update

leetcode-cn/originData/minimum-suffix-flips.json

@@ -12,7 +12,7 @@
             "translatedContent": "<p>给你一个长度为 <code>n</code> 、下标从 <strong>0</strong> 开始的二进制字符串 <code>target</code> 。你自己有另一个长度为 <code>n</code> 的二进制字符串 <code>s</code> ，最初每一位上都是 0 。你想要让 <code>s</code> 和 <code>target</code> 相等。</p>\n\n<p>在一步操作，你可以选择下标 <code>i</code>（<code>0 &lt;= i &lt; n</code>）并翻转在 <strong>闭区间</strong> <code>[i, n - 1]</code> 内的所有位。翻转意味着 <code>'0'</code> 变为 <code>'1'</code> ，而 <code>'1'</code> 变为 <code>'0'</code> 。</p>\n\n<div class=\"original__bRMd\">\n<div>\n<p>返回使<em> </em><code>s</code><em> </em>与<em> </em><code>target</code> 相等需要的最少翻转次数。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>target = \"10111\"\n<strong>输出：</strong>3\n<strong>解释：</strong>最初，s = \"00000\" 。\n选择下标 i = 2: \"00<em><strong>000</strong></em>\" -&gt; \"00<em><strong>111</strong></em>\"\n选择下标 i = 0: \"<em><strong>00111</strong></em>\" -&gt; \"<em><strong>11000</strong></em>\"\n选择下标 i = 1: \"1<em><strong>1000</strong></em>\" -&gt; \"1<em><strong>0111</strong></em>\"\n要达成目标，需要至少 3 次翻转。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>target = \"101\"\n<strong>输出：</strong>3\n<strong>解释：</strong>最初，s = \"000\" 。\n选择下标 i = 0: \"<em><strong>000</strong></em>\" -&gt; \"<em><strong>111</strong></em>\"\n选择下标 i = 1: \"1<em><strong>11</strong></em>\" -&gt; \"1<em><strong>00</strong></em>\"\n选择下标 i = 2: \"10<em><strong>0</strong></em>\" -&gt; \"10<em><strong>1</strong></em>\"\n要达成目标，需要至少 3 次翻转。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>target = \"00000\"\n<strong>输出：</strong>0\n<strong>解释：</strong>由于 s 已经等于目标，所以不需要任何操作\n</pre>\n</div>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == target.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>target[i]</code> 为 <code>'0'</code> 或 <code>'1'</code></li>\n</ul>\n",
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@@ -143,7 +143,7 @@
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             "hints": [
                 "Consider a strategy where the choice of bulb with number i is increasing. In such a strategy, you no longer need to worry about bulbs that have been set to the left."
             ],