2942. Find Words Containing Character

Problem Description

You are given an array of strings, where each string is indexed starting from 0 - this is what is meant by a "0-indexed" array. Your task is to find and return the indices of all strings in this array that contain a specific character x. The character x is provided to you, and you need to search through each string in the array to check if x is included in that string. If it is, you keep track of that string's index. Once you're done checking all the strings, you produce an array of these indices. It's important to note that the order of these indices in the final array does not matter.

Intuition

The intuition behind the solution is pretty straightforward. Since we only need to know whether a character exists in the words, we can simply iterate through each word in the list and check for the presence of the character x. This can be done easily with the help of Python's in keyword, which checks for the presence of a substring in a string.

For efficiency, we use a list comprehension which allows us to iterate through the list of words and at the same time, keep track of the index of each word. We leverage enumerate() here, which returns both the index and the corresponding word. The condition if x in w inside the list comprehension checks if the character x is in the current word w. If it is, the index i gets added to the output list. This process continues for each word in the list words.

By the end of the iteration, the list comprehension will have generated a list containing the indices of all the words that contain the character x, satisfying the problem requirement.

Solution Approach

For the given problem, the solution uses a simple yet effective approach that does not require complex algorithms or data structures. The approach follows a sequential procedure using basic operations provided by Python:

1. Iteration with Enumeration:

   • Python's built-in enumerate() function is used, which is a common pattern when you need to traverse a list and also keep track of indices.
   • enumerate(words) gives us a neat way to loop over words while having access to both the index i and word w at the same time.

2. Membership Testing:

   • Inside the loop, we use the membership operator in which checks whether the character x is present in the word w. This operation is performed for every word in the list.
   • The condition x in w evaluates to either True if x is found inside the string w, or False otherwise.

3. List Comprehension:

   • A list comprehension is utilized to build the resulting list of indices in a clean and efficient one-liner.
   • This eliminates the need for initializing an empty list beforehand and appending indices one by one inside a for-loop.

4. Conditional Expression in List Comprehension:

   • The list comprehension has a form of [expression for item in iterable if condition], where the expression is simply the index i.
   • The condition in our case is if x in w, and it ensures that the index i is included in the final list only if the word w contains the character x.

The function findWordsContaining returns the list that is generated by the list comprehension, which includes all indices i where the condition x in w holds true.

In summary, the solution achieves the goal by using iteration to go through each word, using the membership test to find words containing the character x, and a list comprehension to build the list of indices efficiently. This is a direct and practical approach that leverages Python's language features to solve the problem succinctly.

Example Walkthrough

Let's walk through an example to illustrate the solution approach described above.

Imagine we are given an array of strings words = ["apple", "banana", "cherry", "date"] and the character x = 'a'. We want to find all the indices of strings containing the character 'a'.

Following the solution approach:

1. Iteration with Enumeration: We start by enumerating the list ["apple", "banana", "cherry", "date"]. This means we get pairs of indices and words like (0, "apple"), (1, "banana"), (2, "cherry"), and (3, "date").

2. Membership Testing:

   • For (0, "apple"), we check if 'a' is in "apple". It is, so this index (0) could be part of our resulting list.
   • Then for (1, "banana"), 'a' is also found in "banana". Index (1) should also be part of our resulting list.
   • For (2, "cherry"), 'a' is not found, so we skip this index.
   • Finally, for (3, "date"), 'a' is again not found, so we also skip this index.

3. List Comprehension: Using list comprehension, we combine the steps of enumeration and membership testing into one line. It would look like this:

   1[i for i, w in enumerate(words) if x in w]

   Executing this line with our example list would produce [0, 1] since those are the indices of the words containing the letter 'a'.

4. Conditional Expression in List Comprehension: The conditional (if x in w) ensures that only indices of words containing 'a' are included in the final list.

Putting it all together, the function findWordsContaining(['apple', 'banana', 'cherry', 'date'], 'a') returns the list [0, 1], successfully identifying the indices of words that contain the character 'a'.

Solution Implementation

Time and Space Complexity

The time complexity of the given code snippet can be understood by considering each operation in the list comprehension. The function iterates through each word in the list words and checks if the substring x is present in each word w. This operation involves checking each character in each word against the characters in x until a match is found or the word is exhausted. Thus, if we let L be the sum of the lengths of all the strings in the list words, the worst-case scenario is when x is not present in any word, resulting in a time complexity of O(L).

The space complexity of the function, aside from the output list, is O(1). This is because the only extra space used is for the loop iteration variable and the index i, which do not depend on the input size. Therefore, the space used for the function's operations other than the storage for the return list is constant.

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