Count Prefix and Suffix Pairs (LeetCode)

Problem Description

Problem Statement	Solution Link	LeetCode Profile
Count Prefix and Suffix Pairs	Count Prefix and Suffix Pairs Solution on LeetCode	vaishu_1904

Problem 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.

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.

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.

Constraints

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

Approach

To solve this problem, we need to check for each pair (i, j) if words[i] is both a prefix and a suffix of words[j]. We can do this by iterating over all pairs and checking the conditions for each pair.

Solution Code

Python

class Solution:
    def isPrefixAndSuffix(self, str1, str2):
        return str2.startswith(str1) and str2.endswith(str1)
    
    def countPrefixAndSuffixPairs(self, words):
        count = 0
        for i in range(len(words)):
            for j in range(i + 1, len(words)):
                if self.isPrefixAndSuffix(words[i], words[j]):
                    count += 1
        return count

C++

#include <vector>
#include <string>
using namespace std;

class Solution {
public:
    bool isPrefixAndSuffix(const string& str1, const string& str2) {
        return str2.compare(0, str1.size(), str1) == 0 && str2.compare(str2.size() - str1.size(), str1.size(), str1) == 0;
    }
    
    int countPrefixAndSuffixPairs(vector<string>& words) {
        int count = 0;
        for (int i = 0; i < words.size(); i++) {
            for (int j = i + 1; j < words.size(); j++) {
                if (isPrefixAndSuffix(words[i], words[j])) {
                    count++;
                }
            }
        }
        return count;
    }
};

Java

class Solution {
    private boolean isPrefixAndSuffix(String str1, String str2) {
        return str2.startsWith(str1) && str2.endsWith(str1);
    }
    
    public int countPrefixAndSuffixPairs(String[] words) {
        int count = 0;
        for (int i = 0; i < words.length; i++) {
            for (int j = i + 1; j < words.length; j++) {
                if (isPrefixAndSuffix(words[i], words[j])) {
                    count++;
                }
            }
        }
        return count;
    }
}

Conclusion

The above solutions iterate over all pairs of strings in the input list and check if the first string in the pair is both a prefix and a suffix of the second string. The count of such valid pairs is returned as the result.