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            "questionId": "3364",
            "questionFrontendId": "3117",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2734243,
            "title": "Minimum Sum of Values by Dividing Array",
            "titleSlug": "minimum-sum-of-values-by-dividing-array",
            "content": "<p>You are given two arrays <code>nums</code> and <code>andValues</code> of length <code>n</code> and <code>m</code> respectively.</p>\n\n<p>The <strong>value</strong> of an array is equal to the <strong>last</strong> element of that array.</p>\n\n<p>You have to divide <code>nums</code> into <code>m</code> <strong>disjoint contiguous</strong> <span data-keyword=\"subarray-nonempty\">subarrays</span> such that for the <code>i<sup>th</sup></code> subarray <code>[l<sub>i</sub>, r<sub>i</sub>]</code>, the bitwise <code>AND</code> of the subarray elements is equal to <code>andValues[i]</code>, in other words, <code>nums[l<sub>i</sub>] &amp; nums[l<sub>i</sub> + 1] &amp; ... &amp; nums[r<sub>i</sub>] == andValues[i]</code> for all <code>1 &lt;= i &lt;= m</code>, where <code>&amp;</code> represents the bitwise <code>AND</code> operator.</p>\n\n<p>Return <em>the <strong>minimum</strong> possible sum of the <strong>values</strong> of the </em><code>m</code><em> subarrays </em><code>nums</code><em> is divided into</em>. <em>If it is not possible to divide </em><code>nums</code><em> into </em><code>m</code><em> subarrays satisfying these conditions, return</em> <code>-1</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [1,4,3,3,2], andValues = [0,3,3,2]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">12</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The only possible way to divide <code>nums</code> is:</p>\n\n<ol>\n\t<li><code>[1,4]</code> as <code>1 &amp; 4 == 0</code>.</li>\n\t<li><code>[3]</code> as the bitwise <code>AND</code> of a single element subarray is that element itself.</li>\n\t<li><code>[3]</code> as the bitwise <code>AND</code> of a single element subarray is that element itself.</li>\n\t<li><code>[2]</code> as the bitwise <code>AND</code> of a single element subarray is that element itself.</li>\n</ol>\n\n<p>The sum of the values for these subarrays is <code>4 + 3 + 3 + 2 = 12</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [2,3,5,7,7,7,5], andValues = [0,7,5]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">17</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>There are three ways to divide <code>nums</code>:</p>\n\n<ol>\n\t<li><code>[[2,3,5],[7,7,7],[5]]</code> with the sum of the values <code>5 + 7 + 5 == 17</code>.</li>\n\t<li><code>[[2,3,5,7],[7,7],[5]]</code> with the sum of the values <code>7 + 7 + 5 == 19</code>.</li>\n\t<li><code>[[2,3,5,7,7],[7],[5]]</code> with the sum of the values <code>7 + 7 + 5 == 19</code>.</li>\n</ol>\n\n<p>The minimum possible sum of the values is <code>17</code>.</p>\n</div>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">nums = [1,2,3,4], andValues = [2]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">-1</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p>The bitwise <code>AND</code> of the entire array <code>nums</code> is <code>0</code>. As there is no possible way to divide <code>nums</code> into a single subarray to have the bitwise <code>AND</code> of elements <code>2</code>, return <code>-1</code>.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == nums.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= m == andValues.length &lt;= min(n, 10)</code></li>\n\t<li><code>1 &lt;= nums[i] &lt; 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= andValues[j] &lt; 10<sup>5</sup></code></li>\n</ul>\n",
            "translatedTitle": "划分数组得到最小的值之和",
