Check if a string is concatenation of another given string
Last Updated :
19 Oct, 2023
Given two strings str1 and str2 of length N and M respectively, the task is to check if the string str1 can be formed by concatenating the string str2 repetitively or not.
Examples:
Input: str1 = “abcabcabc”, str2 = “abc”
Output: Yes
Explanation:
Concatenating the string str2 thrice generates the string (“abc” + “abc” + “abc” = ) “abcabcabc”.
Therefore, the required output is Yes.
Input: str1 = “abcabcab”, str2 = “abc”
Output: No
Approach: Follow the steps below to solve the problem:
- Traverse the strings str1 and str2.
- For each character of str1 and str2, check if str1[i] == str2[i % M] or not.
- If found to be false for any character, print "No".
- Otherwise, print "Yes".
Below is the implementation of the above approach:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if a string is
// concatenation of another string
bool checkConcat(string str1,
string str2)
{
// Stores the length of str2
int N = str1.length();
// Stores the length of str1
int M = str2.length();
// If M is not multiple of N
if (N % M != 0) {
return false;
}
// Traverse both the strings
for (int i = 0; i < N; i++) {
// If str1 is not concatenation
// of str2
if (str1[i] != str2[i % M]) {
return false;
}
}
return true;
}
// Driver Code
int main()
{
string str1 = "abcabcabc";
string str2 = "abc";
if (checkConcat(str1, str2)) {
cout << "Yes";
}
else {
cout << "No";
}
}
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to check if a String is
// concatenation of another String
static boolean checkConcat(String str1,
String str2)
{
// Stores the length of str2
int N = str1.length();
// Stores the length of str1
int M = str2.length();
// If M is not multiple of N
if (N % M != 0)
{
return false;
}
// Traverse both the Strings
for(int i = 0; i < N; i++)
{
// If str1 is not concatenation
// of str2
if (str1.charAt(i) !=
str2.charAt(i % M))
{
return false;
}
}
return true;
}
// Driver Code
public static void main(String[] args)
{
String str1 = "abcabcabc";
String str2 = "abc";
if (checkConcat(str1, str2))
{
System.out.print("Yes");
}
else
{
System.out.print("No");
}
}
}
// This code is contributed by Amit Katiyar
# Python3 program to implement
# the above approach
# Function to check if a is
# concatenation of another string
def checkConcat(str1, str2):
# Stores the length of str2
N = len(str1)
# Stores the length of str1
M = len(str2)
# If M is not multiple of N
if (N % M != 0):
return False
# Traverse both the strings
for i in range(N):
# If str1 is not concatenation
# of str2
if (str1[i] != str2[i % M]):
return False
return True
# Driver Code
if __name__ == '__main__':
str1 = "abcabcabc"
str2 = "abc"
if (checkConcat(str1, str2)):
print("Yes")
else:
print("No")
# This code is contributed by mohit kumar 29
// C# program to implement
// the above approach
using System;
class GFG{
// Function to check if a String is
// concatenation of another String
static bool checkConcat(String str1,
String str2)
{
// Stores the length
// of str2
int N = str1.Length;
// Stores the length
// of str1
int M = str2.Length;
// If M is not multiple
// of N
if (N % M != 0)
{
return false;
}
// Traverse both the Strings
for(int i = 0; i < N; i++)
{
// If str1 is not
// concatenation of str2
if (str1[i] !=
str2[i % M])
{
return false;
}
}
return true;
}
// Driver Code
public static void Main(String[] args)
{
String str1 = "abcabcabc";
String str2 = "abc";
if (checkConcat(str1, str2))
{
Console.Write("Yes");
}
else
{
Console.Write("No");
}
}
}
// This code is contributed by 29AjayKumar
<script>
// javascript program for the
// above approach
// Function to check if a String is
// concatenation of another String
function checkConcat(str1, str2)
{
// Stores the length of str2
let N = str1.length;
// Stores the length of str1
let M = str2.length;
// If M is not multiple of N
if (N % M != 0)
{
return false;
}
// Traverse both the Strings
for(let i = 0; i < N; i++)
{
// If str1 is not concatenation
// of str2
if (str1[i] !=
str2[i % M])
{
return false;
}
}
return true;
}
// Driver Code
let str1 = "abcabcabc";
let str2 = "abc";
if (checkConcat(str1, str2))
{
document.write("Yes");
}
else
{
document.write("No");
}
</script>
Time Complexity: O(N)
Auxiliary Space: O(1)
Using a loop to check all possible concatenations in python:
Approach:
- Define a function named is_concatenation that takes two string arguments str1 and str2.
