3042. Count Prefix and Suffix Pairs I

Problem Description

In this problem, we're given an array of 0-indexed strings and we need to count the number of pairs of indices (i, j) for which the string at index i is both a prefix and a suffix of the string at index j, and i must be less than j. A prefix of a string is a leading substring of the original string, while a suffix is a trailing substring.

For clarity, let's consider what constitutes a prefix and a suffix:

• A prefix of a string is the start of that string. For example, "cat" is a prefix of "cater".
• A suffix is the end of the string. For example, "ter" is a suffix of "cater".

Therefore, if a given string is both a prefix and a suffix of another, it appears at both the beginning and the end of the other string. The function isPrefixAndSuffix(str1, str2) is supposed to return true if str1 is both a prefix and suffix of str2, otherwise false.

The task is to count how many such distinct index pairs (i, j) exist in the given array where i < j.

Intuition

To solve this problem, we need to look at every possible pair of strings where the first string could be a potential prefix and suffix for the second string. We can do this through a nested loop where we iterate over each string and compare it with every other string that comes after it in the array.

When comparing the strings, we use two important string operations: startswith() and endswith(). These methods return true if the first string is a prefix or suffix of the second string, respectively:

1. The method startswith(s) will check if the string under consideration starts with the substring s.
2. Likewise, endswith(s) will verify if the string ends with the substring s.

By ensuring that both these conditions are true for a pair of strings, we confirm that the first string is both a prefix and a suffix of the second one.

We only consider pairs where the first string index is less than the second (i < j), to comply with the problem statement. For each such valid pair, we increment our answer. This ensures that we count all the unique pairs that satisfy the condition. The summed total gives us the desired output.

Learn more about Trie patterns.

Solution Approach

The implementation of the solution utilizes a straightforward brute force approach, where we simply iterate through every possible pair of strings and check if one string is both a prefix and a suffix of the other.

The algorithm can be described as follows:

1. Initialize a counter (let's call it ans) and set it to 0. This will hold the count of valid pairs (i, j) we find.
2. Use nested loops to iterate through the array of strings:
   • The outer loop will go through each string s in the array, starting from the first element up to the second-to-last. We use enumerate to get the index i alongside the string s.
   • The inner loop will iterate through the remaining elements in the array, starting from the index i + 1 and continuing to the end of the array. These will be the strings t against which we will compare s.
3. For each pair of strings (s, t), where s is from the outer loop and t is from the inner loop:
   • We check if string t starts with s using t.startswith(s).
   • Simultaneously, we check if string t ends with s using t.endswith(s).
   • If both conditions are true, meaning s is both a prefix and suffix of t, increment ans by 1.
4. After the loops complete, return the count ans which now contains the number of index pairs (i, j) where isPrefixAndSuffix(words[i], words[j]) is true.

There are no advanced data structures utilized in this solution; it's a simple application of nested iteration over the given array and string comparison methods. The efficiency of the algorithm is not optimized; it operates with a time complexity of O(n^2 * m) where n is the length of the words array, and m is the average string length. This is because for every pair of strings, we potentially check each character up to the length of the shorter string twice (once for prefix, once for suffix).

Even though the solution is not the most optimal, it works well for small datasets and serves as a clear example of how to implement such a count conditionally on a collection of strings.

Example Walkthrough

Let's consider a simple example to illustrate the solution approach. Suppose we are given the following array of strings:

1words = ["one", "oneon", "neon", "oneonne"]

We want to find the count of pairs of indices (i, j) such that the string at index i is both a prefix and a suffix of the string at index j, with i < j.

Here's how we apply the solution step by step:

1. Initialize ans to 0.
2. Iterate over the array to consider every potential string s:
   • At i = 0, s = "one".
     • Now we compare s with every other string t with index greater than i.
     • At j = 1, t = "oneon". We find that t.startswith(s) is True but t.endswith(s) is False. Since both conditions are not met, we do not increment ans.
     • At j = 2, t = "neon". Neither t.startswith(s) nor t.endswith(s) are True, so we continue.
     • At j = 3, t = "oneonne". Both t.startswith(s) and t.endswith(s) are True. So, we increment ans by 1. Now, ans = 1.
   • Move to i = 1, s = "oneon". Repeat the steps, but there are no strings after s that satisfy the conditions.
   • Continue to i = 2, s = "neon". There's only one string after this, and it doesn't satisfy the conditions.
3. No further pairs can be considered because i should be less than j, and we have exhausted all possibilities.
4. Finally, return the count ans, which in this case is 1.

Therefore, there is only 1 valid pair (0, 3) where the first string is both a prefix and a suffix of the second string, and i is less than j.

Solution Implementation

Time and Space Complexity

Time Complexity

The provided function iterates over each word using a nested loop structure. The outer loop runs n times where n is the length of the words list. For each iteration of the outer loop, the inner loop runs n - i - 1 times, where i is the current index of the outer loop. There, each comparison between two strings, for startswith and endswith, can take up to m operations where m is the maximum length of the strings in the words list. Hence, the worst-case time complexity is O(n^2 * m).

Space Complexity

The function mainly uses constant extra space, only storing the ans variable and loop indices. It operates directly on the input and does not allocate additional space that depends on the input size. Therefore, the space complexity is O(1) as it remains constant regardless of the input size.

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