leetcode-problemset/leetcode-cn/originData/maximum-number-of-ways-to-partition-an-array.json

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        "question": {
            "questionId": "2135",
            "questionFrontendId": "2025",
            "categoryTitle": "Algorithms",
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            "title": "Maximum Number of Ways to Partition an Array",
            "titleSlug": "maximum-number-of-ways-to-partition-an-array",
            "content": "<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code>. The number of ways to <strong>partition</strong> <code>nums</code> is the number of <code>pivot</code> indices that satisfy both conditions:</p>\n\n<ul>\n\t<li><code>1 &lt;= pivot &lt; n</code></li>\n\t<li><code>nums[0] + nums[1] + ... + nums[pivot - 1] == nums[pivot] + nums[pivot + 1] + ... + nums[n - 1]</code></li>\n</ul>\n\n<p>You are also given an integer <code>k</code>. You can choose to change the value of <strong>one</strong> element of <code>nums</code> to <code>k</code>, or to leave the array <strong>unchanged</strong>.</p>\n\n<p>Return <em>the <strong>maximum</strong> possible number of ways to <strong>partition</strong> </em><code>nums</code><em> to satisfy both conditions after changing <strong>at most</strong> one element</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,-1,2], k = 3\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> One optimal approach is to change nums[0] to k. The array becomes [<strong><u>3</u></strong>,-1,2].\nThere is one way to partition the array:\n- For pivot = 2, we have the partition [3,-1 | 2]: 3 + -1 == 2.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [0,0,0], k = 1\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> The optimal approach is to leave the array unchanged.\nThere are two ways to partition the array:\n- For pivot = 1, we have the partition [0 | 0,0]: 0 == 0 + 0.\n- For pivot = 2, we have the partition [0,0 | 0]: 0 + 0 == 0.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [22,4,-25,-20,-15,15,-16,7,19,-10,0,-13,-14], k = -33\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> One optimal approach is to change nums[2] to k. The array becomes [22,4,<u><strong>-33</strong></u>,-20,-15,15,-16,7,19,-10,0,-13,-14].\nThere are four ways to partition the array.\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>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>-10<sup>5</sup> &lt;= k, nums[i] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
            "translatedTitle": "分割数组的最多方案数",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始且长度为 <code>n</code>&nbsp;的整数数组&nbsp;<code>nums</code>&nbsp;。<strong>分割</strong>&nbsp;数组 <code>nums</code>&nbsp;的方案数定义为符合以下两个条件的 <code>pivot</code>&nbsp;数目：</p>\n\n<ul>\n\t<li><code>1 &lt;= pivot &lt; n</code></li>\n\t<li><code>nums[0] + nums[1] + ... + nums[pivot - 1] == nums[pivot] + nums[pivot + 1] + ... + nums[n - 1]</code></li>\n</ul>\n\n<p>同时给你一个整数&nbsp;<code>k</code>&nbsp;。你可以将&nbsp;<code>nums</code>&nbsp;中&nbsp;<strong>一个</strong>&nbsp;元素变为&nbsp;<code>k</code>&nbsp;或&nbsp;<strong>不改变</strong>&nbsp;数组。</p>\n\n<p>请你返回在 <strong>至多</strong>&nbsp;改变一个元素的前提下，<strong>最多</strong>&nbsp;有多少种方法 <strong>分割</strong>&nbsp;<code>nums</code>&nbsp;使得上述两个条件都满足。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>nums = [2,-1,2], k = 3\n<b>输出：</b>1\n<b>解释：</b>一个最优的方案是将 nums[0] 改为 k&nbsp;。数组变为 [<em><strong>3</strong></em>,-1,2] 。\n有一种方法分割数组：\n- pivot = 2 ，我们有分割 [3,-1 | 2]：3 + -1 == 2 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>nums = [0,0,0], k = 1\n<b>输出：</b>2\n<b>解释：</b>一个最优的方案是不改动数组。\n有两种方法分割数组：\n- pivot = 1 ，我们有分割 [0 | 0,0]：0 == 0 + 0 。\n- pivot = 2 ，我们有分割 [0,0 | 0]: 0 + 0 == 0 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><b>输入：</b>nums = [22,4,-25,-20,-15,15,-16,7,19,-10,0,-13,-14], k = -33\n<b>输出：</b>4\n<b>解释：</b>一个最优的方案是将 nums[2] 改为 k 。数组变为 [22,4,<em><strong>-33</strong></em>,-20,-15,15,-16,7,19,-10,0,-13,-14] 。\n有四种方法分割数组。\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>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>-10<sup>5</sup> &lt;= k, nums[i] &lt;= 10<sup>5</sup></code></li>\n</ul>\n",
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                "A pivot point splits the array into equal prefix and suffix. If no change is made to the array, the goal is to find the number of pivot p such that prefix[p-1] == suffix[p].",
                "Consider how prefix and suffix will change when we change a number nums[i] to k.",
                "When sweeping through each element, can you find the total number of pivots where the difference of prefix and suffix happens to equal to the changes of k-nums[i]."
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