Question

2900. Longest Unequal Adjacent Groups Subsequence I

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Description

You are given a string array words and a binary array groups both of length n, where words[i] is associated with groups[i].

Your task is to select the longest alternating subsequence from words. A subsequence of words is alternating if for any two consecutive strings in the sequence, their corresponding elements in the binary array groups differ. Essentially, you are to choose strings such that adjacent elements have non-matching corresponding bits in the groups array.

Formally, you need to find the longest subsequence of an array of indices [0, 1, ..., n - 1] denoted as [i0, i1, ..., ik-1], such that groups[ij] != groups[ij+1] for each 0 <= j < k - 1 and then find the words corresponding to these indices.

Return _the selected subsequence. If there are multiple answers, return any of them._

Note: The elements in words are distinct.

 

Example 1:

Input: words = ["e","a","b"], groups = [0,0,1]

Output: ["e","b"]

Explanation: A subsequence that can be selected is ["e","b"] because groups[0] != groups[2]. Another subsequence that can be selected is ["a","b"] because groups[1] != groups[2]. It can be demonstrated that the length of the longest subsequence of indices that satisfies the condition is 2.

Example 2:

Input: words = ["a","b","c","d"], groups = [1,0,1,1]

Output: ["a","b","c"]

Explanation: A subsequence that can be selected is ["a","b","c"] because groups[0] != groups[1] and groups[1] != groups[2]. Another subsequence that can be selected is ["a","b","d"] because groups[0] != groups[1] and groups[1] != groups[3]. It can be shown that the length of the longest subsequence of indices that satisfies the condition is 3.

 

Constraints:

• 1 <= n == words.length == groups.length <= 100
• 1 <= words[i].length <= 10
• groups[i] is either 0 or 1.
• words consists of distinct strings.
• words[i] consists of lowercase English letters.

Solutions

Solution 1: Greedy

We can traverse the array $groups$, and for the current index $i$, if $i=0$ or $groups[i] \neq groups[i - 1]$, we add $words[i]$ to the answer array.

The time complexity is $O(n)$, where $n$ is the length of the array $groups$. The space complexity is $O(n)$.

class Solution:
  def getWordsInLongestSubsequence(
    self, n: int, words: List[str], groups: List[int]
  ) -> List[str]:
    return [words[i] for i, x in enumerate(groups) if i == 0 or x != groups[i - 1]]

class Solution {
  public List<String> getWordsInLongestSubsequence(int n, String[] words, int[] groups) {
    List<String> ans = new ArrayList<>();
    for (int i = 0; i < n; ++i) {
      if (i == 0 || groups[i] != groups[i - 1]) {
        ans.add(words[i]);
      }
    }
    return ans;
  }
}

class Solution {
public:
  vector<string> getWordsInLongestSubsequence(int n, vector<string>& words, vector<int>& groups) {
    vector<string> ans;
    for (int i = 0; i < n; ++i) {
      if (i == 0 || groups[i] != groups[i - 1]) {
        ans.emplace_back(words[i]);
      }
    }
    return ans;
  }
};

func getWordsInLongestSubsequence(n int, words []string, groups []int) (ans []string) {
  for i, x := range groups {
    if i == 0 || x != groups[i-1] {
      ans = append(ans, words[i])
    }
  }
  return
}

function getWordsInLongestSubsequence(n: number, words: string[], groups: number[]): string[] {
  const ans: string[] = [];
  for (let i = 0; i < n; ++i) {
    if (i === 0 || groups[i] !== groups[i - 1]) {
      ans.push(words[i]);
    }
  }
  return ans;
}

impl Solution {
  pub fn get_words_in_longest_subsequence(
    n: i32,
    words: Vec<String>,
    groups: Vec<i32>
  ) -> Vec<String> {
    let mut ans = vec![];

    for i in 0..n {
      if i == 0 || groups[i as usize] != groups[(i - 1) as usize] {
        ans.push(words[i as usize].clone());
      }
    }

    ans
  }
}