2901. Longest Unequal Adjacent Groups Subsequence II

Problem Description

The problem asks us to select the longest subsequence from an array of indices that represent words and groups. A subsequence in this context is an ordered sequence that may skip elements from the original sequence but maintains the relative order of the remaining elements. There are two primary requirements for the sequence:

1. For any pair of adjacent indices in our subsequence (say i_j and i_(j + 1)), the groups that correspond to these indices must be different. In other words, groups[i_j] must not be equal to groups[i_(j + 1)].
2. The words at these indices must have the same length and a Hamming distance of 1. The Hamming distance is defined as the number of positions at which the two strings of equal length differ.

For example, if we have words ["wheel", "pearl", "world", "word"] and corresponding groups [1, 2, 1, 3], a valid subsequence that follows our conditions might be [0, 2, 3] which corresponds to the words ["wheel", "world", "word"].

The problem is to identify such the longest valid subsequence and return the words corresponding to that subsequence.

Intuition

The solution approach uses dynamic programming, which is a method of solving complex problems by breaking them down into simpler subproblems. The idea is to create an array f where each f[i] stores the length of the longest valid subsequence ending at index i. Moreover, g is an array where each g[i] stores the index of the predecessor in the subsequence ending with i.

Initially, we assume that the longest subsequence that ends with each word has a length of 1 (because a subsequence with a single word is always valid) — this is our base case.

The algorithm then checks each word against all the words before it. For two words at index i and j, if they belong to different groups and the Hamming distance between them is 1, it means that the word at index i could potentially follow the word at index j in a valid subsequence.

The algorithm goes through all possible pairs of words and updates f[i], g[i], and a variable mx (which represents the length of the longest subsequence found so far) accordingly. After considering all words, mx will hold the length of the longest valid subsequence.

Finally, by tracing back from an index i with the maximum length mx in f, using the g array to find the predecessor indices, the algorithm constructs the subsequence in reverse order. Reversing this sequence gives us the longest required subsequence.

This approach uses the concept of comparing all possible pairs of words (hence the O(n^2 * L) time complexity), and utilizes additional arrays to store the subsequence lengths and the predecessor indices (which accounts for the O(n) space complexity).

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Solution Approach

The solution approach employs dynamic programming (DP) to build up a solution by evaluating all sub-problems, represented by each possible subsequence ending at any index in the given arrays. The primary idea is to keep track of two things for each position i: the length of the longest valid subsequence that ends with the word at position i, and the index of the word that came just before i in that subsequence.

Dynamic Arrays: f and g

Two arrays are used in the DP algorithm:

• f: An array where f[i] contains the length of the longest subsequence ending at index i.
• g: An array where g[i] contains the index of the predecessor of i in the subsequence.

Both arrays f and g are vital for reconstructing the longest subsequence once we've completed the iteration process. We start by initializing each f[i] value to 1, because a sequence containing only the i-th element is obviously valid, with a length of 1.

Updating the DP Arrays

We then iterate through all pairs of indices using nested loops. For each index i, we look at all indices j that come before it, i.e., j < i. Two conditions must be satisfied to consider extending the subsequence ending at j to include i:

• The groups must be different: groups[i] != groups[j].
• The Hamming distance between words[i] and words[j] must be exactly 1: sum(a != b for a, b in zip(words[i], words[j])) == 1.

When both conditions are met, we check if the length of the subsequence at j plus one is greater than the current length at i (f[i] < f[j] + 1). If so, we update f[i] to f[j] + 1 and record the index j in g[i]. We also update mx to keep track of the current maximum length of any subsequence we've found.

Reconstructing the Longest Subsequence

Once we've processed all indices, we can then find out which index corresponds to the length mx. Starting from this index, we trace back through the g array, effectively reconstructing the longest subsequence in reverse order. This reverse order is necessary because we start from the end and go backwards to the beginning of the subsequence, following the stored predecessors.

Finally, we reverse the reconstructed subsequence to present it in the correct order, from the first to the last index of the subsequence.

Time Complexity: O(n^2 * L)

The time complexity is O(n^2 * L), where n is the number of elements in the words and groups arrays and L is the maximum length of the strings in words. This time complexity arises because we perform a nested iteration over all pairs of elements (O(n^2)) and, for each pair, potentially compare every character of the two strings involved (O(L)).

