Given two strings A and B, the task is to find the shortest string which is some multiple of both A and B. A string X is said to be a multiple of string Y if string X can be formed by the concatenation of multiple occurrences of string Y.
Example:
Input: A = “aaa”, B= “aa”
Output: aaaaaa
Explanation: Multiplying A two times and B three times will result in aaaaaaInput: A=”cold”, B =”old”
Output: -1
Approach: The given problem can be solved based on the observation that the length of the smallest string that can be a multiple of both the strings A and B must be equal to the LCM of the length of A and length of B. Therefore, the given problem can be solved using the following steps:
- Create a variable lcm, which stores the LCM value of the length of A and length of B.
- Create a string C which is formed after concatenating string A, lcm / (length of A) times.
- Similarly, create a string D which is formed after concatenating string B, lcm / (length of B) times.
- Check if strings C and D are equal or not. If they are, print any of them otherwise print -1.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to find GCD of two numbersint gcd(int a, int b){ if (a == 0) { return b; } return gcd(b % a, a);}// Function to find shortest string that// is a multiple of string A and Bvoid ShortestString(string A, string B){ int n1 = A.length(); int n2 = B.length(); int g = gcd(n1, n2); // Stores the Lcm of length // of string A and B int lcm = (n1 * n2) / g; // Stores multiple of string A string C = ""; // Stores multiple of string B string D = ""; // Concatenate A, lcm / n1 times for (int i = 0; i < (lcm / n1); i++) { C = C + A; } // Concatenate B, lcm / n2 times for (int i = 0; i < (lcm / n2); i++) { D = D + B; } // If both strings are equal // to each other if (C == D) { cout << C; } else { cout << -1; }}// Driver Codeint main(){ string A = "aaa"; string B = "aa"; ShortestString(A, B); return 0;} |
Java
// Java program for the above approachimport java.util.*;class GFG{// Function to find GCD of two numbersstatic int gcd(int a, int b){ if (a == 0) { return b; } return gcd(b % a, a);}// Function to find shortest String that// is a multiple of String A and Bstatic void ShortestString(String A, String B){ int n1 = A.length(); int n2 = B.length(); int g = gcd(n1, n2); // Stores the Lcm of length // of String A and B int lcm = (n1 * n2) / g; // Stores multiple of String A String C = ""; // Stores multiple of String B String D = ""; // Concatenate A, lcm / n1 times for (int i = 0; i < (lcm / n1); i++) { C = C + A; } // Concatenate B, lcm / n2 times for (int i = 0; i < (lcm / n2); i++) { D = D + B; } // If both Strings are equal // to each other if (C.equals(D)) { System.out.print(C); } else { System.out.print(-1); }}// Driver Codepublic static void main(String[] args){ String A = "aaa"; String B = "aa"; ShortestString(A, B);}}// This code is contributed by 29AjayKumar |
Python3
# Python code for the above approach# Function to find GCD of two numbersdef gcd(a, b): if (a == 0): return b return gcd(b % a, a)# Function to find shortest string that# is a multiple of string A and Bdef ShortestString(A, B): n1 = len(A) n2 = len(B) g = gcd(n1, n2) # Stores the Lcm of length # of string A and B lcm = (n1 * n2) / g # Stores multiple of string A C = "" # Stores multiple of string B D = "" # Concatenate A, lcm / n1 times for i in range(0, (int)(lcm//n1)): C = C + A # Concatenate B, lcm / n2 times for i in range((int)(lcm // n2)): D = D + B # If both strings are equal # to each other if (C == D): print(C) else: print(-1)# Driver CodeA = "aaa"B = "aa"ShortestString(A, B)# This code is contributed by Saurabh Jaiswal |
C#
// C# program for the above approachusing System;class GFG{// Function to find GCD of two numbersstatic int gcd(int a, int b){ if (a == 0) { return b; } return gcd(b % a, a);}// Function to find shortest String that// is a multiple of String A and Bstatic void ShortestString(string A, string B){ int n1 = A.Length; int n2 = B.Length; int g = gcd(n1, n2); // Stores the Lcm of length // of String A and B int lcm = (n1 * n2) / g; // Stores multiple of String A string C = ""; // Stores multiple of String B string D = ""; // Concatenate A, lcm / n1 times for (int i = 0; i < (lcm / n1); i++) { C = C + A; } // Concatenate B, lcm / n2 times for (int i = 0; i < (lcm / n2); i++) { D = D + B; } // If both Strings are equal // to each other if (C.Equals(D)) { Console.Write(C); } else { Console.Write(-1); }}// Driver Codepublic static void Main(){ string A = "aaa"; string B = "aa"; ShortestString(A, B);}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript code for the above approach // Function to find GCD of two numbers function gcd(a, b) { if (a == 0) { return b; } return gcd(b % a, a); } // Function to find shortest string that // is a multiple of string A and B function ShortestString(A, B) { let n1 = A.length; let n2 = B.length; let g = gcd(n1, n2); // Stores the Lcm of length // of string A and B let lcm = (n1 * n2) / g; // Stores multiple of string A let C = ""; // Stores multiple of string B let D = ""; // Concatenate A, lcm / n1 times for (let i = 0; i < (lcm / n1); i++) { C = C + A; } // Concatenate B, lcm / n2 times for (let i = 0; i < (lcm / n2); i++) { D = D + B; } // If both strings are equal // to each other if (C == D) { document.write(C); } else { document.write(-1); } } // Driver Code let A = "aaa"; let B = "aa"; ShortestString(A, B);// This code is contributed by Potta Lokesh </script> |
Output
aaaaaa
Time Complexity: O(N)
Auxiliary Space: O(N)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
