leetcode-problemset/leetcode-cn/originData/separate-black-and-white-balls.json

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            "title": "Separate Black and White Balls",
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            "content": "<p>There are <code>n</code> balls on a table, each ball has a color black or white.</p>\n\n<p>You are given a <strong>0-indexed</strong> binary string <code>s</code> of length <code>n</code>, where <code>1</code> and <code>0</code> represent black and white balls, respectively.</p>\n\n<p>In each step, you can choose two adjacent balls and swap them.</p>\n\n<p>Return <em>the <strong>minimum</strong> number of steps to group all the black balls to the right and all the white balls to the left</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;101&quot;\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> We can group all the black balls to the right in the following way:\n- Swap s[0] and s[1], s = &quot;011&quot;.\nInitially, 1s are not grouped together, requiring at least 1 step to group them to the right.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;100&quot;\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> We can group all the black balls to the right in the following way:\n- Swap s[0] and s[1], s = &quot;010&quot;.\n- Swap s[1] and s[2], s = &quot;001&quot;.\nIt can be proven that the minimum number of steps needed is 2.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;0111&quot;\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> All the black balls are already grouped to the right.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == s.length &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": "区分黑球与白球",
            "translatedContent": "<p>桌子上有 <code>n</code> 个球，每个球的颜色不是黑色，就是白色。</p>\n\n<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>s</code>，其中 <code>1</code> 和 <code>0</code> 分别代表黑色和白色的球。</p>\n\n<p>在每一步中，你可以选择两个相邻的球并交换它们。</p>\n\n<p>返回「将所有黑色球都移到右侧，所有白色球都移到左侧所需的 <strong>最小步数</strong>」。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"101\"\n<strong>输出：</strong>1\n<strong>解释：</strong>我们可以按以下方式将所有黑色球移到右侧：\n- 交换 s[0] 和 s[1]，s = \"011\"。\n最开始，1 没有都在右侧，需要至少 1 步将其移到右侧。</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"100\"\n<strong>输出：</strong>2\n<strong>解释：</strong>我们可以按以下方式将所有黑色球移到右侧：\n- 交换 s[0] 和 s[1]，s = \"010\"。\n- 交换 s[1] 和 s[2]，s = \"001\"。\n可以证明所需的最小步数为 2 。\n</pre>\n\n<p><strong class=\"example\">示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"0111\"\n<strong>输出：</strong>0\n<strong>解释：</strong>所有黑色球都已经在右侧。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n == s.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s[i]</code> 不是 <code>'0'</code>，就是 <code>'1'</code>。</li>\n</ul>\n",
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