Space Complexity: O(n)

The space complexity is O(n) due to the additional arrays f and g that we maintain throughout the algorithm.

Example Walkthrough

Let's use a small example to illustrate the solution approach with words ["ab", "bc", "ac", "ad"] and corresponding groups [1, 1, 2, 3].

Step 1: Initialize DP arrays

• f: [1, 1, 1, 1] (Each word can be a subsequence by itself)
• g: [-1, -1, -1, -1] (No predecessors yet)

Step 2: Evaluating Pairs

• Compare "ab" and "bc":
  • Different groups? No, both in group 1. Do not update f or g.
• Compare "ab" and "ac":
  • Different groups? Yes.
  • Hamming distance = 1? Yes ("ab" vs "ac").
  • Update: f[2] = f[0] + 1 = 2, g[2]= 0.
• Update f: [1, 1, 2, 1], Update g: [-1, -1, 0, -1]
• Compare "ab" and "ad":
  • Different groups? Yes.
  • Hamming distance = 1? Yes ("ab" vs "ad").
  • Update: f[3] = f[0] + 1 = 2, g[3] = 0.
• Update f: [1, 1, 2, 2], Update g: [-1, -1, 0, 0]
• Compare "bc" with all previous entries, no updates since "bc" is in the same group as "ab".
• Compare "ac" and "ad":
  • Different groups? Yes.
  • Hamming distance = 1? Yes ("ac" vs "ad").
  • Update: f[3] = f[2] + 1 = 3, g[3] = 2 (since f[2] > f[0]).
• Final update f: [1, 1, 2, 3], g: [-1, -1, 0, 2]

Step 3: Reconstruct the Longest Subsequence

• mx is 3, and it's at index 3, word "ad".
• Trace back predecessors: for "ad" (g[3] = 2), the predecessor is "ac" (at index 2).
• For "ac" (g[2] = 0), the predecessor is "ab" (at index 0).
• We construct the subsequence in reverse order: [3, 2, 0].

Step 4: Reverse the Subsequence

• After reversing, we get [0, 2, 3].

Step 5: Return Words of Longest Subsequence

• Corresponding words: ["ab", "ac", "ad"].

Through this example, the solution approach successfully identified and constructed the longest valid subsequence where adjacent words are from different groups and have a Hamming distance of 1.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code is O(n^2 * L). The code consists of two nested loops where n is the number of words, and it compares pairs of words where each word has a length of L. The function check is called for each pair, which operates in O(L) time to compare two words and check if their Hamming distance is exactly 1.

However, as mentioned in the reference answer, we can optimize the process by using a wildcard hash table. When using a wildcard hash table, we create all possible variants of a word by replacing each letter once with a wildcard. We use these variants as keys in a hash table that map to indices of words in the original list. This operation takes O(L) time for each word (because we're iterating through each letter and creating variants) and is done for all n words, leading to O(n * L) for this preprocessing step. Subsequently, for each word, we look up its variants in the hash table in O(L) time to find all potential matches with a Hamming distance of 1. Since each lookup yields a constant number of potential matches on average (assuming close to uniform distribution of words across the different wildcard variants), the average time complexity for processing all words drops.

Thus, while the worst-case time complexity remains the same, in practice, we expect the average complexity to be lower due to fewer pairs of words being compared.

Space Complexity

The space complexity of the original solution is O(n). The code creates two lists, f and g, both of size n. There are no additional data structures that grow with the input size; therefore, the space complexity is linear with respect to the number of words.

With the optimization mentioned in the reference answer, we would have to account for the space taken by the wildcard hash table. The hash table would have potentially O(n * L) keys (since each of n words contributes L wildcard variants). However, because each word leads to at most L keys and the space for indices is shared among keys, the size of the list of indices is O(n).

Thus, the optimized solution has a space complexity of O(n + L * n), which simplifies to O(n * L) since L could be larger than 1 and in this case, dominates the n term. This represents a trade-off compared to the original solution: better average time efficiency at the cost of increased space usage.

Learn more about how to find time and space complexity quickly using problem constraints.