You are given an integer n, a 0-indexed string array words, and a 0-indexed array groups, both arrays having length n.
The hamming distance between two strings of equal length is the number of positions at which the corresponding characters are different.
\n\nYou need to select the longest subsequence from an array of indices [0, 1, ..., n - 1], such that for the subsequence denoted as [i0, i1, ..., ik - 1] having length k, the following holds:
groups[ij] != groups[ij + 1], for each j where 0 < j + 1 < k.words[ij] and words[ij + 1] are equal in length, and the hamming distance between them is 1, where 0 < j + 1 < k, for all indices in the subsequence.Return a string array containing the words corresponding to the indices (in order) in the selected subsequence. If there are multiple answers, return any of them.
\n\nA subsequence of an array is a new array that is formed from the original array by deleting some (possibly none) of the elements without disturbing the relative positions of the remaining elements.
\n\nNote: strings in words may be unequal in length.
\n
Example 1:
\n\n\nInput: n = 3, words = ["bab","dab","cab"], groups = [1,2,2]\nOutput: ["bab","cab"]\nExplanation: A subsequence that can be selected is [0,2].\n- groups[0] != groups[2]\n- words[0].length == words[2].length, and the hamming distance between them is 1.\nSo, a valid answer is [words[0],words[2]] = ["bab","cab"].\nAnother subsequence that can be selected is [0,1].\n- groups[0] != groups[1]\n- words[0].length == words[1].length, and the hamming distance between them is 1.\nSo, another valid answer is [words[0],words[1]] = ["bab","dab"].\nIt can be shown that the length of the longest subsequence of indices that satisfies the conditions is 2.\n\n
Example 2:
\n\n\nInput: n = 4, words = ["a","b","c","d"], groups = [1,2,3,4]\nOutput: ["a","b","c","d"]\nExplanation: We can select the subsequence [0,1,2,3].\nIt satisfies both conditions.\nHence, the answer is [words[0],words[1],words[2],words[3]] = ["a","b","c","d"].\nIt has the longest length among all subsequences of indices that satisfy the conditions.\nHence, it is the only answer.\n\n\n
\n
Constraints:
\n\n1 <= n == words.length == groups.length <= 10001 <= words[i].length <= 101 <= groups[i] <= nwords consists of distinct strings.words[i] consists of lowercase English letters.给你一个整数 n 和一个下标从 0 开始的字符串数组 words ，和一个下标从 0 开始的数组 groups ，两个数组长度都是 n 。
两个长度相等字符串的 汉明距离 定义为对应位置字符 不同 的数目。
\n\n你需要从下标 [0, 1, ..., n - 1] 中选出一个 最长子序列 ，将这个子序列记作长度为 k 的 [i0, i1, ..., ik - 1] ，它需要满足以下条件：
groups 值 不同。即，对于所有满足 0 < j + 1 < k 的 j 都有 groups[ij] != groups[ij + 1] 。0 < j + 1 < k 的下标 j ，都满足 words[ij] 和 words[ij + 1] 的长度 相等 ，且两个字符串之间的 汉明距离 为 1 。请你返回一个字符串数组，它是下标子序列 依次 对应 words 数组中的字符串连接形成的字符串数组。如果有多个答案，返回任意一个。
子序列 指的是从原数组中删掉一些（也可能一个也不删掉）元素，剩余元素不改变相对位置得到的新的数组。
\n\n注意：words 中的字符串长度可能 不相等 。
\n\n
示例 1：
\n\n\n输入：n = 3, words = [\"bab\",\"dab\",\"cab\"], groups = [1,2,2]\n输出：[\"bab\",\"cab\"]\n解释：一个可行的子序列是 [0,2] 。\n- groups[0] != groups[2]\n- words[0].length == words[2].length 且它们之间的汉明距离为 1 。\n所以一个可行的答案是 [words[0],words[2]] = [\"bab\",\"cab\"] 。\n另一个可行的子序列是 [0,1] 。\n- groups[0] != groups[1]\n- words[0].length = words[1].length 且它们之间的汉明距离为 1 。\n所以另一个可行的答案是 [words[0],words[1]] = [\"bab\",\"dab\"] 。\n符合题意的最长子序列的长度为 2 。\n\n
示例 2：
\n\n\n输入：n = 4, words = [\"a\",\"b\",\"c\",\"d\"], groups = [1,2,3,4]\n输出：[\"a\",\"b\",\"c\",\"d\"]\n解释：我们选择子序列 [0,1,2,3] 。\n它同时满足两个条件。\n所以答案为 [words[0],words[1],words[2],words[3]] = [\"a\",\"b\",\"c\",\"d\"] 。\n它是所有下标子序列里最长且满足所有条件的。\n所以它是唯一的答案。\n\n\n
\n\n
提示：
\n\n1 <= n == words.length == groups.length <= 10001 <= words[i].length <= 101 <= groups[i] <= nwords 中的字符串 互不相同 。words[i] 只包含小写英文字母。dp[i] represent the length of the longest subsequence ending with words[i] that satisfies the conditions.",
"dp[i] = (maximum value of dp[j]) + 1 for indices j < i, where groups[i] != groups[j], words[i] and words[j] are equal in length, and the hamming distance between words[i] and words[j] is exactly 1.",
"Keep track of the j values used to achieve the maximum dp[i] for each index i.",
"The expected array's length is max(dp[0:n]), and starting from the index having the maximum value in dp, we can trace backward to get the words."
],
"solution": null,
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