leetcode-problemset/leetcode-cn/originData/minimum-cost-to-make-all-characters-equal.json

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            "title": "Minimum Cost to Make All Characters Equal",
            "titleSlug": "minimum-cost-to-make-all-characters-equal",
            "content": "<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> of length <code>n</code> on which you can apply two types of operations:</p>\n\n<ul>\n\t<li>Choose an index <code>i</code> and invert all characters from&nbsp;index <code>0</code> to index <code>i</code>&nbsp;(both inclusive), with a cost of <code>i + 1</code></li>\n\t<li>Choose an index <code>i</code> and invert all characters&nbsp;from&nbsp;index <code>i</code> to index <code>n - 1</code>&nbsp;(both inclusive), with a cost of <code>n - i</code></li>\n</ul>\n\n<p>Return <em>the <strong>minimum cost </strong>to make all characters of the string <strong>equal</strong></em>.</p>\n\n<p><strong>Invert</strong> a character means&nbsp;if its value is &#39;0&#39; it becomes &#39;1&#39; and vice-versa.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;0011&quot;\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> Apply the second operation with <code>i = 2</code> to obtain <code>s = &quot;0000&quot; for a cost of 2</code>. It can be shown that 2 is the minimum cost to make all characters equal.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;010101&quot;\n<strong>Output:</strong> 9\n<strong>Explanation:</strong> Apply the first operation with i = 2 to obtain s = &quot;101101&quot; for a cost of 3.\nApply the first operation with i = 1 to obtain s = &quot;011101&quot; for a cost of 2. \nApply the first operation with i = 0 to obtain s = &quot;111101&quot; for a cost of 1. \nApply the second operation with i = 4 to obtain s = &quot;111110&quot; for a cost of 2.\nApply the second operation with i = 5 to obtain s = &quot;111111&quot; for a cost of 1. \nThe total cost to make all characters equal is 9. It can be shown that 9 is the minimum cost to make all characters equal.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length == n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> is either <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code></li>\n</ul>\n",
            "translatedTitle": "使所有字符相等的最小成本",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始、长度为 <code>n</code> 的二进制字符串 <code>s</code> ，你可以对其执行两种操作：</p>\n\n<ul>\n\t<li>选中一个下标 <code>i</code> 并且反转从下标 <code>0</code> 到下标 <code>i</code>（包括下标 <code>0</code> 和下标 <code>i</code> ）的所有字符，成本为 <code>i + 1</code> 。</li>\n\t<li>选中一个下标 <code>i</code> 并且反转从下标 <code>i</code> 到下标 <code>n - 1</code>（包括下标 <code>i</code> 和下标 <code>n - 1</code> ）的所有字符，成本为 <code>n - i</code> 。</li>\n</ul>\n\n<p>返回使字符串内所有字符 <strong>相等</strong> 需要的 <strong>最小成本</strong> 。</p>\n\n<p><strong>反转</strong> 字符意味着：如果原来的值是 '0' ，则反转后值变为 '1' ，反之亦然。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"0011\"\n<strong>输出：</strong>2\n<strong>解释：</strong>执行第二种操作，选中下标 <code>i = 2</code> ，可以得到 <code>s = \"0000\" ，成本为 2</code> 。可以证明 2 是使所有字符相等的最小成本。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"010101\"\n<strong>输出：</strong>9\n<strong>解释：</strong>执行第一种操作，选中下标 i = 2 ，可以得到 s = \"101101\" ，成本为 3 。\n执行第一种操作，选中下标 i = 1 ，可以得到 s = \"011101\" ，成本为 2 。\n执行第一种操作，选中下标 i = 0 ，可以得到 s = \"111101\" ，成本为 1 。\n执行第二种操作，选中下标 i = 4 ，可以得到 s = \"111110\" ，成本为 2 。\n执行第一种操作，选中下标 i = 5 ，可以得到 s = \"111111\" ，成本为 1 。\n使所有字符相等的总成本等于 9 。可以证明 9 是使所有字符相等的最小成本。 </pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length == n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> 为 <code>'0'</code> 或 <code>'1'</code></li>\n</ul>\n",
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                "For every index i, calculate the number of operations required to make the prefix [0, i - 1] equal to the character at index i, denoted prefix[i].",
                "For every index i, calculate the number of operations required to make the suffix [i + 1, n - 1] equal to the character at index i, denoted suffix[i].",
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