leetcode-problemset/leetcode-cn/originData/shift-distance-between-two-strings.json

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    "data": {
        "question": {
            "questionId": "3591",
            "questionFrontendId": "3361",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2997029,
            "title": "Shift Distance Between Two Strings",
            "titleSlug": "shift-distance-between-two-strings",
            "content": "<p>You are given two strings <code>s</code> and <code>t</code> of the same length, and two integer arrays <code>nextCost</code> and <code>previousCost</code>.</p>\n\n<p>In one operation, you can pick any index <code>i</code> of <code>s</code>, and perform <strong>either one</strong> of the following actions:</p>\n\n<ul>\n\t<li>Shift <code>s[i]</code> to the next letter in the alphabet. If <code>s[i] == &#39;z&#39;</code>, you should replace it with <code>&#39;a&#39;</code>. This operation costs <code>nextCost[j]</code> where <code>j</code> is the index of <code>s[i]</code> in the alphabet.</li>\n\t<li>Shift <code>s[i]</code> to the previous letter in the alphabet. If <code>s[i] == &#39;a&#39;</code>, you should replace it with <code>&#39;z&#39;</code>. This operation costs <code>previousCost[j]</code> where <code>j</code> is the index of <code>s[i]</code> in the alphabet.</li>\n</ul>\n\n<p>The <strong>shift distance</strong> is the <strong>minimum</strong> total cost of operations required to transform <code>s</code> into <code>t</code>.</p>\n\n<p>Return the <strong>shift distance</strong> from <code>s</code> to <code>t</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">s = &quot;abab&quot;, t = &quot;baba&quot;, nextCost = [100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], previousCost = [1,100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">2</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ul>\n\t<li>We choose index <code>i = 0</code> and shift <code>s[0]</code> 25 times to the previous character for a total cost of 1.</li>\n\t<li>We choose index <code>i = 1</code> and shift <code>s[1]</code> 25 times to the next character for a total cost of 0.</li>\n\t<li>We choose index <code>i = 2</code> and shift <code>s[2]</code> 25 times to the previous character for a total cost of 1.</li>\n\t<li>We choose index <code>i = 3</code> and shift <code>s[3]</code> 25 times to the next character for a total cost of 0.</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<div class=\"example-block\">\n<p><strong>Input:</strong> <span class=\"example-io\">s = &quot;leet&quot;, t = &quot;code&quot;, nextCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], previousCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]</span></p>\n\n<p><strong>Output:</strong> <span class=\"example-io\">31</span></p>\n\n<p><strong>Explanation:</strong></p>\n\n<ul>\n\t<li>We choose index <code>i = 0</code> and shift <code>s[0]</code> 9 times to the previous character for a total cost of 9.</li>\n\t<li>We choose index <code>i = 1</code> and shift <code>s[1]</code> 10 times to the next character for a total cost of 10.</li>\n\t<li>We choose index <code>i = 2</code> and shift <code>s[2]</code> 1 time to the previous character for a total cost of 1.</li>\n\t<li>We choose index <code>i = 3</code> and shift <code>s[3]</code> 11 times to the next character for a total cost of 11.</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length == t.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s</code> and <code>t</code> consist only of lowercase English letters.</li>\n\t<li><code>nextCost.length == previousCost.length == 26</code></li>\n\t<li><code>0 &lt;= nextCost[i], previousCost[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "两个字符串的切换距离",
