Find the least common multiple (LCM) of given Strings

Given two strings str1 and str2 of lengths N and M respectively. The task is to find the least common multiple (LCM) of both the strings and if it doesn't exist print -1.

Note: LCM of two strings is a string that can be formed by concatenating the strings to themselves and is shortest in length.

Examples:

Input: str1 = "baba", str2 = "ba"
Output: baba
Explanation: "baba" is the smallest string that can be obtained from both str1 and str2.
Notice that "babababa" can be also be formed but the length of the string is not least.
 

Input: str1 = "aba", str2 = "ab"
Output: -1
Explanation: No such string can be formed.

Approach: This problem is similar to finding shortest string formed by concatenating A x times and B y times. The task is solved by making lengths of both the strings equal and then checking whether both strings become the same or not. Follow the below steps to solve the problem:

• Iterate until the lengths of both strings are unequal.
• If the length of str1 is less than str2, append str1 to itself in else do the same for str2.
• After the loop, if both are equal, print any string, else print -1.

Below is the implementation of the above approach:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the lcm of both strings
string findLcm(string str1, string str2)
{
    string x = str1, y = str2;

    // While their length is unequal
    while (x.length() != y.length()) {
        if (x.length() < y.length())
            x += str1;
        else
            y += str2;
    }

    if (x == y)
        return x;
    return "";
}

// Driver Code
int main()
{
    string str1 = "ba", str2 = "baba";

    string ans = findLcm(str1, str2);
    if (ans != "")
        cout << ans;
    else {
        cout << -1;
    }
    return 0;
}

// Java program for the above approach
import java.util.*;
class GFG{

  // Function to find the lcm of both Strings
  static String findLcm(String str1, String str2)
  {
    String x = str1, y = str2;

    // While their length is unequal
    while (x.length() != y.length()) {
      if (x.length() < y.length())
        x += str1;
      else
        y += str2;
    }

    if (x.equals(y))
      return x;
    return "";
  }

  // Driver Code
  public static void main(String[] args)
  {
    String str1 = "ba", str2 = "baba";

    String ans = findLcm(str1, str2);
    if (ans != "")
      System.out.print(ans);
    else {
      System.out.print(-1);
    }
  }
}

// This code is contributed by shikhasingrajput 

# Python code for the above approach 

# Function to find the lcm of both strings
def findLcm(str1, str2):
    x = str1
    y = str2

    # While their length is unequal
    while (len(x) != len(y)):
        if (len(x) < len(y)):
            x += str1;
        else:
            y += str2;

    if (x == y):
        return x;
    return "";

# Driver Code
str1 = "ba"
str2 = "baba"

ans = findLcm(str1, str2)
if (ans != ""):
    print(ans);
else:
    print(-1);

# This code is contributed by gfgking

// C# program for the above approach
using System;
class GFG
{
  // Function to find the lcm of both strings
  static string findLcm(string str1, string str2)
  {
    string x = str1, y = str2;

    // While their length is unequal
    while (x.Length != y.Length) {
      if (x.Length < y.Length)
        x += str1;
      else
        y += str2;
    }

    if (x == y)
      return x;
    return "";
  }

  // Driver Code
  public static void Main()
  {
    string str1 = "ba", str2 = "baba";

    string ans = findLcm(str1, str2);
    if (ans != "")
      Console.Write(ans);
    else {
      Console.Write(-1);
    }
  }
}

// This code is contributed by Samim Hossain Mondal.

  <script>
        // JavaScript code for the above approach 

        // Function to find the lcm of both strings
        function findLcm(str1, str2) {
            let x = str1, y = str2;

            // While their length is unequal
            while (x.length != y.length) {
                if (x.length < y.length)
                    x += str1;
                else
                    y += str2;
            }

            if (x == y)
                return x;
            return "";
        }

        // Driver Code

        let str1 = "ba", str2 = "baba";

        let ans = findLcm(str1, str2);
        if (ans != "")
            document.write(ans);
        else {
            document.write(-1);
        }

       // This code is contributed by Potta Lokesh
    </script>

baba

Time Complexity: O(N + M)
Auxiliary Space: O(N + M)