Prefix and Suffix Pair Count

Given an array of strings arr[], count the number of pairs (i, j) such that i < j and the string arr[i] is both a prefix and a suffix of the string arr[j].

Examples:

Input: arr[] = ["gfg", "geeks", "gfg", "geeksforgeeks"]
Output: 2
Explanation: Valid pairs: [["gfg", "gfg"], ["geeks", "geeksforgeeks"]]

Input: arr[] = ["ab", "abcab", "a", "aaab"]
Output: 1
Explanation: Valid pairs: ["ab", "abcab"]

[Approach]: Brute Force Prefix-Suffix Check

The idea is to iterate over all pairs of strings (i, j) such that i < j and check whether arr[i] matches both the starting and ending part of arr[j]. For each pair, we compare characters to verify prefix and suffix conditions, and if both match, we increment the count of valid pairs.

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

// Function to check if a is both prefix and suffix of b
bool isPrefixSuffix(string &a, string &b) {
    int n = a.size();
    int m = b.size();

    if (n > m) return false;

    string str1 = b.substr(0, n);           
    string str2 = b.substr(m - n, n);       

    return (str1 == a && str2 == a);
}

int countPairs(vector<string>& arr) {
    int n = arr.size();
    int count = 0;

    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (isPrefixSuffix(arr[i], arr[j])) {
                count++;
            }
        }
    }

    return count;
}

int main() {
    vector<string> arr = {"gfg", "geeks", "gfg", "geeksforgeeks"};
    
    cout << countPairs(arr);
    return 0;
}

import java.util.*;

class GFG {
    
    // Function to check if a is both prefix and suffix of b
    static boolean isPrefixSuffix(String a, String b) {
    int n = a.length();
    int m = b.length();

    if (n > m) return false;

    String str1 = b.substring(0, n);       
    String str2 = b.substring(m - n);     

    return str1.equals(a) && str2.equals(a);
}

    static int countPairs(String[] arr) {
        int n = arr.length;
        int count = 0;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (isPrefixSuffix(arr[i], arr[j])) {
                    count++;
                }
            }
        }

        return count;
    }

    public static void main(String[] args) {
        String[] arr = {"gfg", "geeks", "gfg", "geeksforgeeks"};
        System.out.println(countPairs(arr));
    }
}

# Function to check if a is both prefix and suffix of b
def isPrefixSuffix(a, b):
    n = len(a)
    m = len(b)

    if n > m:
        return False

    str1 = b[:n]       
    str2 = b[m - n:]   

    return str1 == a and str2 == a

def countPairs(arr):
    n = len(arr)
    count = 0

    for i in range(n):
        for j in range(i + 1, n):
            if isPrefixSuffix(arr[i], arr[j]):
                count += 1

    return count

if __name__ == "__main__":
    arr = ["gfg", "geeks", "gfg", "geeksforgeeks"]
    print(countPairs(arr))

using System;

class GFG {

    // Function to check if a is both prefix and suffix of b
   static bool isPrefixSuffix(string a, string b) {
    int n = a.Length;
    int m = b.Length;

    if (n > m) return false;

    string str1 = b.Substring(0, n);
    string str2 = b.Substring(m - n);

    return str1.Equals(a) && str2.Equals(a);
    }

    static int countPairs(string[] arr) {
        int n = arr.Length;
        int count = 0;

        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                if (isPrefixSuffix(arr[i], arr[j])) {
                    count++;
                }
            }
        }

        return count;
    }

    static void Main() {
        string[] arr = {"gfg", "geeks", "gfg", "geeksforgeeks"};
        Console.WriteLine(countPairs(arr));
    }
}

// Function to check if a is both prefix and suffix of b
function isPrefixSuffix(a, b) {
    let n = a.length;
    let m = b.length;

    if (n > m) return false;

    let str1 = b.slice(0, n);
    let str2 = b.slice(m - n);

    return str1 === a && str2 === a;
}

function countPairs(arr) {
    let n = arr.length;
    let count = 0;

    for (let i = 0; i < n; i++) {
        for (let j = i + 1; j < n; j++) {
            if (isPrefixSuffix(arr[i], arr[j])) {
                count++;
            }
        }
    }

    return count;
}

// Driver Code
let arr = ["gfg", "geeks", "gfg", "geeksforgeeks"];
console.log(countPairs(arr));

2

Time Complexity: O(n^2 * m) There are a total of (n(n − 1) / 2) pairs and for each pair we compare up to 'm' characters (length of the smaller string) to check both prefix and suffix.
Auxiliary Space: O(1)