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3305. Count Prefix And Suffix Pairs II

Array String Trie Rolling Hash String Matching Hash Function

Problem - Count Prefix And Suffix Pairs II

Hard

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 <= 105
  • 1 <= words[i].length <= 105
  • words[i] consists only of lowercase English letters.
  • The sum of the lengths of all words[i] does not exceed 5 * 105.

Solutions

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class Trie:
    def __init__(self):
        self.count = 0
        self.children = {}

class Solution:
    def countPrefixSuffixPairs(self, words: List[str]) -> int:
        result = 0
        trie = Trie()
        for word in words:
            node = trie
            for cantidate in zip(word, reversed(word)):
                if cantidate not in node.children:
                    node.children[cantidate] = Trie()
                node = node.children[cantidate]
                result += node.count
            node.count += 1

        return result

Submission Stats:

  • Runtime: 1568 ms (70.33%)
  • Memory: 210.9 MB (34.93%)