Check if any permutation of string is a K times repeated string

Last Updated : 20 Aug, 2020

Given a string S and an integer K, the task is to check that if any permutation of the string can be formed by K times repeating any other string.

Examples:

Input: S = “abba”, K = 2
Output: Yes
Explanation:
Permutations of given string –
{“aabb”, “abab”, “abba”, “baab”, “baba”, “bbaa”}
As “abab” is repeating string of “ab”+”ab” = “abab”, which is also permutation of string.

Input: S = “abcabd”, K = 2
Output: No
Explanation:
There is no such repeating string in all permutations of the given string.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to find the frequency of each character of the string and check that the frequency of the character is a multiple of the given integer K. If the frequency of all characters of the string is divisible by K, then there is a string which is a permutation of the given string and also a K times repeated string.

Below is the implementation of the above approach:

C++

// C++ implementation to check that 
// the permutation of the given string 
// is K times repeated string 
  
#include <bits/stdc++.h> 
  
using namespace std; 
  
// Function to check that permutation 
// of the given string is a  
// K times repeating String 
bool repeatingString(string s,  
                 int n, int k) 
{ 
    // if length of string is  
    // not divisible by K 
    if (n % k != 0) { 
        return false; 
    } 
      
    // Frequency Array 
    int frequency[123]; 
      
    // Intially frequency of each 
    // character is 0 
    for (int i = 0; i < 123; i++) { 
        frequency[i] = 0; 
    } 
      
    // Computing the frequency of  
    // each character in the string 
    for (int i = 0; i < n; i++) { 
        frequency[s[i]]++; 
    } 
  
    int repeat = n / k; 
      
    // Loop to check that frequency of  
    // every character of the string 
    // is divisible by K 
    for (int i = 0; i < 123; i++) { 
  
        if (frequency[i] % repeat != 0) { 
            return false; 
        } 
    } 
  
    return true; 
} 
  
// Driver Code 
int main() 
{ 
    string s = "abcdcba"; 
    int n = s.size(); 
    int k = 3; 
  
    if (repeatingString(s, n, k)) { 
        cout << "Yes" << endl; 
    } 
    else { 
        cout << "No" << endl; 
    } 
  
    return 0; 
} 

Java

// Java implementation to check that  
// the permutation of the given String  
// is K times repeated String  
class GFG{  
  
// Function to check that permutation  
// of the given String is a  
// K times repeating String  
static boolean repeatingString(String s,  
                int n, int k)  
{  
    // if length of String is  
    // not divisible by K  
    if (n % k != 0) {  
        return false;  
    }  
      
    // Frequency Array  
    int []frequency = new int[123];  
      
    // Intially frequency of each  
    // character is 0  
    for (int i = 0; i < 123; i++) {  
        frequency[i] = 0;  
    }  
      
    // Computing the frequency of  
    // each character in the String  
    for (int i = 0; i < n; i++) {  
        frequency[s.charAt(i)]++;  
    }  
  
    int repeat = n / k;  
      
    // Loop to check that frequency of  
    // every character of the String  
    // is divisible by K  
    for (int i = 0; i < 123; i++) {  
  
        if (frequency[i] % repeat != 0) {  
            return false;  
        }  
    }  
  
    return true;  
}  
  
// Driver Code  
public static void main(String[] args)  
{  
    String s = "abcdcba";  
    int n = s.length();  
    int k = 3;  
  
    if (repeatingString(s, n, k)) {  
        System.out.print("Yes" +"\n");  
    }  
    else {  
        System.out.print("No" +"\n");  
    }  
}  
}  
  
// This code is contributed by PrinciRaj1992 

Python3

# Python3 implementation to check that 
# the permutation of the given string 
# is K times repeated string 
  
# Function to check that permutation 
# of the given string is a  
# K times repeating String 
def repeatingString(s, n, k): 
      
    # If length of string is  
    # not divisible by K 
    if (n % k != 0): 
        return False
  
    # Frequency Array 
    frequency = [0 for i in range(123)] 
  
    # Intially frequency of each 
    # character is 0 
    for i in range(123): 
        frequency[i] = 0
      
    # Computing the frequency of  
    # each character in the string 
    for i in range(n): 
        frequency[s[i]] += 1
  
    repeat = n // k 
      
    # Loop to check that frequency of  
    # every character of the string 
    # is divisible by K 
    for i in range(123): 
        if (frequency[i] % repeat != 0): 
            return False
  
    return True
  
# Driver Code 
if __name__ == '__main__': 
      
    s = "abcdcba"
    n = len(s) 
    k = 3
  
    if (repeatingString(s, n, k)): 
        print("Yes") 
    else: 
        print("No") 
          
# This code is contributed by Samarth 

C#

// C# implementation to check that  
// the permutation of the given String  
// is K times repeated String  
using System; 
  
class GFG{  
   
// Function to check that permutation  
// of the given String is a  
// K times repeating String  
static bool repeatingString(String s,  
                int n, int k)  
{  
    // if length of String is  
    // not divisible by K  
    if (n % k != 0) {  
        return false;  
    }  
       
    // Frequency Array  
    int []frequency = new int[123];  
       
    // Intially frequency of each  
    // character is 0  
    for (int i = 0; i < 123; i++) {  
        frequency[i] = 0;  
    }  
       
    // Computing the frequency of  
    // each character in the String  
    for (int i = 0; i < n; i++) {  
        frequency[s[i]]++;  
    }  
   
    int repeat = n / k;  
       
    // Loop to check that frequency of  
    // every character of the String  
    // is divisible by K  
    for (int i = 0; i < 123; i++) {  
   
        if (frequency[i] % repeat != 0) {  
            return false;  
        }  
    }  
   
    return true;  
}  
   
// Driver Code  
public static void Main(String[] args)  
{  
    String s = "abcdcba";  
    int n = s.Length;  
    int k = 3;  
   
    if (repeatingString(s, n, k)) {  
        Console.Write("Yes" +"\n");  
    }  
    else {  
        Console.Write("No" +"\n");  
    }  
}  
}  
   
// This code is contributed by Rajput-Ji 

Output:

No

Performance Analysis:

Time Complexity O(N)

Auxiliary Space: O(123)

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#