- Get the length of str2 and store it in the variable n.
- Use a for loop to iterate through the range from 0 to n-1.
- Inside the loop, create a string named concat which is equal to str2 multiplied by (n-i) times.
- Check if str1 is equal to concat concatenated with the first i characters of str2.
- If the condition is True, return True. Otherwise, continue the loop.
- If the loop completes without returning True, return False.
C++
#include <iostream>
#include <string>
bool isConcatenation(std::string str1, std::string str2)
{
int n = str2.length();
for (int i = 0; i < n; i++) {
// Create a string by repeating str2 to form a
// potential concatenation
std::string concat = str2;
for (int j = 1; j < (n - i); j++) {
concat += str2;
}
// Check if str1 is equal to the concatenation of
// concat and the beginning of str2
if (str1 == concat + str2.substr(0, i)) {
return true;
}
}
return false;
}
int main()
{
std::string str1 = "abcabcabc";
std::string str2 = "abc";
std::cout << std::boolalpha
<< isConcatenation(str1, str2)
<< std::endl; // Output: true
str1 = "abcabcab";
str2 = "abc";
std::cout << std::boolalpha
<< isConcatenation(str1, str2)
<< std::endl; // Output: false
return 0;
}
public class ConcatenationCheck {
public static boolean isConcatenation(String str1, String str2) {
int n = str2.length();
for (int i = 0; i < n; i++) {
String concat = str2.repeat(n - i);
if (str1.equals(concat + str2.substring(0, i))) {
return true;
}
}
return false;
}
public static void main(String[] args) {
String str1 = "abcabcabc";
String str2 = "abc";
System.out.println(isConcatenation(str1, str2)); // Output: true
str1 = "abcabcab";
str2 = "abc";
System.out.println(isConcatenation(str1, str2)); // Output: false
}
}
def is_concatenation(str1, str2):
n = len(str2)
for i in range(n):
concat = str2 * (n - i)
if str1 == concat + str2[:i]:
return True
return False
# Testing the function
str1 = "abcabcabc"
str2 = "abc"
print(is_concatenation(str1, str2)) # Output: True
str1 = "abcabcab"
str2 = "abc"
print(is_concatenation(str1, str2)) # Output: False
using System;
class Program {
// Function to check if str1 is a concatenation of str2
static bool IsConcatenation(string str1, string str2)
{
int n = str2.Length;
for (int i = 0; i < n; i++) {
// Create a string by repeating str2 to form a
// potential concatenation
string concat = str2;
for (int j = 1; j < n - i; j++) {
concat += str2;
}
// Check if str1 is equal to the concatenation
// of concat and the beginning of str2
if (str1 == concat + str2.Substring(0, i)) {
return true;
}
}
return false;
}
static void Main()
{
string str1 = "abcabcabc";
string str2 = "abc";
Console.WriteLine(
IsConcatenation(str1, str2)); // Output: True
str1 = "abcabcab";
str2 = "abc";
Console.WriteLine(
IsConcatenation(str1, str2)); // Output: False
}
}
function is_concatenation(str1, str2) {
const n = str2.length;
for (let i = 0; i < n; i++) {
const concat = str2.repeat(n - i);
if (str1 === concat + str2.slice(0, i)) {
return true;
}
}
return false;
}
// Testing the function
const str1 = "abcabcabc";
const str2 = "abc";
console.log(is_concatenation(str1, str2)); // Output: true
const str1_2 = "abcabcab";
const str2_2 = "abc";
console.log(is_concatenation(str1_2, str2_2)); // Output: false
Time Complexity: O(n^2)
Space Complexity: O(1)
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