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"categoryTitle": "Algorithms",
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"title": "Find the String with LCP",
"titleSlug": "find-the-string-with-lcp",
"content": "<p>We define the <code>lcp</code> matrix of any <strong>0-indexed</strong> string <code>word</code> of <code>n</code> lowercase English letters as an <code>n x n</code> grid such that:</p>\n\n<ul>\n\t<li><code>lcp[i][j]</code> is equal to the length of the <strong>longest common prefix</strong> between the substrings <code>word[i,n-1]</code> and <code>word[j,n-1]</code>.</li>\n</ul>\n\n<p>Given an&nbsp;<code>n x n</code> matrix <code>lcp</code>, return the alphabetically smallest string <code>word</code> that corresponds to <code>lcp</code>. If there is no such string, return an empty string.</p>\n\n<p>A string <code>a</code> is lexicographically smaller than a string <code>b</code> (of the same length) if in the first position where <code>a</code> and <code>b</code> differ, string <code>a</code> has a letter that appears earlier in the alphabet than the corresponding letter in <code>b</code>. For example, <code>&quot;aabd&quot;</code> is lexicographically smaller than <code>&quot;aaca&quot;</code> because the first position they differ is at the third letter, and <code>&#39;b&#39;</code> comes before <code>&#39;c&#39;</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> lcp = [[4,0,2,0],[0,3,0,1],[2,0,2,0],[0,1,0,1]]\n<strong>Output:</strong> &quot;abab&quot;\n<strong>Explanation:</strong> lcp corresponds to any 4 letter string with two alternating letters. The lexicographically smallest of them is &quot;abab&quot;.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]\n<strong>Output:</strong> &quot;aaaa&quot;\n<strong>Explanation:</strong> lcp corresponds to any 4 letter string with a single distinct letter. The lexicographically smallest of them is &quot;aaaa&quot;. \n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,3]]\n<strong>Output:</strong> &quot;&quot;\n<strong>Explanation:</strong> lcp[3][3] cannot be equal to 3 since word[3,...,3] consists of only a single letter; Thus, no answer exists.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n ==&nbsp;</code><code>lcp.length == </code><code>lcp[i].length</code>&nbsp;<code>&lt;= 1000</code></li>\n\t<li><code><font face=\"monospace\">0 &lt;= lcp[i][j] &lt;= n</font></code></li>\n</ul>\n",
"translatedTitle": "找出对应 LCP 矩阵的字符串",
"translatedContent": "<p>对任一由 <code>n</code> 个小写英文字母组成的字符串 <code>word</code> ,我们可以定义一个 <code>n x n</code> 的矩阵,并满足:</p>\n\n<ul>\n\t<li><code>lcp[i][j]</code> 等于子字符串&nbsp;<code>word[i,...,n-1]</code> 和 <code>word[j,...,n-1]</code> 之间的最长公共前缀的长度。</li>\n</ul>\n\n<p>给你一个 <code>n x n</code> 的矩阵 <code>lcp</code> 。返回与 <code>lcp</code> 对应的、按字典序最小的字符串&nbsp;<code>word</code> 。如果不存在这样的字符串,则返回空字符串。</p>\n\n<p>对于长度相同的两个字符串 <code>a</code> 和 <code>b</code> ,如果在 <code>a</code> 和 <code>b</code> 不同的第一个位置,字符串 <code>a</code> 的字母在字母表中出现的顺序先于 <code>b</code> 中的对应字母,则认为字符串 <code>a</code> 按字典序比字符串 <code>b</code> 小。例如,<code>\"aabd\"</code> 在字典上小于 <code>\"aaca\"</code> ,因为二者不同的第一位置是第三个字母,而&nbsp;<code>'b'</code> 先于 <code>'c'</code> 出现。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>lcp = [[4,0,2,0],[0,3,0,1],[2,0,2,0],[0,1,0,1]]\n<strong>输出:</strong>\"abab\"\n<strong>解释:</strong>lcp 对应由两个交替字母组成的任意 4 字母字符串,字典序最小的是 \"abab\" 。\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,1]]\n<strong>输出:</strong>\"aaaa\"\n<strong>解释:</strong>lcp 对应只有一个不同字母的任意 4 字母字符串,字典序最小的是 \"aaaa\" 。 \n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre>\n<strong>输入:</strong>lcp = [[4,3,2,1],[3,3,2,1],[2,2,2,1],[1,1,1,3]]\n<strong>输出:</strong>\"\"\n<strong>解释:</strong>lcp[3][3] 无法等于 3 ,因为 word[3,...,3] 仅由单个字母组成;因此,不存在答案。\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 ==&nbsp;</code><code>lcp.length == </code><code>lcp[i].length</code>&nbsp;<code>&lt;= 1000</code></li>\n\t<li><code><font face=\"monospace\">0 &lt;= lcp[i][j] &lt;= n</font></code></li>\n</ul>\n",
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