Problem Description

The problem provides two arrays of strings, words1 and words2. The goal is to determine which strings from words1 are considered "universal." A string is universal if every string in words2 is a subset of it. A string b is a subset of another string a if a contains all the letters of b in the same number or more. That means for every character and its count in b, this character must appear at least the same number of times in a. For example, 'wrr' is a subset of 'warrior' but not of 'world'. The task is to return an array of universal strings from words1, and the order of results does not matter.

Intuition

To find the universal strings efficiently, rather than comparing each string in words1 with all the strings in words2, we can create a letter frequency count that represents the maximum frequency of each character across all strings in words2. This way, we only need to compare each string in words1 with this combined frequency count.

Firstly, we create a Counter (which is a dictionary subclass for counting hashable objects) to hold the maximum frequency of characters required from words2. For each string in words2, we count its characters and update our maximum frequency Counter. This counter will tell us the least number of times each character should appear in any string from words1 for it to be considered universal.

After this, we iterate through each string in words1 and create a frequency count for it. Then, we check if this count satisfies the requirements of our precomputed maximum frequency counter. The all() function helps in determining whether all conditions are true for each character's count.

If a string a from words1 has at least as many occurrences of each character as computed in the maximum frequency Counter, then it is universal. We append such strings to our answer list ans.

This solution approach minimizes the number of total comparisons needed to identify the universal strings, and therefore is optimized for the task.

Solution Approach

To solve this problem, the code implementation makes use of Python's Counter class from the collections module. The Counter class is a specialized dictionary used for counting hashable objects, in our case, characters in strings. It's particularly useful because it allows us to easily compare the frequencies of letters in different strings.

Here’s a step-by-step walkthrough of the implementation:

1. Initialize a new Counter object called cnt, which will be used to store the maximum frequency requirements for each letter, derived from words2.

   cnt = Counter()

2. Loop over each string b in the array words2. For each string:

   a. Create a temporary counter t from b.

   b. Update the cnt counter with the maximum frequency of characters from t. This means cnt[c] = max(cnt[c], v) for each character c and its count v in t. This ensures that cnt holds the maximum number of times any character needs to appear for a string from words1 to be universal.

   for b in words2:
       t = Counter(b)
       for c, v in t.items():
           cnt[c] = max(cnt[c], v)

3. Initialize an empty list ans, which will hold all the universal strings from words1.

   ans = []

4. Loop over each string a in words1. For each string:

   a. Create a temporary counter t from a.

   b. Use the all function to check if a contains at least as many of each character as required by the cnt counter. The comparison v <= t[c] for c, v in cnt.items() ensures that a meets the criteria for all characters c with their respective counts v in cnt.

   c. If a meets the criteria, append it to the ans list.

   for a in words1:
       t = Counter(a)
       if all(v <= t[c] for c, v in cnt.items()):
           ans.append(a)

5. Finally, return the list ans, which now contains all the universal strings from words1.

       return ans

The primary algorithm patterns used in this solution include frequency counting and iterating with condition checking. By using the Counter class with maximum frequency logic, the code efficiently reduces what could be an O(n * m) problem (checking each of n strings in words1 against each of m strings in words2) to a more manageable O(n + m) problem by minimizing repetitive checks.

Example Walkthrough

Consider the following example to illustrate the solution approach:

Let words1 be ["ecology", "universal", "computer"] and words2 be ["cool", "envy", "yon", "yes"].

1. The maximum frequency counters (cnt) are determined from words2. For "cool", it will be {'c': 1, 'o': 2, 'l': 1}, for "envy", it will be {'e': 1, 'n': 1, 'v': 1, 'y': 1} and so forth. Ultimately, after comparing all words2 strings, cnt would be:

   cnt = {'c': 1, 'o': 2, 'l': 1, 'e': 1, 'n': 1, 'v': 1, 'y': 2, 's': 1}

   This cnt means for a string in words1 to be universal; it must have at least one 'c', two 'o's, one 'l', one 'e', one 'n', one 'v', two 'y's, and one 's'.

2. Now we check each word in words1 against cnt.

   • "ecology": It's counter is {'e': 1, 'c': 1, 'o': 1, 'l': 1, 'g': 1, 'y': 1}. While this word has all the required characters, it fails to meet the count for 'o' and 'y'. Thus, "ecology" is not universal.

   • "universal": Counter is {'u': 1, 'n': 1, 'i': 1, 'v': 1, 'e': 1, 'r': 1, 's': 1, 'a': 1, 'l': 1}. Despite having all characters, it does not have enough 'o's and 'y's. So, "universal" is not universal.

   • "computer": Counter is {'c': 1, 'o': 1, 'm': 1, 'p': 1, 'u': 1, 't': 1, 'e': 1, 'r': 1}. Even though "computer" meets or exceeds the count for 'c', 'e', 't', 'r', it doesn’t contain any 'l', 'v', 'y', or sufficient 'o's. Therefore, it's not universal either.

3. Applying this logic, none of the words in words1 are universal, as none meet the frequency criteria established by words2. The final answer ans list would be empty.

The result is an efficient check that precisely tells us that there are no strings in words1 that are universal with respect to words2. Thus, the function would return an empty list [].

Solution Implementation

Time and Space Complexity

Time Complexity

The given code has two main sections contributing to the time complexity:

1. Building the Universal Counter:

   • Iterating over each word in words2 and updating the cnt counter for each character.
   • If M is the average length of words in words2 and N2 is the number of words in words2, this part is O(M * N2) in the worst case.

2. Checking if Words in words1 are Universal:

   • For each word in words1, a counter is created and checked against the cnt counter.
   • If L is the average length of words in words1 and N1 is the number of words in words1, and if K is the number of unique characters in cnt (bounded by the alphabet size), then checking each word has complexity O(L + K), and doing this for all words in words1 gives O(N1 * (L + K)).

Combining the two we get a total time complexity of O(M * N2) + O(N1 * (L + K)).

• Note that K is bounded by the alphabet size which can be considered a constant, hence it sometimes can be omitted from the Big-O notation, simplifying to O(M * N2 + N1 * L).

Space Complexity

The space complexity of the code comprises the following factors:

1. The Universal Counter cnt:

   • The maximum space required is the size of the alphabet, let's denote it as a, which is constant. Hence, this is O(a).

2. Temporary Counter for Each Word in words1:

   • At most, the size of the alphabet a for each word. As this is temporary and not used simultaneously, it's still O(a).

3. The ans List:

   • At most, it can be as big as N1 in case all words1 are universal. So the space for it would be O(N1).

Considering all, the total space complexity is O(a) + O(N1) which simplifies to O(N1) because a is constant and generally considered negligible compared to N1.

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