Longest Unequal Adjacent Groups Subsequence I

Problem 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.

Examples

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.

Approach

To solve this problem, we can use a greedy approach to construct the longest alternating subsequence. The idea is to start with the first element and then continue adding elements to the subsequence only if the current element has a different group value than the previous element in the subsequence.

This approach ensures that we build the longest possible alternating subsequence in a single pass through the array, making it both simple and efficient.

Time Complexity

The time complexity is $O(n)$. where n is the length of the words array. This is because we are iterating through the array once.

Code Implementation

C++

Written by @ananyag309

class Solution {
public:
    vector<string> getLongestSubsequence(vector<string>& words, vector<int>& groups) {
        vector<string> result;
        int n = words.size();
        if (n == 0) return result;

        result.push_back(words[0]);
        for (int i = 1; i < n; ++i) {
            if (groups[i] != groups[i - 1]) {
                result.push_back(words[i]);
            }
        }
        return result;
    }
};

Java

Written by @ananyag309

class Solution {
    public List<String> getLongestSubsequence(String[] words, int[] groups) {
        List<String> result = new ArrayList<>();
        int n = words.length;
        if (n == 0) return result;

        result.add(words[0]);
        for (int i = 1; i < n; ++i) {
            if (groups[i] != groups[i - 1]) {
                result.add(words[i]);
            }
        }
        return result;
    }
}

Python

Written by @ananyag309

class Solution:
    def getLongestSubsequence(self, words: List[str], groups: List[int]) -> List[str]:
        result = []
        n = len(words)
        if n == 0:
            return result

        result.append(words[0])
        for i in range(1, n):
            if groups[i] != groups[i - 1]:
                result.append(words[i])

        return result

# Example usage
if __name__ == "__main__":
    solution = Solution()
    print(solution.getLongestSubsequence(["a", "b", "c", "d"], [1, 0, 1, 1]))  # Expected output: ["a", "b", "c"]
    print(solution.getLongestSubsequence(["e", "a", "b"], [0, 0, 1]))  # Expected output: ["e", "b"]

Solution Logic

1. A number n is a power of 2 if it has exactly one bit set in its binary representation.
2. To check this, you can use the property: n & (n - 1) == 0.
3. This expression is true only for powers of 2.

Time Complexity

• The function operates in constant time, $O(1)$.

Space Complexity

• The function uses constant space, $O(1)$.

References

• LeetCode Problem: Longest Unequal Adjacent Groups Subsequence I
• Greedy Algorithm: GeeksforGeeks Greedy Algorithm