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"categoryTitle": "Algorithms",
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"title": "Split With Minimum Sum",
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"content": "<p>Given a positive integer <code>num</code>, split it into two non-negative integers <code>num1</code> and <code>num2</code> such that:</p>\n\n<ul>\n\t<li>The concatenation of <code>num1</code> and <code>num2</code> is a permutation of <code>num</code>.\n\n\t<ul>\n\t\t<li>In other words, the sum of the number of occurrences of each digit in <code>num1</code> and <code>num2</code> is equal to the number of occurrences of that digit in <code>num</code>.</li>\n\t</ul>\n\t</li>\n\t<li><code>num1</code> and <code>num2</code> can contain leading zeros.</li>\n</ul>\n\n<p>Return <em>the <strong>minimum</strong> possible sum of</em> <code>num1</code> <em>and</em> <code>num2</code>.</p>\n\n<p><strong>Notes:</strong></p>\n\n<ul>\n\t<li>It is guaranteed that <code>num</code> does not contain any leading zeros.</li>\n\t<li>The order of occurrence of the digits in <code>num1</code> and <code>num2</code> may differ from the order of occurrence of <code>num</code>.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> num = 4325\n<strong>Output:</strong> 59\n<strong>Explanation:</strong> We can split 4325 so that <code>num1 </code>is 24 and num2<code> is </code>35, giving a sum of 59. We can prove that 59 is indeed the minimal possible sum.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> num = 687\n<strong>Output:</strong> 75\n<strong>Explanation:</strong> We can split 687 so that <code>num1</code> is 68 and <code>num2 </code>is 7, which would give an optimal sum of 75.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>10 &lt;= num &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
"translatedTitle": "最小和分割",
"translatedContent": "<p>给你一个正整数&nbsp;<code>num</code>&nbsp;,请你将它分割成两个非负整数&nbsp;<code>num1</code> 和&nbsp;<code>num2</code>&nbsp;,满足:</p>\n\n<ul>\n\t<li><code>num1</code> 和&nbsp;<code>num2</code>&nbsp;直接连起来,得到&nbsp;<code>num</code>&nbsp;各数位的一个排列。\n\n\t<ul>\n\t\t<li>换句话说,<code>num1</code> 和&nbsp;<code>num2</code>&nbsp;中所有数字出现的次数之和等于&nbsp;<code>num</code>&nbsp;中所有数字出现的次数。</li>\n\t</ul>\n\t</li>\n\t<li><code>num1</code> 和&nbsp;<code>num2</code>&nbsp;可以包含前导 0 。</li>\n</ul>\n\n<p>请你返回&nbsp;<code>num1</code> 和 <code>num2</code>&nbsp;可以得到的和的 <strong>最小</strong> 值。</p>\n\n<p><strong>注意:</strong></p>\n\n<ul>\n\t<li><code>num</code>&nbsp;保证没有前导 0 。</li>\n\t<li><code>num1</code> 和&nbsp;<code>num2</code>&nbsp;中数位顺序可以与&nbsp;<code>num</code>&nbsp;中数位顺序不同。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<b>输入:</b>num = 4325\n<b>输出:</b>59\n<b>解释:</b>我们可以将 4325 分割成 <code>num1 </code>= 24 和 <code>num2 </code>= 35 ,和为 59 59 是最小和。\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<b>输入:</b>num = 687\n<b>输出:</b>75\n<b>解释:</b>我们可以将 687 分割成 <code>num1</code> = 68 和 <code>num2 </code>= 7 ,和为最优值 75 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>10 &lt;= num &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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