leetcode-problemset/leetcode-cn/originData/apply-operations-to-make-two-strings-equal.json

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        "question": {
            "questionId": "3033",
            "questionFrontendId": "2896",
            "categoryTitle": "Algorithms",
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            "title": "Apply Operations to Make Two Strings Equal",
            "titleSlug": "apply-operations-to-make-two-strings-equal",
            "content": "<p>You are given two <strong>0-indexed</strong> binary strings <code>s1</code> and <code>s2</code>, both of length <code>n</code>, and a positive integer <code>x</code>.</p>\n\n<p>You can perform any of the following operations on the string <code>s1</code> <strong>any</strong> number of times:</p>\n\n<ul>\n\t<li>Choose two indices <code>i</code> and <code>j</code>, and flip both <code>s1[i]</code> and <code>s1[j]</code>. The cost of this operation is <code>x</code>.</li>\n\t<li>Choose an index <code>i</code> such that <code>i &lt; n - 1</code> and flip both <code>s1[i]</code> and <code>s1[i + 1]</code>. The cost of this operation is <code>1</code>.</li>\n</ul>\n\n<p>Return <em>the <strong>minimum</strong> cost needed to make the strings </em><code>s1</code><em> and </em><code>s2</code><em> equal, or return </em><code>-1</code><em> if it is impossible.</em></p>\n\n<p><strong>Note</strong> that flipping a character means changing it from <code>0</code> to <code>1</code> or vice-versa.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s1 = &quot;1100011000&quot;, s2 = &quot;0101001010&quot;, x = 2\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> We can do the following operations:\n- Choose i = 3 and apply the second operation. The resulting string is s1 = &quot;110<u><strong>11</strong></u>11000&quot;.\n- Choose i = 4 and apply the second operation. The resulting string is s1 = &quot;1101<strong><u>00</u></strong>1000&quot;.\n- Choose i = 0 and j = 8 and apply the first operation. The resulting string is s1 = &quot;<u><strong>0</strong></u>1010010<u><strong>1</strong></u>0&quot; = s2.\nThe total cost is 1 + 1 + 2 = 4. It can be shown that it is the minimum cost possible.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s1 = &quot;10110&quot;, s2 = &quot;00011&quot;, x = 4\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> It is not possible to make the two strings equal.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == s1.length == s2.length</code></li>\n\t<li><code>1 &lt;= n, x &lt;= 500</code></li>\n\t<li><code>s1</code> and <code>s2</code> consist only of the characters <code>&#39;0&#39;</code> and <code>&#39;1&#39;</code>.</li>\n</ul>\n",
            "translatedTitle": "执行操作使两个字符串相等",
            "translatedContent": "<p>给你两个下标从 <strong>0</strong>&nbsp;开始的二进制字符串&nbsp;<code>s1</code> 和&nbsp;<code>s2</code>&nbsp;，两个字符串的长度都是&nbsp;<code>n</code>&nbsp;，再给你一个正整数&nbsp;<code>x</code>&nbsp;。</p>\n\n<p>你可以对字符串 <code>s1</code>&nbsp;执行以下操作 <strong>任意次</strong>&nbsp;：</p>\n\n<ul>\n\t<li>选择两个下标&nbsp;<code>i</code>&nbsp;和&nbsp;<code>j</code>&nbsp;，将&nbsp;<code>s1[i]</code> 和&nbsp;<code>s1[j]</code>&nbsp;都反转，操作的代价为&nbsp;<code>x</code>&nbsp;。</li>\n\t<li>选择满足 <code>i &lt; n - 1</code>&nbsp;的下标&nbsp;<code>i</code>&nbsp;，反转&nbsp;<code>s1[i]</code> 和&nbsp;<code>s1[i + 1]</code>&nbsp;，操作的代价为&nbsp;<code>1</code>&nbsp;。</li>\n</ul>\n\n<p>请你返回使字符串&nbsp;<code>s1</code>&nbsp;和&nbsp;<code>s2</code>&nbsp;相等的&nbsp;<strong>最小</strong>&nbsp;操作代价之和，如果无法让二者相等，返回&nbsp;<code>-1</code>&nbsp;。</p>\n\n<p><strong>注意</strong>&nbsp;，反转字符的意思是将&nbsp;<code>0</code>&nbsp;变成&nbsp;<code>1</code>&nbsp;，或者 <code>1</code>&nbsp;变成 <code>0</code>&nbsp;。</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>s1 = \"1100011000\", s2 = \"0101001010\", x = 2\n<b>输出：</b>4\n<b>解释：</b>我们可以执行以下操作：\n- 选择 i = 3 执行第二个操作。结果字符串是 s1 = \"110<em><strong>11</strong></em>11000\" 。\n- 选择 i = 4 执行第二个操作。结果字符串是 s1 = \"1101<em><strong>00</strong></em>1000\" 。\n- 选择 i = 0 和 j = 8 ，执行第一个操作。结果字符串是 s1 = \"<em><strong>0</strong></em>1010010<em><strong>1</strong></em>0\" = s2 。\n总代价是 1 + 1 + 2 = 4 。这是最小代价和。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>s1 = \"10110\", s2 = \"00011\", x = 4\n<b>输出：</b>-1\n<b>解释：</b>无法使两个字符串相等。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == s1.length == s2.length</code></li>\n\t<li><code>1 &lt;= n, x &lt;= 500</code></li>\n\t<li><code>s1</code> 和&nbsp;<code>s2</code>&nbsp;只包含字符&nbsp;<code>'0'</code> 和&nbsp;<code>'1'</code> 。</li>\n</ul>\n",
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