leetcode-problemset/leetcode-cn/originData/minimum-number-of-operations-to-make-x-and-y-equal.json

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            "categoryTitle": "Algorithms",
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            "title": "Minimum Number of Operations to Make X and Y Equal",
            "titleSlug": "minimum-number-of-operations-to-make-x-and-y-equal",
            "content": "<p>You are given two positive integers <code>x</code> and <code>y</code>.</p>\n\n<p>In one operation, you can do one of the four following operations:</p>\n\n<ol>\n\t<li>Divide <code>x</code> by <code>11</code> if <code>x</code> is a multiple of <code>11</code>.</li>\n\t<li>Divide <code>x</code> by <code>5</code> if <code>x</code> is a multiple of <code>5</code>.</li>\n\t<li>Decrement <code>x</code> by <code>1</code>.</li>\n\t<li>Increment <code>x</code> by <code>1</code>.</li>\n</ol>\n\n<p>Return <em>the <strong>minimum</strong> number of operations required to make </em> <code>x</code> <i>and</i> <code>y</code> equal.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> x = 26, y = 1\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> We can make 26 equal to 1 by applying the following operations: \n1. Decrement x by 1\n2. Divide x by 5\n3. Divide x by 5\nIt can be shown that 3 is the minimum number of operations required to make 26 equal to 1.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> x = 54, y = 2\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> We can make 54 equal to 2 by applying the following operations: \n1. Increment x by 1\n2. Divide x by 11 \n3. Divide x by 5\n4. Increment x by 1\nIt can be shown that 4 is the minimum number of operations required to make 54 equal to 2.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> x = 25, y = 30\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> We can make 25 equal to 30 by applying the following operations: \n1. Increment x by 1\n2. Increment x by 1\n3. Increment x by 1\n4. Increment x by 1\n5. Increment x by 1\nIt can be shown that 5 is the minimum number of operations required to make 25 equal to 30.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= x, y &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
            "translatedTitle": "使 X 和 Y 相等的最少操作次数",
            "translatedContent": "<p>给你两个正整数&nbsp;<code>x</code> 和&nbsp;<code>y</code>&nbsp;。</p>\n\n<p>一次操作中，你可以执行以下四种操作之一：</p>\n\n<ol>\n\t<li>如果 <code>x</code>&nbsp;是 <code>11</code>&nbsp;的倍数，将&nbsp;<code>x</code>&nbsp;除以&nbsp;<code>11</code>&nbsp;。</li>\n\t<li>如果 <code>x</code>&nbsp;是 <code>5</code>&nbsp;的倍数，将 <code>x</code>&nbsp;除以 <code>5</code>&nbsp;。</li>\n\t<li>将&nbsp;<code>x</code> 减&nbsp;<code>1</code>&nbsp;。</li>\n\t<li>将&nbsp;<code>x</code>&nbsp;加&nbsp;<code>1</code>&nbsp;。</li>\n</ol>\n\n<p>请你返回让 <code>x</code>&nbsp;和 <code>y</code>&nbsp;相等的 <strong>最少</strong>&nbsp;操作次数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>x = 26, y = 1\n<b>输出：</b>3\n<b>解释</b><strong>：</strong>我们可以通过以下操作将 26 变为 1 ：\n1. 将 x 减 1\n2. 将 x 除以 5\n3. 将 x 除以 5\n将 26 变为 1 最少需要 3 次操作。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>x = 54, y = 2\n<b>输出：</b>4\n<b>解释：</b>我们可以通过以下操作将 54 变为 2 ：\n1. 将 x 加 1\n2. 将 x 除以 11\n3. 将 x 除以 5\n4. 将 x 加 1\n将 54 变为 2 最少需要 4 次操作。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<b>输入：</b>x = 25, y = 30\n<b>输出：</b>5\n<b>解释：</b>我们可以通过以下操作将 25 变为 30 ：\n1. 将 x 加 1\n2. 将 x 加 1\n3. 将 x 加 1\n4. 将 x 加 1\n5. 将 x 加 1\n将 25 变为 30 最少需要 5 次操作。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= x, y &lt;= 10<sup>4</sup></code></li>\n</ul>\n",
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            "hints": [
                "The only way to make <code>x</code> larger is to increase it by <code>1</code> so if <code>y >= x</code> the answer is <code>y - x</code>.",
                "For <code>y < x</code>, <code>x - y</code> is always a candidate answer since we can repeatedly decrease <code>x</code> by one to reach <code>y</code>.",
                "We can also increase <code>x</code> and then use the division operations. For example, if <code>x = 10</code> and <code>y = 1</code>, we can increment <code>x</code> by <code>1</code> then divide it by <code>11</code>.",
                "Find an upper bound <code>U</code> on the maximum value of <code>x</code> we will reach an optimal solution. Since all values of <code>x</code> will be in the range <code>[1, U]</code>, we can use BFS to find the answer.",
                "One possible upper bound on <code>x</code> is <code>U = x + (x - y) </code>. To reach any number strictly greater than <code>U</code> from <code>x</code>, we will need more than <code>x - y</code> operations which is not optimal since we can always reach <code>y</code> in <code>x - y</code> operations."
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