNums</code>&nbsp;替换 数组&nbsp;<code>nums</code>&nbsp;。</li>
	<li>从步骤 1 开始&nbsp;重复&nbsp;整个过程。</li>
</ol>

请你返回&nbsp;<code>nums</code>&nbsp;的三角和。

&nbsp;

示例 1：

<img alt="" src="https://assets.leetcode.com/uploads/2022/02/22/ex1drawio.png" style="width: 250px; height: 250px;" />

<pre>
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。</pre>

示例 2：

<pre>
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。</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 9</code></li>
</ul>