leetcode-problemset/leetcode-cn/originData/extra-characters-in-a-string.json

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        "question": {
            "questionId": "2755",
            "questionFrontendId": "2707",
            "categoryTitle": "Algorithms",
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            "title": "Extra Characters in a String",
            "titleSlug": "extra-characters-in-a-string",
            "content": "<p>You are given a <strong>0-indexed</strong> string <code>s</code> and a dictionary of words <code>dictionary</code>. You have to break <code>s</code> into one or more <strong>non-overlapping</strong> substrings such that each substring is present in <code>dictionary</code>. There may be some <strong>extra characters</strong> in <code>s</code> which are not present in any of the substrings.</p>\n\n<p>Return <em>the <strong>minimum</strong> number of extra characters left over if you break up </em><code>s</code><em> optimally.</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;leetscode&quot;, dictionary = [&quot;leet&quot;,&quot;code&quot;,&quot;leetcode&quot;]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong> We can break s in two substrings: &quot;leet&quot; from index 0 to 3 and &quot;code&quot; from index 5 to 8. There is only 1 unused character (at index 4), so we return 1.\n\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;sayhelloworld&quot;, dictionary = [&quot;hello&quot;,&quot;world&quot;]\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> We can break s in two substrings: &quot;hello&quot; from index 3 to 7 and &quot;world&quot; from index 8 to 12. The characters at indices 0, 1, 2 are not used in any substring and thus are considered as extra characters. Hence, we return 3.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 50</code></li>\n\t<li><code>1 &lt;= dictionary.length &lt;= 50</code></li>\n\t<li><code>1 &lt;= dictionary[i].length &lt;= 50</code></li>\n\t<li><code>dictionary[i]</code>&nbsp;and <code>s</code> consists of only lowercase English letters</li>\n\t<li><code>dictionary</code> contains distinct words</li>\n</ul>\n",
            "translatedTitle": "字符串中的额外字符",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong>&nbsp;开始的字符串&nbsp;<code>s</code>&nbsp;和一个单词字典&nbsp;<code>dictionary</code>&nbsp;。你需要将&nbsp;<code>s</code>&nbsp;分割成若干个 <strong>互不重叠</strong>&nbsp;的子字符串，每个子字符串都在&nbsp;<code>dictionary</code>&nbsp;中出现过。<code>s</code>&nbsp;中可能会有一些&nbsp;<strong>额外的字符</strong>&nbsp;不在任何子字符串中。</p>\n\n<p>请你采取最优策略分割 <code>s</code>&nbsp;，使剩下的字符 <strong>最少</strong>&nbsp;。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><b>输入：</b>s = \"leetscode\", dictionary = [\"leet\",\"code\",\"leetcode\"]\n<b>输出：</b>1\n<b>解释：</b>将 s 分成两个子字符串：下标从 0 到 3 的 \"leet\" 和下标从 5 到 8 的 \"code\" 。只有 1 个字符没有使用（下标为 4），所以我们返回 1 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><b>输入：</b>s = \"sayhelloworld\", dictionary = [\"hello\",\"world\"]\n<b>输出：</b>3\n<b>解释：</b>将 s 分成两个子字符串：下标从 3 到 7 的 \"hello\" 和下标从 8 到 12 的 \"world\" 。下标为 0 ，1 和 2 的字符没有使用，所以我们返回 3 。\n</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 &lt;= 50</code></li>\n\t<li><code>1 &lt;= dictionary.length &lt;= 50</code></li>\n\t<li><code>1 &lt;= dictionary[i].length &lt;= 50</code></li>\n\t<li><code>dictionary[i]</code>&nbsp;和&nbsp;<code>s</code>&nbsp;只包含小写英文字母。</li>\n\t<li><code>dictionary</code>&nbsp;中的单词互不相同。</li>\n</ul>\n",
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                "Can we use Dynamic Programming here?",
                "Define DP[i] as the min extra character if breaking up s[0:i] optimally."
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