3042

3042. Count Prefix and Suffix Pairs I (Easy)

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.

Similar Questions:

• Implement Trie (Prefix Tree) (Medium)
• Design Add and Search Words Data Structure (Medium)

Hints:

• Iterate through all index pairs (i, j), such that i < j, and check isPrefixAndSuffix(words[i], words[j]).
• The answer is the total number of pairs where isPrefixAndSuffix(words[i], words[j]) == true.

Solution 1.

// OJ: https://leetcode.com/problems/count-prefix-and-suffix-pairs-i
// Author: github.com/lzl124631x
// Time: O(N^2 * W)
// Space: O(W)
class Solution {
    bool isPrefixAndSuffix(string &a, string &b) {
        return a.size() <= b.size() && b.substr(0, a.size()) == a && b.substr(b.size() - a.size()) == a;
    }
public:
    int countPrefixSuffixPairs(vector<string>& A) {
        int N = A.size(), ans = 0;
        for (int i = 0; i < N; ++i) {
            for (int j = i + 1; j < N; ++j) {
                if (isPrefixAndSuffix(A[i], A[j])) ++ ans;
            }
        }
        return ans;
    }
};