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2901. Longest Unequal Adjacent Groups Subsequence II
Description
You are given an integer n
, a 0indexed string array words
, and a 0indexed 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.
You need to select the longest subsequence from an array of indices [0, 1, ..., n  1]
, such that for the subsequence denoted as [i_{0}, i_{1}, ..., i_{k  1}]
having length k
, the following holds:
 For adjacent indices in the subsequence, their corresponding groups are unequal, i.e.,
groups[i_{j}] != groups[i_{j + 1}]
, for eachj
where0 < j + 1 < k
. words[i_{j}]
andwords[i_{j + 1}]
are equal in length, and the hamming distance between them is1
, where0 < 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.
A 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.
Note: strings in words
may be unequal in length.
Example 1:
Input: n = 3, words = ["bab","dab","cab"], groups = [1,2,2] Output: ["bab","cab"] Explanation: A subsequence that can be selected is [0,2].  groups[0] != groups[2]  words[0].length == words[2].length, and the hamming distance between them is 1. So, a valid answer is [words[0],words[2]] = ["bab","cab"]. Another subsequence that can be selected is [0,1].  groups[0] != groups[1]  words[0].length == words[1].length, and the hamming distance between them is 1. So, another valid answer is [words[0],words[1]] = ["bab","dab"]. It can be shown that the length of the longest subsequence of indices that satisfies the conditions is 2.
Example 2:
Input: n = 4, words = ["a","b","c","d"], groups = [1,2,3,4] Output: ["a","b","c","d"] Explanation: We can select the subsequence [0,1,2,3]. It satisfies both conditions. Hence, the answer is [words[0],words[1],words[2],words[3]] = ["a","b","c","d"]. It has the longest length among all subsequences of indices that satisfy the conditions. Hence, it is the only answer.
Constraints:
1 <= n == words.length == groups.length <= 1000
1 <= words[i].length <= 10
1 <= groups[i] <= n
words
consists of distinct strings.words[i]
consists of lowercase English letters.
Solutions
Solution 1: Dynamic Programming
We define $f[i]$ as the length of the longest adjacent nonequal subsequence ending with the $i$th word, and $g[i]$ as the predecessor index of the longest adjacent nonequal subsequence ending with the $i$th word. Initially, we set $f[i] = 1$ and $g[i] = 1$.
In addition, we define a variable $mx$ to represent the length of the longest adjacent nonequal subsequence.
We traverse $i$ and $j \in [0, i)$, and if $groups[i] \neq groups[j]$, $f[i] \lt f[j] + 1$, and the Hamming distance between $words[i]$ and $words[j]$ is $1$, we update $f[i] = f[j] + 1$, $g[i] = j$, and update $mx = \max(mx, f[i])$.
Finally, we find the index $i$ corresponding to the maximum value in the $f$ array, and then continuously search backwards from $i$ until we find $g[i] = 1$, which gives us the longest adjacent nonequal subsequence.
The time complexity is $O(n^2 \times L)$, and the space complexity is $O(n)$. Here, $L$ represents the maximum length of a word.
Optimization: Space for Time
In Solution 1, we need to enumerate all $i$ and $j$ combinations, a step that can be optimized by maintaining a wildcard hash table. For each string $word[i]$, we enumerate each character, replace it with a wildcard, and then use the replaced string as the key and add its subscript to the list which is the value in the hash table. This allows us to find all $word[j]$ with a Hamming distance of 1 from $word[i]$ in $O(L)$ time. Although the time complexity is still $O(n^2 \times L)$, the average complexity is reduced.

class Solution { public List<String> getWordsInLongestSubsequence(int n, String[] words, int[] groups) { int[] f = new int[n]; int[] g = new int[n]; Arrays.fill(f, 1); Arrays.fill(g, 1); int mx = 1; for (int i = 0; i < n; ++i) { for (int j = 0; j < i; ++j) { if (groups[i] != groups[j] && f[i] < f[j] + 1 && check(words[i], words[j])) { f[i] = f[j] + 1; g[i] = j; mx = Math.max(mx, f[i]); } } } List<String> ans = new ArrayList<>(); for (int i = 0; i < n; ++i) { if (f[i] == mx) { for (int j = i; j >= 0; j = g[j]) { ans.add(words[j]); } break; } } Collections.reverse(ans); return ans; } private boolean check(String s, String t) { if (s.length() != t.length()) { return false; } int cnt = 0; for (int i = 0; i < s.length(); ++i) { if (s.charAt(i) != t.charAt(i)) { ++cnt; } } return cnt == 1; } }

