leetcode-problemset/leetcode-cn/originData/maximum-sum-circular-subarray.json

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        "question": {
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            "categoryTitle": "Algorithms",
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            "title": "Maximum Sum Circular Subarray",
            "titleSlug": "maximum-sum-circular-subarray",
            "content": "<p>Given a <strong>circular integer array</strong> <code>nums</code> of length <code>n</code>, return <em>the maximum possible sum of a non-empty <strong>subarray</strong> of </em><code>nums</code>.</p>\n\n<p>A <strong>circular array</strong> means the end of the array connects to the beginning of the array. Formally, the next element of <code>nums[i]</code> is <code>nums[(i + 1) % n]</code> and the previous element of <code>nums[i]</code> is <code>nums[(i - 1 + n) % n]</code>.</p>\n\n<p>A <strong>subarray</strong> may only include each element of the fixed buffer <code>nums</code> at most once. Formally, for a subarray <code>nums[i], nums[i + 1], ..., nums[j]</code>, there does not exist <code>i &lt;= k1</code>, <code>k2 &lt;= j</code> with <code>k1 % n == k2 % n</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,-2,3,-2]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> Subarray [3] has maximum sum 3.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [5,-3,5]\n<strong>Output:</strong> 10\n<strong>Explanation:</strong> Subarray [5,5] has maximum sum 5 + 5 = 10.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [-3,-2,-3]\n<strong>Output:</strong> -2\n<strong>Explanation:</strong> Subarray [-2] has maximum sum -2.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>\n\t<li><code>-3 * 10<sup>4</sup> &lt;= nums[i] &lt;= 3 * 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "环形子数组的最大和",
            "translatedContent": "<p>给定一个长度为 <code>n</code> 的<strong>环形整数数组</strong>&nbsp;<code>nums</code>&nbsp;，返回<em>&nbsp;<code>nums</code>&nbsp;的非空 <strong>子数组</strong> 的最大可能和&nbsp;</em>。</p>\n\n<p><strong>环形数组</strong><em>&nbsp;</em>意味着数组的末端将会与开头相连呈环状。形式上， <code>nums[i]</code> 的下一个元素是 <code>nums[(i + 1) % n]</code> ， <code>nums[i]</code>&nbsp;的前一个元素是 <code>nums[(i - 1 + n) % n]</code> 。</p>\n\n<p><strong>子数组</strong> 最多只能包含固定缓冲区&nbsp;<code>nums</code>&nbsp;中的每个元素一次。形式上，对于子数组&nbsp;<code>nums[i], nums[i + 1], ..., nums[j]</code>&nbsp;，不存在&nbsp;<code>i &lt;= k1, k2 &lt;= j</code>&nbsp;其中&nbsp;<code>k1 % n == k2 % n</code>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,-2,3,-2]\n<strong>输出：</strong>3\n<strong>解释：</strong>从子数组 [3] 得到最大和 3\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [5,-3,5]\n<strong>输出：</strong>10\n<strong>解释：</strong>从子数组 [5,5] 得到最大和 5 + 5 = 10\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [3,-2,2,-3]\n<strong>输出：</strong>3\n<strong>解释：</strong>从子数组 [3] 和 [3,-2,2] 都可以得到最大和 3\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>\n\t<li><code>-3 * 10<sup>4</sup>&nbsp;&lt;= nums[i] &lt;= 3 * 10<sup>4</sup></code>​​​​​​​</li>\n</ul>\n",
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                "For those of you who are familiar with the <b>Kadane's algorithm</b>, think in terms of that. For the newbies, Kadane's algorithm is used to finding the maximum sum subarray from a given array. This problem is a twist on that idea and it is advisable to read up on that algorithm first before starting this problem. Unless you already have a great algorithm brewing up in your mind in which case, go right ahead!",
                "What is an alternate way of representing a circular array so that it appears to be a straight array?\r\nEssentially, there are two cases of this problem that we need to take care of. Let's look at the figure below to understand those two cases:\r\n\r\n<br>\r\n<img src=\"https://assets.leetcode.com/uploads/2019/10/20/circular_subarray_hint_1.png\" width=\"700\"/>",
                "The first case can be handled by the good old Kadane's algorithm. However, is there a smarter way of going about handling the second case as well?"
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