leetCode 3042.Count Prefix and Suffix Pairs I

leetCode 3042.Count Prefix and Suffix Pairs I.

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.

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.

Constraints:

1 <= words.length <= 50
1 <= words[i].length <= 10
words[i] consists only of lowercase English letters.

Approach

I used the includes() method on my first try. I passed the test cases, but i didn't passed one of use cases. Then I found out that words[j] was supposed to end with words[i] string too. I missed that part. I searched for methods and found startsWith() and endsWitdh().

Solution

var countPrefixSuffixPairs = function (words) {
  let sum = 0

  for (let i = 0; i < words.length; i++) {
    for (let j = i + 1; j < words.length; j++) {
      if (words[j].startsWith(words[i]) && words[j].endsWith(words[i])) {
        sum++
      }
    }
  }

  return sum
}