            "translatedContent": "<p>给你两个长度相同的字符串&nbsp;<code>s</code> 和&nbsp;<code>t</code>&nbsp;，以及两个整数数组&nbsp;<code>nextCost</code> 和&nbsp;<code>previousCost</code>&nbsp;。</p>\n\n<p>一次操作中，你可以选择 <code>s</code>&nbsp;中的一个下标 <code>i</code>&nbsp;，执行以下操作 <strong>之一</strong>&nbsp;：</p>\n\n<ul>\n\t<li>将&nbsp;<code>s[i]</code>&nbsp;切换为字母表中的下一个字母，如果&nbsp;<code>s[i] == 'z'</code>&nbsp;，切换后得到&nbsp;<code>'a'</code>&nbsp;。操作的代价为&nbsp;<code>nextCost[j]</code>&nbsp;，其中&nbsp;<code>j</code>&nbsp;表示&nbsp;<code>s[i]</code>&nbsp;在字母表中的下标。</li>\n\t<li>将&nbsp;<code>s[i]</code>&nbsp;切换为字母表中的上一个字母，如果&nbsp;<code>s[i] == 'a'</code>&nbsp;，切换后得到&nbsp;<code>'z'</code>&nbsp;。操作的代价为&nbsp;<code>previousCost[j]</code>&nbsp;，其中&nbsp;<code>j</code> 是&nbsp;<code>s[i]</code>&nbsp;在字母表中的下标。</li>\n</ul>\n\n<p><strong>切换距离</strong>&nbsp;指的是将字符串 <code>s</code>&nbsp;变为字符串 <code>t</code>&nbsp;的 <strong>最少</strong>&nbsp;操作代价总和。</p>\n\n<p>请你返回从 <code>s</code>&nbsp;到 <code>t</code>&nbsp;的 <strong>切换距离</strong>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>s = \"abab\", t = \"baba\", nextCost = [100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0], previousCost = [1,100,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>2</span></p>\n\n<p><b>解释：</b></p>\n\n<ul>\n\t<li>选择下标&nbsp;<code>i = 0</code>&nbsp;并将&nbsp;<code>s[0]</code>&nbsp;向前切换 25 次，总代价为 1 。</li>\n\t<li>选择下标&nbsp;<code>i = 1</code>&nbsp;并将&nbsp;<code>s[1]</code>&nbsp;向后切换 25 次，总代价为 0 。</li>\n\t<li>选择下标&nbsp;<code>i = 2</code>&nbsp;并将&nbsp;<code>s[2]</code>&nbsp;向前切换 25 次，总代价为 1 。</li>\n\t<li>选择下标&nbsp;<code>i = 3</code>&nbsp;并将&nbsp;<code>s[3]</code>&nbsp;向后切换 25 次，总代价为 0 。</li>\n</ul>\n</div>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<div class=\"example-block\">\n<p><span class=\"example-io\"><b>输入：</b>s = \"leet\", t = \"code\", nextCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], previousCost = [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1]</span></p>\n\n<p><span class=\"example-io\"><b>输出：</b>31</span></p>\n\n<p><b>解释：</b></p>\n\n<ul>\n\t<li>选择下标&nbsp;<code>i = 0</code>&nbsp;并将&nbsp;<code>s[0]</code>&nbsp;向前切换 9 次，总代价为 9 。</li>\n\t<li>选择下标&nbsp;<code>i = 1</code>&nbsp;并将&nbsp;<code>s[1]</code>&nbsp;向后切换 10 次，总代价为 10 。</li>\n\t<li>选择下标&nbsp;<code>i = 2</code> 并将&nbsp;<code>s[2]</code>&nbsp;向前切换 1 次，总代价为 1 。</li>\n\t<li>选择下标 <code>i = 3</code> 并将&nbsp;<code>s[3]</code>&nbsp;向后切换 11 次，总代价为 11 。</li>\n</ul>\n</div>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length == t.length &lt;= 10<sup>5</sup></code></li>\n\t<li><code>s</code> 和&nbsp;<code>t</code>&nbsp;都只包含小写英文字母。</li>\n\t<li><code>nextCost.length == previousCost.length == 26</code></li>\n\t<li><code>0 &lt;= nextCost[i], previousCost[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    long long shiftDistance(string s, string t, vector<int>& nextCost, vector<int>& previousCost) {\n        \n    }\n};",
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                "- For every unordered pair of characters <code>(a, b)</code>, the cost of turning <code>a</code> into <code>b</code> is equal to the minimum between: \r\n<ul>\r\n<li>If <code>i < j</code>, <code>nextCost[i] + nextCost[i + 1] + … + nextCost[j - 1]</code>, and <code>nextCost[i] + nextCost[i + 1] + … + nextCost[25] + nextCost[0] + … + nextCost[j - 1]</code> otherwise.</li>\r\n    \r\n    <li>If <code>i < j</code>, <code>prevCost[i] + prevCost[i - 1] + … + prevCost[0] + prevCost[25] + … + prevCost[j + 1]</code>, and <code>prevCost[i] + prevCost[i - 1] + … + prevCost[j + 1]</code> otherwise.</li>\r\n    </ul>\r\n    Where <code>i</code> and <code>j</code> are the indices of <code>a</code> and <code>b</code> in the alphabet.",
                "The shift distance is the sum of costs of turning <code>s[i]</code> into <code>t[i]</code>."
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<code>use-after-free<\\/code> \\u9519\\u8bef\\u3002<\\/p>\\r\\n\\r\\n<p>\\u4e3a\\u4e86\\u4f7f\\u7528\\u65b9\\u4fbf\\uff0c\\u5927\\u90e8\\u5206\\u6807\\u51c6\\u5e93\\u7684\\u5934\\u6587\\u4ef6\\u5df2\\u7ecf\\u88ab\\u81ea\\u52a8\\u5bfc\\u5165\\u3002<\\/p>\\r\\n\\r\\n<p>\\u5982\\u60f3\\u4f7f\\u7528\\u54c8\\u5e0c\\u8868\\u8fd0\\u7b97, \\u60a8\\u53ef\\u4ee5\\u4f7f\\u7528 <a href=\\\"https:\\/\\/troydhanson.github.io\\/uthash\\/\\\" target=\\\"_blank\\\">uthash<\\/a>\\u3002 \\\"uthash.h\\\"\\u5df2\\u7ecf\\u9ed8\\u8ba4\\u88ab\\u5bfc\\u5165\\u3002\\u8bf7\\u770b\\u5982\\u4e0b\\u793a\\u4f8b:<\\/p>\\r\\n\\r\\n<p><b>1. \\u5f80\\u54c8\\u5e0c\\u8868\\u4e2d\\u6dfb\\u52a0\\u4e00\\u4e2a\\u5bf9\\u8c61\\uff1a<\\/b>\\r\\n<pre>\\r\\nstruct hash_entry {\\r\\n    int id;            \\/* we'll use this field as the key *\\/\\r\\n    char name[10];\\r\\n    UT_hash_handle hh; \\/* makes this structure hashable *\\/\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry 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target=\\\"_blank\\\">\\u300c\\u4ed3\\u9889\\u7f16\\u7a0b\\u8bed\\u8a00\\u5f00\\u53d1\\u6307\\u5357\\u300d<\\/a><\\/p>\"]}",
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