3042. Count Prefix and Suffix Pairs I

Description

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.

 

Solutions

Solution: Brute Force

• Time complexity: O(n²)
• Space complexity: O(1)

 

JavaScript

/**
 * @param {string[]} words
 * @return {number}
 */
const countPrefixSuffixPairs = function (words) {
  const n = words.length;
  let result = 0;

  for (let a = 0; a < n - 1; a++) {
    const wordA = words[a];

    for (let b = a + 1; b < n; b++) {
      const wordB = words[b];

      if (!wordB.startsWith(wordA) || !wordB.endsWith(wordA)) continue;

      result += 1;
    }
  }
  return result;
};