CountPrefixAndSuffixPairs.py

'''
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.
'''

class Solution:
    def countPrefixSuffixPairs(self, words: List[str]) -> int:
        def isGood(i, j):
            return words[j].startswith(words[i]) and words[j].endswith(words[i])
        
        count = 0
        for i in range(len(words)):
            for j in range(i+1, len(words)):
                if isGood(i, j):
                    count += 1
        return count
    
'''
Runtime 61 ms Beats 33.72% of users with Python3
Memory 16.46 MB Beats 86.81% of users with Python3
'''