Count the number of common divisors of the given strings

Given two strings a and b, the task is to count the number of common divisors of both the strings. A string s is a divisor of string t if t can be generated by repeating s a number of times.

Examples:

Input: a = “xaxa”, b = “xaxaxaxa” 
Output: 2 
The common divisors are “xa” and “xaxa”

Input: a = “bbbb”, b = “bbb” 
Output:1 
The only common divisor is “b” 

Approach: For a string s to be a candidate divisor of string t, the following conditions must be fulfilled: 

1. s must be a prefix of t.
2. len(t) % len(s) = 0

Initialize count = 0 and starting from the first character as the ending character of the prefix, check whether the length of the prefix divides the length of both the strings and also if the prefix is same in both the strings. If yes then update count = count + 1. Repeat these steps for all the possible prefixes. Print the value of count in the end.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function that returns true if sub-string
// s[0...k] is repeated a number of times
// to generate string s
int check(string s, int k)
{
    for (int i = 0; i < s.length(); i++)
        if (s[i] != s[i % k])
            return false;

    return true;
}

// Function to return the count of common divisors
int countCommonDivisors(string a, string b)
{
    int ct = 0;
    int n = a.size(), m = b.size();
    for (int i = 1; i <= min(n, m); i++) {

        // If the length of the sub-string
        // divides length of both the strings
        if (n % i == 0 && m % i == 0)

            // If prefixes match in both the strings
            if (a.substr(0, i) == b.substr(0, i))

                // If both the strings can be generated
                // by repeating the current prefix
                if (check(a, i) && check(b, i))
                    ct++;
    }
    return ct;
}

// Driver code
int main()
{
    string a = "xaxa", b = "xaxaxaxa";

    cout << countCommonDivisors(a, b);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

    // Function that returns true if sub-string
    // s[0...k] is repeated a number of times
    // to generate String s
    static boolean check(String s, int k)
    {
        for (int i = 0; i < s.length(); i++)
        {
            if (s.charAt(i) != s.charAt(i % k))
            {
                return false;
            }
        }
        return true;
    }

    // Function to return the
    // count of common divisors
    static int countCommonDivisors(String a, String b)
    {
        int ct = 0;
        int n = a.length(), m = b.length();
        for (int i = 1; i <= Math.min(n, m); i++)
        {

            // If the length of the sub-string
            // divides length of both the strings
            if (n % i == 0 && m % i == 0)
            {
                // If prefixes match in both the strings
                if (a.substring(0, i).equals(b.substring(0, i)))
                // by repeating the current prefix
                {
                    // If both the strings can be generated
                    if (check(a, i) && check(b, i))
                    {
                        ct++;
                    }
                }
            }
        }
        return ct;
    }

    // Driver code
    public static void main(String[] args)
    {
        String a = "xaxa", b = "xaxaxaxa";
        System.out.println(countCommonDivisors(a, b));
    }
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the above approach

# Function that returns true if sub-string
# s[0...k] is repeated a number of times
# to generate String s
def check(s, k):

    for i in range (0, len(s)):

        if (s[i] != s[i % k]):

            return False

    return True

# Function to return the
# count of common divisors
def countCommonDivisors(a, b):

    ct = 0
    n = len(a)
    m = len(b)
    for i in range(1, min(n, m) + 1):

        # If the length of the sub-string
        # divides length of both the strings
        if (n % i == 0 and m % i == 0):

            # If prefixes match in both the strings
            if (a[0 : i] == b[0 : i]) :

                # by repeating the current prefix

                # If both the strings can be generated
                if (check(a, i) and check(b, i)) :

                    ct = ct + 1

    return ct

# Driver code
a = "xaxa"
b = "xaxaxaxa"
print(countCommonDivisors(a, b))

# This code is contributed by ihritik

C#

// C# implementation of the above approach
using System;

class GFG
{

// Function that returns true if sub-string
// s[0...k] is repeated a number of times
// to generate String s
static bool check(string s, int k)
{
    for (int i = 0; i < s.Length; i++)
    {
        if (s[i] != s[i % k])
        {
            return false;
        }
    }
    return true;
}

// Function to return the count of
// common divisors
static int countCommonDivisors(string a,
                               string b)
{
    int ct = 0;
    int n = a.Length, m = b.Length;
    for (int i = 1;
             i <= Math.Min(n, m); i++)
    {

        // If the length of the sub-string
        // divides length of both the strings
        if (n % i == 0 && m % i == 0)
        {
            // If prefixes match in both the strings
            if (a.Substring(0, i) == (b.Substring(0, i)))

            // by repeating the current prefix
            {
                // If both the strings can be generated
                if (check(a, i) && check(b, i))
                {
                    ct++;
                }
            }
        }
    }
    return ct;
}

// Driver code
public static void Main()
{
    string a = "xaxa", b = "xaxaxaxa";
    Console.WriteLine(countCommonDivisors(a, b));
}
}

// This code is contributed by ihritik

Javascript

<script>

// Javascript implementation of the approach

// Function that returns true if sub-string
// s[0...k] is repeated a number of times
// to generate string s
function check(s, k)
{
    for (var i = 0; i < s.length; i++)
        if (s[i] != s[i % k])
            return false;

    return true;
}

// Function to return the count of common divisors
function countCommonDivisors(a, b)
{
    var ct = 0;
    var n = a.length, m = b.length;
    for (var i = 1; i <= Math.min(n, m); i++) {

        // If the length of the sub-string
        // divides length of both the strings
        if (n % i == 0 && m % i == 0)

            // If prefixes match in both the strings
            if (a.substring(0, i) == b.substring(0, i))

                // If both the strings can be generated
                // by repeating the current prefix
                if (check(a, i) && check(b, i))
                    ct++;
    }
    return ct;
}

// Driver code
var a = "xaxa", b = "xaxaxaxa";
document.write( countCommonDivisors(a, b));

// This code is contributed by famously.
</script>

Output

2

Complexity Analysis:

• Time Complexity: O(N * M)
• Auxiliary Space: O(1)