Problem Description

The problem presents us with a list of strings called words, and another string s. We are asked to determine if s is an acronym for the words. An acronym in this context is defined as a string that can be formed by taking the first character of each string in words, in their respective order. For instance, given words = ["keep", "it", "simple", "stupid"], the string s = "kiss" would be an acronym of words because taking the first letter of each word in order produces "kiss".

To decide whether s is an acronym for words, we need to follow this rule and check if the concatenation of the first letter of each word in the words array is equal to s.

Intuition

To solve this problem, the intuition is straightforward: we will sequentially take the first character from each string in the words array and concatenate them into a new string. Then, all we need to do is to compare this newly formed string to the given string s.

Here's the step-by-step intuition:

1. Initialize an empty string that will hold the concatenated first characters, or simply prepare to perform a comparison on-the-fly without creating a new string.

2. Traverse the words list, and for each word, take its first character.

3. As we get each first character, either add it to the initialized string or compare it directly if not storing it.

4. After processing all words, we either compare the created acronym string to s or if we've compared on-the-fly, ensure all characters matched.

If the acronym we assembled matches s, we return true, indicating that s is an acronym of words. If not, we return false.

The provided Python solution uses a generator expression to create the acronym string by concatenating the first character of each word in words and then compares it directly to s. This approach is concise, memory-efficient, and takes advantage of Python's expressive syntax.

Solution Approach

The implementation of the solution in Python is quite simple and utilizes few advanced concepts, but it demonstrates some of the core principles of programming — iteration and comparison.

• Algorithm: The algorithm followed here is a linear scan of the words list to create a string made up of the first character of each word.

• Data Structures: In the solution, the data structure used is the list (List[str]). This list holds the input strings that are processed.

• Patterns: The pattern used here is a common Python idiom of using a generator expression within a string join method. A generator expression is an efficient way to iterate over data without the need for explicitly writing a loop or using additional memory for a list. It is commonly used for its concise syntax and because it generates elements on the fly; it is memory efficient.

The implementation in Python can be walked through step-by-step:

1. String Join and Generator Expression:

   return "".join(w[0] for w in words) == s

   In this line of Python code, we have "".join(...), which is a method that takes an iterable and concatenates its elements separated by the string it is called on—in this case, an empty string. This means no separator between the elements.

2. Generator Expression:

   The part inside the join method w[0] for w in words is a generator expression. It goes through each element w in the list words. For each element, it takes the first character w[0].

3. Comparison:

   Finally, the resulting string after concatenating all first characters is compared to the input string s. The operator == evaluates to True if both strings are equal, else False. This is the returned value of the isAcronym method.

The elegance of this solution comes from its use of Python's generator expressions and the join method, which both play well together to create a compact and efficient line of code. The approach is direct and specific to checking whether the acronym matches the input string without any unnecessary steps or calculations.

Example Walkthrough

Let's walk through a small example using the solution approach to understand it better. Suppose we have the input list of words ["national", "aeronautics", "space", "administration"] and we want to verify whether the string s = "nasa" is an acronym for this list of words.

Here's how the solution algorithm would process this example:

1. We start with an empty string (conceptually, as per the provided solution we do not explicitly create it) to collect the first letters.

2. We look at the first word "national", extract its first character "n", and add it to the acronym string. Our acronym string is now "n".

3. Next, we take the first character of the second word "aeronautics" which is "a", and append it to our acronym string. The acronym string becomes "na".

4. We continue with the third word "space" and append its first character "s" to our acronym string, which now becomes "nas".

5. Finally, we extract the first character of the last word "administration" which is "a", and append it to our acronym string. The acronym string is now complete and equals "nasa".

6. We then compare this string to the given string s. Since "nasa" == "nasa", we return True.

Using the actual Python solution from the solution approach:

words = ["national", "aeronautics", "space", "administration"]
s = "nasa"
result = "".join(w[0] for w in words) == s

Upon executing the above, result would be True, showing that s is indeed an acronym for the provided list words.

The elegant part of this algorithm is that we don't even need to store the acronym string. The comparison happens on-the-fly within the generator expression as it generates each first character and is immediately followed by the string concatenation and comparison with s. Thus, the process is both memory- and time-efficient.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the given code snippet can be analyzed as follows:

• There is a list comprehension that iterates through each word in the list words. This operation takes O(n) time, where n is the number of words in the list.
• For each word, we are accessing the first character w[0], which is an O(1) operation.
• The join operation itself takes O(m), where m is the total number of characters in the final acronym string, which in this case is the same as the number of words n since we are only considering the first character of each word.

Therefore, the total time complexity of the given function is O(n), since both iterating through the words and joining the characters are linear operations with respect to the number of words.

Space Complexity

The space complexity of the given code snippet can be analyzed as follows:

• The list comprehension creates a temporary list that holds the first character of each word, which requires O(n) space, where n is the number of words.
• The acronym string created by the join operation also requires O(n) space.

Considering this, the total space complexity of the function is O(n), the space needed to create the temporary list and the acronym string.

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