Problem

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.

Examples

Example 1:

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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:

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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:

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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

Method 1 - Naive

  • The problem requires checking if a word words[i] is both a prefix and suffix of another word words[j] where i < j.
  • A simple way to check this is to use string methods startsWith and endsWith (in Java) or the equivalent operations in Python.
  • We iterate over all possible pairs (i, j) and count the valid pairs.

Video explanation

Here is the video explaining this method in detail. Please check it out:

Code

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class Solution {
    public int countPrefixSuffixPairs(String[] words) {
        int ans = 0;
        for (int i = 0; i < words.length; i++) {
            for (int j = i + 1; j < words.length; j++) {
                if (isPrefixAndSuffix(words[i], words[j])) {
                    ans++;
                }
            }
        }
        return ans;
    }

    private boolean isPrefixAndSuffix(String str1, String str2) {
        return str2.startsWith(str1) && str2.endsWith(str1);
    }
}
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class Solution:
    def countPrefixSuffixPairs(self, words: List[str]) -> int:
        ans: int = 0
        for i in range(len(words)):
            for j in range(i + 1, len(words)):
                if self.isPrefixAndSuffix(words[i], words[j]):
                    ans += 1
        return ans

    def isPrefixAndSuffix(self, str1: str, str2: str) -> bool:
        return str2.startswith(str1) and str2.endswith(str1)

Complexity

  • ⏰ Time complexity: O(n^2 * l),  where n is the length of the words array and l is average length of words, because of the nested loop and starts with and ends with take O(l) time where l is length of prefix and suffix, and is proportional to average length of word..
  • 🧺 Space complexity: O(1)