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

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 1: 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

Javascript

<script>


// JavaScript 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
function repeatingString( s, n, k)
{
    // if length of string is
    // not divisible by K
    if (n % k != 0) {
        return false;
    }

    // Frequency Array
    var frequency = new Array(123);

    // Initially frequency of each
    // character is 0
    for (let i = 0; i < 123; i++) {
        frequency[i] = 0;
    }

    // Computing the frequency of
    // each character in the string
    for (let i = 0; i < n; i++) {
        frequency[s[i]]++;
    }

    var repeat = n / k;

    // Loop to check that frequency of
    // every character of the string
    // is divisible by K
    for (let i = 0; i < 123; i++) {

        if (frequency[i] % repeat != 0) {
            return false;
        }
    }

    return true;
}

// Driver Code

var s = "abcdcba";
var n = s.length;
var k = 3;

if (repeatingString(s, n, k)) {
    console.log("Yes");
}
else {
    console.log("No" );
}

// This code is contributed by ukasp.


</script>

Output:

No

Performance Analysis:
Time Complexity O(N) 
Auxiliary Space: O(1) 

Approach 2 :

We can check for all possible substrings of the given string and then check if any of these substrings can be repeated K times to form a permutation of the original string.

1. For each substring of length l = n/K, we can check if all characters in the substring have the same frequency (count) in the original string S. If this condition is not satisfied, we move to the next substring.
2. We repeat step 1 for all substrings of length l in the original string S. If we find any substring which satisfies the condition, we return “Yes”, else we return “No”.

Below is the code for above approach :

C++

#include <bits/stdc++.h>
using namespace std;

// Function to check if a repeating substring of given length
// exists in the given string
bool isRepeatingSubstring(string s, int len)
{
    int n = s.length();

    // Check if given substring length is greater than string length
    if(len >= n) {
        return false;
    }

    // Map to store the frequency of characters
    unordered_map<char, int> freq;

    // Find the frequency of characters in first len characters of the string
    for(int i = 0; i < len; i++) {
        freq[s[i]]++;
    }

    // Check if the substring consisting of first len characters is repeating
    bool repeating = true;
    for(auto it : freq) {
        if(it.second != n/len) {
            repeating = false;
            break;
        }
    }

    // If the substring is not repeating, check for other substrings
    if(!repeating) {
        for(int i = len; i < n; i++) {
            // Update the frequency map
            freq[s[i-len]]--;
            freq[s[i]]++;

            // Check if the substring consisting of previous len characters is repeating
            if(freq[s[i-len]] == 0) {
                freq.erase(s[i-len]);
            }
            if(freq.size() == 1 && freq.begin()->second == n/len) {
                return true;
            }
        }
    }

    return false;
}

// Driver code
int main() {
    string S = "abba";
    int K = 2;
    if(isRepeatingSubstring(S, K)) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
    return 0;
}

Output :

Yes

Time Complexity : O(n^2)

Auxiliary Space : O(1)