How do you solve Count Prefix and Suffix Pairs II on LeetCode?

Count Prefix and Suffix Pairs II (LeetCode #3045) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: The brute-force approach checks every possible pair of words to see if one is both a prefix and a suffix of the other. This is straightforward but inefficient for larger inputs.

What is the time complexity of Count Prefix and Suffix Pairs II?

The optimal solution for Count Prefix and Suffix Pairs II runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Count Prefix and Suffix Pairs II?

Count Prefix and Suffix Pairs II uses the following concepts: Array, String, Trie, Rolling Hash, String Matching, Hash Function. The recommended patterns to study are: .

#3045

Count Prefix and Suffix Pairs II

Hard

ArrayStringTrieRolling HashString MatchingHash Function

LeetCode ↗

Approaches

💡

Intuition

Time Space

The brute-force approach checks every possible pair of words to see if one is both a prefix and a suffix of the other. This is straightforward but inefficient for larger inputs.

⚙️

Algorithm

5 steps

1Step 1: Initialize a counter to zero.
2Step 2: Loop through each pair of words (i, j) where i < j.
3Step 3: For each pair, check if words[i] is a prefix and a suffix of words[j] using the isPrefixAndSuffix function.
4Step 4: If true, increment the counter.
5Step 5: Return the counter.

solution.py17 lines

1# Full working Python code
2
3def isPrefixAndSuffix(str1, str2):
4    return str2.startswith(str1) and str2.endswith(str1)
5
6def countPairs(words):
7    count = 0
8    n = len(words)
9    for i in range(n):
10        for j in range(i + 1, n):
11            if isPrefixAndSuffix(words[i], words[j]):
12                count += 1
13    return count
14
15# Example usage
16words = ['a', 'aba', 'ababa', 'aa']
17print(countPairs(words))

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.