3042 - 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.

Solution 1: Enumeration

C++

• class Solution {
  public:
      int countPrefixSuffixPairs(vector<string>& words) {
          int ans = 0;
          int n = words.size();
          for (int i = 0; i < n; ++i) {
              string s = words[i];
              for (int j = i + 1; j < n; ++j) {
                  string t = words[j];
                  if (t.find(s) == 0 && t.rfind(s) == t.length() - s.length()) {
                      ++ans;
                  }
              }
          }
          return ans;
      }
  };

JAVA

• class Solution {
      public int countPrefixSuffixPairs(String[] words) {
          int ans = 0;
          int n = words.length;
          for (int i = 0; i < n; ++i) {
              String s = words[i];
              for (int j = i + 1; j < n; ++j) {
                  String t = words[j];
                  if (t.startsWith(s) && t.endsWith(s)) {
                      ++ans;
                  }
              }
          }
          return ans;
      }
  }

PYTHON

• class Solution:
      def countPrefixSuffixPairs(self, words: list) -> int:
          ans = 0
          n = len(words)
          for i in range(n):
              s = words[i]
              for j in range(i + 1, n):
                  t = words[j]
                  if t.startswith(s) and t.endswith(s):
                      ans += 1
          return ans