            "translatedContent": "<p>给你两个数组 <code>nums</code> 和 <code>andValues</code>，长度分别为 <code>n</code> 和 <code>m</code>。</p>\n\n<p>数组的 <strong>值 </strong>等于该数组的 <strong>最后一个 </strong>元素。</p>\n\n<p>你需要将 <code>nums</code> 划分为 <code>m</code> 个 <strong>不相交的连续 </strong><span data-keyword=\"subarray-nonempty\">子数组</span>，对于第 <code>i<sup>th</sup></code> 个子数组 <code>[l<sub>i</sub>, r<sub>i</sub>]</code>，子数组元素的按位&nbsp;<code>AND</code>&nbsp;运算结果等于 <code>andValues[i]</code>，换句话说，对所有的 <code>1 &lt;= i &lt;= m</code>，<code>nums[l<sub>i</sub>] &amp; nums[l<sub>i</sub> + 1] &amp; ... &amp; nums[r<sub>i</sub>] == andValues[i]</code> ，其中 <code>&amp;</code> 表示按位&nbsp;<code>AND</code>&nbsp;运算符。</p>\n\n<p>返回将 <code>nums</code> 划分为 <code>m</code> 个子数组所能得到的可能的 <strong>最小 </strong>子数组 <strong>值</strong> 之和。如果无法完成这样的划分，则返回 <code>-1</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong> <span class=\"example-io\">nums = [1,4,3,3,2], andValues = [0,3,3,2]</span></p>\n\n<p><strong>输出：</strong> <span class=\"example-io\">12</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>唯一可能的划分方法为：</p>\n\n<ol>\n\t<li><code>[1,4]</code> 因为 <code>1 &amp; 4 == 0</code></li>\n\t<li><code>[3]</code> 因为单元素子数组的按位 <code>AND</code> 结果就是该元素本身</li>\n\t<li><code>[3]</code> 因为单元素子数组的按位 <code>AND</code> 结果就是该元素本身</li>\n\t<li><code>[2]</code> 因为单元素子数组的按位 <code>AND</code> 结果就是该元素本身</li>\n</ol>\n\n<p>这些子数组的值之和为 <code>4 + 3 + 3 + 2 = 12</code></p>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong> <span class=\"example-io\">nums = [2,3,5,7,7,7,5], andValues = [0,7,5]</span></p>\n\n<p><strong>输出：</strong> <span class=\"example-io\">17</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>划分 <code>nums</code> 的三种方式为：</p>\n\n<ol>\n\t<li><code>[[2,3,5],[7,7,7],[5]]</code> 其中子数组的值之和为 <code>5 + 7 + 5 = 17</code></li>\n\t<li><code>[[2,3,5,7],[7,7],[5]]</code> 其中子数组的值之和为 <code>7 + 7 + 5 = 19</code></li>\n\t<li><code>[[2,3,5,7,7],[7],[5]]</code> 其中子数组的值之和为 <code>7 + 7 + 5 = 19</code></li>\n</ol>\n\n<p>子数组值之和的最小可能值为 <code>17</code></p>\n</div>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<div class=\"example-block\">\n<p><strong>输入：</strong> <span class=\"example-io\">nums = [1,2,3,4], andValues = [2]</span></p>\n\n<p><strong>输出：</strong> <span class=\"example-io\">-1</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>整个数组 <code>nums</code> 的按位 <code>AND</code> 结果为 <code>0</code>。由于无法将 <code>nums</code> 划分为单个子数组使得元素的按位 <code>AND</code> 结果为 <code>2</code>，因此返回 <code>-1</code>。</p>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == nums.length &lt;= 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= m == andValues.length &lt;= min(n, 10)</code></li>\n\t<li><code>1 &lt;= nums[i] &lt; 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= andValues[j] &lt; 10<sup>5</sup></code></li>\n</ul>\n",
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                "Let <code>dp[i][j]</code> be the optimal answer to split  <code>nums[0..(i - 1)]</code> into the first <code>j</code> andValues.",
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                "The answer is <code>dp[n][m]</code>.",
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<code>use-after-free<\\/code> \\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/troydhanson.github.io\\/uthash\\/\\\" target=\\\"_blank\\\">uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n<p><b>1. \\u5f80\\u54c8\\u5e0c\\u8868\\u4e2d\\u6dfb\\u52a0\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n<pre>\\r\\nstruct hash_entry {\\r\\n    int id;            \\/* we'll use this field as the key *\\/\\r\\n    char name[10];\\r\\n    UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry 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5.1.6<\\/p>\\r\\n\\r\\n<p>Compile Options: --alwaysStrict --strictBindCallApply --strictFunctionTypes --target ES2022<\\/p>\\r\\n\\r\\n<p><a href=\\\"https:\\/\\/lodash.com\\\" target=\\\"_blank\\\">lodash.js<\\/a> \\u5e93\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5305\\u542b\\u3002<\\/p>\\r\\n\\r\\n<p> \\u5982\\u9700\\u4f7f\\u7528\\u961f\\u5217\\/\\u4f18\\u5148\\u961f\\u5217\\uff0c\\u60a8\\u53ef\\u4f7f\\u7528 <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/priority-queue\\/blob\\/v5\\/README.md\\\" target=\\\"_blank\\\"> datastructures-js\\/priority-queue@5.4.0<\\/a>\\uff0c<a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/queue\\/tree\\/v4.2.3\\\" target=\\\"_blank\\\"> datastructures-js\\/queue@4.2.3<\\/a> \\u4ee5\\u53ca <a href=\\\"https:\\/\\/github.com\\/datastructures-js\\/deque\\/tree\\/v1.0.4\\\" target=\\\"_blank\\\"> datastructures-js\\/deque@1.0.4<\\/a>\\u3002<\\/p>\"],\"racket\":[\"Racket\",\"<p><a 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