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# Smallest string divisible by two given strings

• Difficulty Level : Medium
• Last Updated : 30 Jun, 2021

Given two strings S and T of length N and M respectively, the task is to find the smallest string that is divisible by both the two strings. If no such string exists, then print -1.

For any two strings A and B, B divides A if and only if A is the concatenation of B at least once.

Examples:

Input: S = “abab”, T = “ab”
Output: abab
Explanation: The string “abab” is the same as S and twice the concatenation of string T (“abab” = “ab” + “ab” = T + T)

Input: S = “ccc”, T = “cc”
Output: cccccc
Explanation: The string “cccccc” is a concatenation of S and T twice and thrice respectively.
(“cccccc” = “ccc” + “ccc” = S + S)
(“cccccc” = “cc” + “cc” + “cc” = T + T + T)

Approach: The idea is based on the observation that the length of the required string, say, L, must be equal to the least common multiple of N and M. Check if string S concatenated L / N number of times is equal to string T being concatenated L / M number of times or not. If found to be true, print any one of them. Otherwise, print -1. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to calculate``// GCD of two numbers``int` `gcd(``int` `a, ``int` `b)``{``    ``if` `(b == 0)``        ``return` `a;``    ``return` `gcd(b, a % b);``}` `// Function to calculate``// LCM of two numbers``int` `lcm(``int` `a, ``int` `b)``{``    ``return` `(a / gcd(a, b)) * b;``}` `// Function to find the smallest string``// which is divisible by strings S and T``void` `findSmallestString(string s, string t)``{``    ``// Store the length of both strings``    ``int` `n = s.length(), m = t.length();` `    ``// Store LCM of n and m``    ``int` `l = lcm(n, m);` `    ``// Temporary strings to store``    ``// concatenated strings``    ``string s1 = ``""``, t1 = ``""``;` `    ``// Concatenate s1 (l / n) times``    ``for` `(``int` `i = 0; i < l / n; i++) {``        ``s1 += s;``    ``}` `    ``// Concatenate t1 (l / m) times``    ``for` `(``int` `i = 0; i < l / m; i++) {``        ``t1 += t;``    ``}` `    ``// If s1 and t1 are equal``    ``if` `(s1 == t1)``        ``cout << s1;` `    ``// Otherwise, print -1``    ``else``        ``cout << -1;``}` `// Driver Code``int` `main()``{``    ``string S = ``"baba"``, T = ``"ba"``;``    ``findSmallestString(S, T);` `    ``return` `0;``}`

## Java

 `// Java program for above approach``import` `java.io.*;` `class` `GFG``{` `  ``// Function to calculate``  ``// GCD of two numbers``  ``static` `int` `gcd(``int` `a, ``int` `b)``  ``{``    ``if` `(b == ``0``)``      ``return` `a;``    ``return` `gcd(b, a % b);``  ``}` `  ``// Function to calculate``  ``// LCM of two numbers``  ``static` `int` `lcm(``int` `a, ``int` `b)``  ``{``    ``return` `(a / gcd(a, b)) * b;``  ``}` `  ``// Function to find the smallest string``  ``// which is divisible by strings S and T``  ``static` `void` `findSmallestString(String s, String t)``  ``{``    ``// Store the length of both strings``    ``int` `n = s.length(), m = t.length();` `    ``// Store LCM of n and m``    ``int` `l = lcm(n, m);` `    ``// Temporary strings to store``    ``// concatenated strings``    ``String s1 = ``""``, t1 = ``""``;` `    ``// Concatenate s1 (l / n) times``    ``for` `(``int` `i = ``0``; i < l / n; i++) {``      ``s1 += s;``    ``}` `    ``// Concatenate t1 (l / m) times``    ``for` `(``int` `i = ``0``; i < l / m; i++) {``      ``t1 += t;``    ``}` `    ``// If s1 and t1 are equal``    ``if` `(s1.equals(t1)){``      ``System.out.println(s1);``    ``}` `    ``// Otherwise, print -1``    ``else``{``      ``System.out.println(-``1``);``    ``}``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String[] args)``  ``{` `    ``String S = ``"baba"``, T = ``"ba"``;``    ``findSmallestString(S, T);``  ``}``}` `// This code is contributed by susmitakundugoaldanga.`

## Python3

 `# Python3 program for the above approach` `# Function to calculate``# GCD of two numbers``def` `gcd(a, b):``    ``if` `(b ``=``=` `0``):``        ``return` `a``    ``return` `gcd(b, a ``%` `b)` `# Function to calculate``# LCM of two numbers``def` `lcm(a, b):``    ``return` `(a ``/``/` `gcd(a, b)) ``*` `b` `# Function to find the smallest string``# which is divisible by strings S and T``def` `findSmallestString(s, t):``    ``# Store the length of both strings``    ``n, m ``=` `len``(s), ``len``(t)` `    ``# Store LCM of n and m``    ``l ``=` `lcm(n, m)` `    ``# Temporary strings to store``    ``# concatenated strings``    ``s1, t1 ``=` `"``", "``"` `    ``# Concatenate s1 (l / n) times``    ``for` `i ``in` `range``(l``/``/``n):``        ``s1 ``+``=` `s` `    ``# Concatenate t1 (l / m) times``    ``for` `i ``in` `range``(l``/``/``m):``        ``t1 ``+``=` `t` `    ``# If s1 and t1 are equal``    ``if` `(s1 ``=``=` `t1):``        ``print``(s1)` `    ``# Otherwise, pr-1``    ``else``:``        ``print``(``-``1``)` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``S, T ``=` `"baba"``, ``"ba"``    ``findSmallestString(S, T)` `# This code is contributed by mohit kumar 29.`

## C#

 `// C# program for above approach``using` `System;` `public` `class` `GFG``{` `  ``// Function to calculate``  ``// GCD of two numbers``  ``static` `int` `gcd(``int` `a, ``int` `b)``  ``{``    ``if` `(b == 0)``      ``return` `a;``    ``return` `gcd(b, a % b);``  ``}` `  ``// Function to calculate``  ``// LCM of two numbers``  ``static` `int` `lcm(``int` `a, ``int` `b)``  ``{``    ``return` `(a / gcd(a, b)) * b;``  ``}` `  ``// Function to find the smallest string``  ``// which is divisible by strings S and T``  ``static` `void` `findSmallestString(``string` `s, ``string` `t)``  ``{``    ``// Store the length of both strings``    ``int` `n = s.Length, m = t.Length;` `    ``// Store LCM of n and m``    ``int` `l = lcm(n, m);` `    ``// Temporary strings to store``    ``// concatenated strings``    ``string` `s1 = ``""``, t1 = ``""``;` `    ``// Concatenate s1 (l / n) times``    ``for` `(``int` `i = 0; i < l / n; i++) {``      ``s1 += s;``    ``}` `    ``// Concatenate t1 (l / m) times``    ``for` `(``int` `i = 0; i < l / m; i++) {``      ``t1 += t;``    ``}` `    ``// If s1 and t1 are equal``    ``if` `(s1 == t1)``      ``Console.WriteLine(s1);` `    ``// Otherwise, print -1``    ``else``      ``Console.WriteLine(-1);``  ``}` `  ``// Driver code``  ``public` `static` `void` `Main(String[] args)``  ``{``    ``string` `S = ``"baba"``, T = ``"ba"``;``    ``findSmallestString(S, T);``  ``}``}` `// This code is contributed by sanjoy_62.`

## Javascript

 ``
Output:
`baba`

Time Complexity: O(max(N, M))
Auxiliary Space: O(max(N, M))

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