3042. Count Prefix and Suffix Pairs I

Problem Link

Description

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You are given a 0-indexed string array words.

Let's define a boolean function isPrefixAndSuffix that takes two strings, str1 and str2:

• isPrefixAndSuffix(str1, str2) returns true if str1 is both a prefix and a suffix of str2, and false otherwise.

For example, isPrefixAndSuffix("aba", "ababa") is true because "aba" is a prefix of "ababa" and also a suffix, but isPrefixAndSuffix("abc", "abcd") is false.

Return an integer denoting the number of index pairs (i, j) such that i < j, and isPrefixAndSuffix(words[i], words[j]) is true.

 

Example 1:

Input: words = ["a","aba","ababa","aa"]
Output: 4
Explanation: In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix("a", "aba") is true.
i = 0 and j = 2 because isPrefixAndSuffix("a", "ababa") is true.
i = 0 and j = 3 because isPrefixAndSuffix("a", "aa") is true.
i = 1 and j = 2 because isPrefixAndSuffix("aba", "ababa") is true.
Therefore, the answer is 4.

Example 2:

Input: words = ["pa","papa","ma","mama"]
Output: 2
Explanation: In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix("pa", "papa") is true.
i = 2 and j = 3 because isPrefixAndSuffix("ma", "mama") is true.
Therefore, the answer is 2.  

Example 3:

Input: words = ["abab","ab"]
Output: 0
Explanation: In this example, the only valid index pair is i = 0 and j = 1, and isPrefixAndSuffix("abab", "ab") is false.
Therefore, the answer is 0.

 

Constraints:

• 1 <= words.length <= 50
• 1 <= words[i].length <= 10
• words[i] consists only of lowercase English letters.

Solution

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Python3

class Solution:
    def countPrefixSuffixPairs(self, words: List[str]) -> int:
        N = len(words)
        res = 0
 
        def good(s1, s2):
            A, B = len(s1), len(s2)
 
            i = j = 0
            while i < A and j < B and s1[i] == s2[j]:
                i += 1
                j += 1
            
            if i != A: return False
 
            i, j = A - 1, B - 1
            while i >= 0 and j >= 0 and s1[i] == s2[j]:
                i -= 1
                j -= 1
            
            if i != -1: return False
 
            return True
 
        for i in range(N):
            for j in range(i + 1, N):
                if good(words[i], words[j]):
                    res += 1
        
        return res