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 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.

Approach 01:

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

class Solution {
public:
    // Helper function to check if str1 is both a prefix and suffix of str2
    int isPrefixAndSuffix(string& prefixSuffixCandidate, string& targetString) {
        long long candidateLength = prefixSuffixCandidate.length();
        long long targetLength = targetString.length();

        // Extract the prefix and suffix from the target string
        string prefix = targetString.substr(0, candidateLength);
        string suffix = targetString.substr(targetLength - candidateLength, candidateLength);

        // Check if both the prefix and suffix match the candidate string
        if (prefix == prefixSuffixCandidate && suffix == prefixSuffixCandidate) {
            return 1;
        }
        return 0;
    }

    // Count all valid prefix-suffix pairs
    int countPrefixSuffixPairs(vector<string>& words) {
        long long totalWords = words.size();
        long long validPairCount = 0;

        // Iterate over all pairs of strings
        for (int i = 0; i < totalWords - 1; ++i) {
            for (int j = i + 1; j < totalWords; ++j) {
                // Check if the first word can be a prefix and suffix of the second word
                if (words[i].length() <= words[j].length()) {
                    validPairCount += isPrefixAndSuffix(words[i], words[j]);
                }
            }
        }

        return validPairCount;
    }
};

class Solution:
    # Helper function to check if str1 is both a prefix and suffix of str2
    def isPrefixAndSuffix(self, prefixSuffixCandidate, targetString):
        candidateLength = len(prefixSuffixCandidate)
        targetLength = len(targetString)

        # Extract the prefix and suffix from the target string
        prefix = targetString[:candidateLength]
        suffix = targetString[targetLength - candidateLength:]

        # Check if both the prefix and suffix match the candidate string
        if prefix == prefixSuffixCandidate and suffix == prefixSuffixCandidate:
            return 1
        return 0

    # Count all valid prefix-suffix pairs
    def countPrefixSuffixPairs(self, words):
        totalWords = len(words)
        validPairCount = 0

        # Iterate over all pairs of strings
        for i in range(totalWords - 1):
            for j in range(i + 1, totalWords):
                # Check if the first word can be a prefix and suffix of the second word
                if len(words[i]) <= len(words[j]):
                    validPairCount += self.isPrefixAndSuffix(words[i], words[j])

        return validPairCount

Time Complexity:

• Outer Loop:

  The outer loop iterates over all pairs of words, leading to \( O(n^2) \), where \( n \) is the number of words in words.

• Substring Operations:

  For each pair, the substr function is called twice, which takes \( O(m) \), where \( m \) is the length of the candidate string.

• String Comparison:

  Each comparison of prefix and suffix strings also takes \( O(m) \).

• Overall Time Complexity:

  \( O(n^2 \cdot m) \), where \( n \) is the number of words, and \( m \) is the average length of the strings in words.

Space Complexity:

• Auxiliary Space for Strings:

  The substr function creates new strings for prefix and suffix comparisons, requiring \( O(m) \) additional space for each call.

• Overall Space Complexity:

  \( O(m) \), where \( m \) is the length of the longest string being processed.