Longest Unequal Adjacent Groups Subsequence II

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ArraysStringsDynamic Programming

You are given a string array words, and an 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 [i₀, i₁, ..., 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 each j where 0 < j + 1 < k.
words[i_j] and words[i_j+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.

Note: strings in words may be unequal in length.

Example 1:

Input: 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: 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.

Try coding it in LeetCode

Longest Unequal Adjacent Groups Subsequence II

Medium

Asked by:

1 view

Topics:

ArraysStringsDynamic Programming

You are given a string array words, and an 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 [i₀, i₁, ..., 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 each j where 0 < j + 1 < k.
words[i_j] and words[i_j+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.

Note: strings in words may be unequal in length.

Example 1:

Input: 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: 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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values for elements within the `groups` array? Can they be negative, zero, or non-integer?
What is the maximum size of the `words` array and the maximum length of each individual word?
If there are multiple longest subsequences satisfying the condition, can I return any one of them?
If no valid subsequence exists, what should I return? Should I return an empty list or null?
Are the strings in the `words` array case-sensitive? Should I consider case when determining group membership?

Brute Force Solution

Approach

The brute force method to solve this problem involves considering every single possible way to pick a group of words in order. We systematically check if each such group satisfies the condition that adjacent words are from different categories, and then we keep track of the length of the longest valid sequence we can find this way.

Here's how the algorithm would work step-by-step:

Start by picking the first word.
Then consider picking the first and second word, then the first, second, and third word, and so on. At each step, verify if the sequence is valid (that is, adjacent words are not from the same category).
If the sequence is invalid, skip it and move on.
If the sequence is valid, store its length.
Continue this process, starting each time from the first word, then the second word, then the third word, and so on until you have checked all possible starting positions for the sequence.
Also, make sure that you are considering all possible sequence lengths. For example, after picking the first word, you can explore sequences of length 2, 3, 4, ..., all the way up to the total number of words.
After exploring all possible combinations of words, select the longest sequence that was considered valid. That is the answer.

Code Implementation

def longest_unequal_adjacent_groups_subsequence_brute_force(groups):
    number_of_groups = len(groups)
    maximum_length = 0

    for starting_index in range(number_of_groups):
        for subsequence_length in range(1, number_of_groups - starting_index + 1):
            # Iterate over all possible subsequences starting from the current index
            current_subsequence = []
            is_valid = True

            for i in range(subsequence_length):
                current_subsequence.append(groups[starting_index + i])

            # Check for validity
            for i in range(len(current_subsequence) - 1):
                if current_subsequence[i] == current_subsequence[i + 1]:
                    is_valid = False
                    break

            # Update maximum length if a valid subsequence is found
            if is_valid:
                maximum_length = max(maximum_length, len(current_subsequence))

    return maximum_length

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves exploring every possible subsequence of the input array of n words. This means considering each word for inclusion or exclusion in a potential subsequence, resulting in 2^n possible subsequences. For each subsequence, we need to verify if adjacent words belong to different groups, which takes O(n) time in the worst case for a subsequence of length n. However, the dominant factor is the generation of all 2^n subsequences, leading to an overall time complexity of O(2^n). The subsequence validation cost is less significant compared to the exponential cost of generating all subsequences.

Space Complexity

O(1) – The brute force approach outlined primarily iterates through the input words. The algorithm does not utilize any auxiliary data structures that scale with the input size N (number of words). It only keeps track of the length of the longest valid sequence found so far which takes constant space. Therefore, the space complexity is O(1).

Optimal Solution

Approach

This problem asks us to find the longest sequence of words such that no two adjacent words belong to the same group. We can use dynamic programming to efficiently find this sequence by keeping track of the best possible sequence ending at each word, avoiding redundant calculations. We consider all possible previous words and choose the one that extends our sequence the most, as long as it's from a different group.

Here's how the algorithm would work step-by-step:

Imagine we have a table where each entry represents the longest sequence ending at that word.
Start with the first word. The longest sequence ending at this word is just the word itself, so we initialize the table accordingly.
Now, for each subsequent word, we want to find the best sequence ending at that word.
To do this, we look at all the previous words in our list.
If a previous word belongs to a different group than the current word, we can potentially add the current word to the sequence ending at the previous word.
We check all such compatible previous words and choose the one that gives us the longest sequence when we add the current word.
We update our table with the length of this longest sequence.
Finally, after processing all the words, we find the maximum value in the table. This value represents the length of the longest valid subsequence.
To actually get the words in that sequence, we can trace back through the table, following the path of how we built the longest sequence. Start at the word where the longest valid subsequence ends and trace back.

Code Implementation

def longest_unequal_adjacent_groups_subsequence_two(words, groups):
    number_of_words = len(words)
    longest_subsequence_ending_here = [1] * number_of_words

    # Initialize the best subsequence length for each word to 1

    for i in range(1, number_of_words):
        for j in range(i):
            # Only update if the previous word comes from a different group
            if groups[i] != groups[j]:

                longest_subsequence_ending_here[i] = max(
                    longest_subsequence_ending_here[i],
                    longest_subsequence_ending_here[j] + 1,
                )

    # Find and return the maximum subsequence length.
    return max(longest_subsequence_ending_here)

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the 'n' words. For each word, it compares it against all preceding words to find the longest compatible sequence. This comparison involves checking up to 'n-1' previous words. Thus, the nested loop structure results in approximately n * (n-1) operations. Consequently, the time complexity simplifies to O(n²).

Space Complexity

O(N) – The described solution uses a table to store the length of the longest sequence ending at each word. Since there is one entry in the table for each of the N words, the auxiliary space required for this table is proportional to N. The space for tracing back the sequence can also be at most N. Therefore, the overall space complexity is O(N).

Edge Cases

Case	How to Handle
Empty input arrays (values or groups)	Return an empty list as no subsequence can be formed.
Arrays with single element in values and groups	Return the single element from 'values' if it exists, otherwise an empty list.
values and groups have different lengths	Throw an exception or return an empty list, as they must correspond 1:1.
All groups are identical	Return a single element of the values that appears at least once.
values contains duplicate elements that have different adjacent groups	The DP approach considers all valid subsequences, so duplicate values are handled correctly.
Large input size, potential for stack overflow or time limit exceed	Use dynamic programming instead of recursion to avoid stack overflow and optimize time complexity.
Integer overflow in calculations involving values (e.g., if calculating a score)	Use appropriate data types (e.g., long) to prevent integer overflow.
No valid subsequence exists (all adjacent groups are the same)	The algorithm will return an empty list, indicating no valid subsequence could be formed.