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

  • Last Updated : 31 May, 2021

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. 
 

 

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];
     
    // Initially 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];
     
    // Initially 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)]
 
    # Initially 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];
      
    // Initially 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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