class Solution { public: vector<string> getWordsInLongestSubsequence(int n, vector<string>& words, vector<int>& groups) { auto check = [](string& s, string& t) { if (s.size() != t.size()) { return false; } int cnt = 0; for (int i = 0; i < s.size(); ++i) { cnt += s[i] != t[i]; } return cnt == 1; }; vector<int> f(n, 1); vector<int> g(n, 1); int mx = 1; for (int i = 0; i < n; ++i) { for (int j = 0; j < i; ++j) { if (groups[i] != groups[j] && f[i] < f[j] + 1 && check(words[i], words[j])) { f[i] = f[j] + 1; g[i] = j; mx = max(mx, f[i]); } } } vector<string> ans; for (int i = 0; i < n; ++i) { if (f[i] == mx) { for (int j = i; ~j; j = g[j]) { ans.emplace_back(words[j]); } break; } } reverse(ans.begin(), ans.end()); return ans; } };

class Solution: def getWordsInLongestSubsequence( self, n: int, words: List[str], groups: List[int] ) > List[str]: def check(s: str, t: str) > bool: return len(s) == len(t) and sum(a != b for a, b in zip(s, t)) == 1 f = [1] * n g = [1] * n mx = 1 for i, x in enumerate(groups): for j, y in enumerate(groups[:i]): if x != y and f[i] < f[j] + 1 and check(words[i], words[j]): f[i] = f[j] + 1 g[i] = j mx = max(mx, f[i]) ans = [] for i in range(n): if f[i] == mx: j = i while j >= 0: ans.append(words[j]) j = g[j] break return ans[::1]

func getWordsInLongestSubsequence(n int, words []string, groups []int) []string { check := func(s, t string) bool { if len(s) != len(t) { return false } cnt := 0 for i := range s { if s[i] != t[i] { cnt++ } } return cnt == 1 } f := make([]int, n) g := make([]int, n) for i := range f { f[i] = 1 g[i] = 1 } mx := 1 for i, x := range groups { for j, y := range groups[:i] { if x != y && f[i] < f[j]+1 && check(words[i], words[j]) { f[i] = f[j] + 1 g[i] = j if mx < f[i] { mx = f[i] } } } } ans := make([]string, 0, mx) for i, x := range f { if x == mx { for j := i; j >= 0; j = g[j] { ans = append(ans, words[j]) } break } } for i, j := 0, len(ans)1; i < j; i, j = i+1, j1 { ans[i], ans[j] = ans[j], ans[i] } return ans }

function getWordsInLongestSubsequence(n: number, words: string[], groups: number[]): string[] { const f: number[] = Array(n).fill(1); const g: number[] = Array(n).fill(1); let mx = 1; const check = (s: string, t: string) => { if (s.length !== t.length) { return false; } let cnt = 0; for (let i = 0; i < s.length; ++i) { if (s[i] !== t[i]) { ++cnt; } } return cnt === 1; }; for (let i = 0; i < n; ++i) { for (let j = 0; j < i; ++j) { if (groups[i] !== groups[j] && f[i] < f[j] + 1 && check(words[i], words[j])) { f[i] = f[j] + 1; g[i] = j; mx = Math.max(mx, f[i]); } } } const ans: string[] = []; for (let i = 0; i < n; ++i) { if (f[i] === mx) { for (let j = i; ~j; j = g[j]) { ans.push(words[j]); } break; } } return ans.reverse(); }

impl Solution { pub fn get_words_in_longest_subsequence( n: i32, words: Vec<String>, groups: Vec<i32> ) > Vec<String> { fn check(s: &str, t: &str) > bool { s.len() == t.len() && s .chars() .zip(t.chars()) .filter((a, b) a != b) .count() == 1 } let n = n as usize; let mut f = vec![1; n]; let mut g = vec![1; n]; let mut mx = 1; for i in 0..n { let x = groups[i] as usize; for j in 0..i { let y = groups[j] as usize; if x != y && f[i] < f[j] + 1 && check(&words[i], &words[j]) { f[i] = f[j] + 1; g[i] = j as i32; mx = mx.max(f[i]); } } } let mut ans = vec![]; let mut i = n  1; while f[i] != mx { i = 1; } let mut j = i as i32; while j >= 0 { ans.push(words[j as usize].clone()); j = g[j as usize]; } ans.reverse(); ans } }