3309 Count Prefix And Suffix Pairs I

Count Prefix and Suffix Pairs I

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:

        res = 0

        def starts_with(s1, s2):
            if len(s1) > len(s2):
                return False
            for a1, a2 in zip(s1, s2):
                if a1 != a2:
                    return False
            return True

        def ends_with(s1, s2):
            if len(s1) > len(s2):
                return False
            i = len(s1) - 1
            j = len(s2) - 1
            while(i >= 0 and j >= 0):
                if s1[i] != s2[j]:
                    return False
                i -= 1
                j -= 1
            return True

        for idx1, k1 in enumerate(words):
            for idx2, k2 in enumerate(words[idx1+1:]):
                if starts_with(k1,k2) and ends_with(k1, k2):
                    res += 1
        return res