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leetcode-problemset/leetcode-cn/originData/find-triangular-sum-of-an-array.json
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"question": {
"questionId": "2324",
"questionFrontendId": "2221",
"categoryTitle": "Algorithms",
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"title": "Find Triangular Sum of an Array",
"titleSlug": "find-triangular-sum-of-an-array",
"content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>, where <code>nums[i]</code> is a digit between <code>0</code> and <code>9</code> (<strong>inclusive</strong>).</p>\n\n<p>The <strong>triangular sum</strong> of <code>nums</code> is the value of the only element present in <code>nums</code> after the following process terminates:</p>\n\n<ol>\n\t<li>Let <code>nums</code> comprise of <code>n</code> elements. If <code>n == 1</code>, <strong>end</strong> the process. Otherwise, <strong>create</strong> a new <strong>0-indexed</strong> integer array <code>newNums</code> of length <code>n - 1</code>.</li>\n\t<li>For each index <code>i</code>, where <code>0 &lt;= i &lt;&nbsp;n - 1</code>, <strong>assign</strong> the value of <code>newNums[i]</code> as <code>(nums[i] + nums[i+1]) % 10</code>, where <code>%</code> denotes modulo operator.</li>\n\t<li><strong>Replace</strong> the array <code>nums</code> with <code>newNums</code>.</li>\n\t<li><strong>Repeat</strong> the entire process starting from step 1.</li>\n</ol>\n\n<p>Return <em>the triangular sum of</em> <code>nums</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/02/22/ex1drawio.png\" style=\"width: 250px; height: 250px;\" />\n<pre>\n<strong>Input:</strong> nums = [1,2,3,4,5]\n<strong>Output:</strong> 8\n<strong>Explanation:</strong>\nThe above diagram depicts the process from which we obtain the triangular sum of the array.</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [5]\n<strong>Output:</strong> 5\n<strong>Explanation:</strong>\nSince there is only one element in nums, the triangular sum is the value of that element itself.</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 9</code></li>\n</ul>\n",
"translatedTitle": "数组的三角和",
"translatedContent": "<p>给你一个下标从 <strong>0</strong>&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。</p>\n\n<p><code>nums</code>&nbsp;的 <strong>三角和</strong>&nbsp;是执行以下操作以后最后剩下元素的值:</p>\n\n<ol>\n\t<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;<strong>终止</strong>&nbsp;操作。否则,<strong>创建</strong>&nbsp;一个新的下标从&nbsp;<strong>0</strong>&nbsp;开始的长度为 <code>n - 1</code>&nbsp;的整数数组&nbsp;<code>newNums</code>&nbsp;。</li>\n\t<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> <strong>赋值</strong>&nbsp;为&nbsp;<code>(nums[i] + nums[i+1]) % 10</code>&nbsp;<code>%</code>&nbsp;表示取余运算。</li>\n\t<li>将&nbsp;<code>newNums</code>&nbsp;<strong>替换</strong> 数组&nbsp;<code>nums</code>&nbsp;。</li>\n\t<li>从步骤 1 开始&nbsp;<strong>重复</strong>&nbsp;整个过程。</li>\n</ol>\n\n<p>请你返回&nbsp;<code>nums</code>&nbsp;的三角和。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<p><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/02/22/ex1drawio.png\" style=\"width: 250px; height: 250px;\" /></p>\n\n<pre>\n<b>输入:</b>nums = [1,2,3,4,5]\n<b>输出:</b>8\n<strong>解释:</strong>\n上图展示了得到数组三角和的过程。</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<b>输入:</b>nums = [5]\n<b>输出:</b>5\n<b>解释:</b>\n由于 nums 中只有一个元素,数组的三角和为这个元素自己。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>0 &lt;= nums[i] &lt;= 9</code></li>\n</ul>\n",
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"To reduce space, use a temporary array to update nums in every step instead of creating a new array at each step."
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