8_jan

/*
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.
*/

//approch 1
class Solution {
public:
    int countPrefixSuffixPairs(vector<string>& words) {
        int n = words.size();
        int ans = 0;

        for (int i = 0; i < n; i++) {
            string s1 = words[i];

            for (int j = i + 1; j < n; j++) {
                string s2 = words[j];

                // Skip if s2 is shorter than s1
                if (s2.length() < s1.length()) 
                    continue;

                // Extract prefix and suffix
                string pre = s2.substr(0, s1.length());
                string suf = s2.substr(s2.length() - s1.length());

                // Check if both prefix and suffix match s1
                if (pre == s1 && suf == s1) {
                    ans++;
                }
            }
        }
        return ans;
    }
};