leetcode-problemset/leetcode-cn/originData/balanced-k-factor-decomposition.json

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        "question": {
            "questionId": "3947",
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            "categoryTitle": "Algorithms",
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            "title": "Balanced K-Factor Decomposition",
            "titleSlug": "balanced-k-factor-decomposition",
            "content": "<p>Given two integers <code>n</code> and <code>k</code>, split the number <code>n</code> into exactly <code>k</code> positive integers such that the <strong>product</strong> of these integers is equal to <code>n</code>.</p>\n\n<p>Return <em>any</em> <em>one</em> split in which the <strong>maximum</strong> difference between any two numbers is <strong>minimized</strong>. You may return the result in <em>any order</em>.</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\">n = 100, k = 2</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[10,10]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<p data-end=\"157\" data-start=\"0\">The split <code>[10, 10]</code> yields <code>10 * 10 = 100</code> and a max-min difference of 0, which is minimal.</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\">n = 44, k = 3</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">[2,2,11]</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ul>\n\t<li data-end=\"46\" data-start=\"2\">Split <code>[1, 1, 44]</code> yields a difference of 43</li>\n\t<li data-end=\"93\" data-start=\"49\">Split <code>[1, 2, 22]</code> yields a difference of 21</li>\n\t<li data-end=\"140\" data-start=\"96\">Split <code>[1, 4, 11]</code> yields a difference of 10</li>\n\t<li data-end=\"186\" data-start=\"143\">Split <code>[2, 2, 11]</code> yields a difference of 9</li>\n</ul>\n\n<p data-end=\"264\" data-is-last-node=\"\" data-is-only-node=\"\" data-start=\"188\">Therefore, <code>[2, 2, 11]</code> is the optimal split with the smallest difference 9.</p>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li data-end=\"54\" data-start=\"37\"><code data-end=\"52\" data-start=\"37\">4 &lt;= n &lt;= 10<sup><span style=\"font-size: 10.8333px;\">5</span></sup></code></li>\n\t<li data-end=\"71\" data-start=\"57\"><code data-end=\"69\" data-start=\"57\">2 &lt;= k &lt;= 5</code></li>\n\t<li data-end=\"145\" data-is-last-node=\"\" data-start=\"74\"><code data-end=\"77\" data-start=\"74\">k</code> is strictly less than the total number of positive divisors of <code data-end=\"144\" data-is-only-node=\"\" data-start=\"141\">n</code>.</li>\n</ul>\n",
            "translatedTitle": "K 因数分解",
            "translatedContent": "<p>给你两个整数 <code>n</code> 和 <code>k</code>，将数字 <code>n</code> 恰好分割成 <code>k</code> 个正整数，使得这些整数的&nbsp;<strong>乘积&nbsp;</strong>等于 <code>n</code>。</p>\n\n<p>返回一个分割方案，使得这些数字中&nbsp;<strong>最大值&nbsp;</strong>和&nbsp;<strong>最小值&nbsp;</strong>之间的&nbsp;<strong>差值&nbsp;</strong>最小化。结果可以以&nbsp;<strong>任意顺序</strong>&nbsp;返回。</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\">n = 100, k = 2</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">[10,10]</span></p>\n\n<p><strong>解释：</strong></p>\n\n<p>分割方案 <code>[10, 10]</code> 的结果是 <code>10 * 10 = 100</code>，且最大值与最小值的差值为 0，这是最小可能值。</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\">n = 44, k = 3</span></p>\n\n<p><strong>输出：</strong><span class=\"example-io\">[2,2,11]</span></p>\n\n<p><strong>解释：</strong></p>\n\n<ul>\n\t<li>分割方案 <code>[1, 1, 44]</code> 的差值为 43</li>\n\t<li>分割方案 <code>[1, 2, 22]</code> 的差值为 21</li>\n\t<li>分割方案 <code>[1, 4, 11]</code> 的差值为 10</li>\n\t<li>分割方案 <code>[2, 2, 11]</code> 的差值为 9</li>\n</ul>\n\n<p>因此，<code>[2, 2, 11]</code> 是最优分割方案，其差值最小，为 9。</p>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>4 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>2 &lt;= k &lt;= 5</code></li>\n\t<li><code>k</code> 严格小于 <code>n</code> 的正因数的总数。</li>\n</ul>\n",
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                "First, compute all positive divisors of <code>n</code> and sort them into a list <code>divs</code>.",
                "Use a recursive search <code>dfs(start, picked, prod, path)</code> that picks the next divisor from <code>divs[start...]</code>, multiplies it into <code>prod</code>, and appends to <code>path</code> until you have <code>k</code> factors whose product is <code>n</code>.",
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